Solve the following system of equations, 3x +5y+2-0

Answer:

  linear

Step-by-step explanation:

The x-values all differ by 1, which is to say they are equally-spaced. The corresponding y-values all differ by -3. When (first) differences of equally-spaced values of y are constant, the function is of first degree, which is to say it is linear.

___

If second differences are non-zero and constant, the function is of second degree, quadratic.

Answer:

line

Step-by-step explanation:

The graphing option sounds nice...

But  lines have the same slope no matter what two points you choose.

You can see that x is going up by the same number (plus 1) each time and the y's are going down by the same number each time (minus 3) so this says no matter what two points you choose you will have the same slope which means it is a line.

Answer:

[tex]x_2 \approx -1.769[/tex]

Step-by-step explanation:

Let [tex]f(x)=x^3+x+7[/tex]

So [tex]f'(x)=3x^2+1[/tex]

[tex]x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}[/tex]

Let [tex]x_1=-2[/tex]

We are going to find [tex]x_2[/tex]

So we are evaluating [tex]-2-\frac{f(-2)}{f'(-2)}[/tex]

First step find f(-2)

Second step find f'(-2)

Third step plug in those values and apply PEMDAS!

[tex]f(-2)=(-2)^3+(-2)+7=-8-2+7=-10+7=-3[/tex]

[tex]f'(-2)=3(-2)^2+1=3(4)+1=12+1=13[/tex]

So

[tex]x_2=-2-\frac{-3}{13} \\\\ x_2=\frac{-26+3}{13} \\\\ x_2=\frac{-23}{13} \\\\ x_2 \approx -1.769[/tex]

The second approximation x2 using Newton's method for the equation x3 + x + 7 = 0 with an initial approximation of x1 = -2 is -2.2764.

In order to find the second approximation x2 using

Newton's method

, we need to use the definition of Newton's method, which states that: x

n+1

= x

n

- f(x

n

)/f'(x

n

). Here, our function f(x) is x

3

+ x + 7. The derivative, f'(x), is 3x

2

+ 1. If our initial approximation, x1, is -2, we can substitute these values into our method to find x2. So, x2 = x1 - f(x1)/f'(x1) = -2 - ((-2)^3 + (-2) + 7) / (3*(-2)^2 + 1) = -2.2764 (rounded to four decimal places).

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Try this option:

1] P(S)=0.84;

2] P(S or C, but not both)=0.4;

3] P(C)=0.76;

4] P(S∪C)=0.6;

5] P(C, but not S)=0.16;

6] P(S∩C)=1.00.

Answer:

We are given with a Venn diagram.

In Venn Diagram,

S represent Swam

C represent Built Sandcastles.

n( S - (S∩C) ) = 6

n( C - (S∩C) ) = 4

n( S ∩ C ) = 15

To find: P(S) , P(C) , P(S or C, but not Both) = P((S∪C) - (S∩C)) , P( S ∪ C ) ,

P(S ∩ C) , P(C , but not S ) =  P(C - (S∩C))

n(S) = n( S - (S∩C) ) + n(S∩C) = 6 + 15 = 21

n(C) = n( C - (S∩C) ) + n(S∩C) = 4 + 15 = 19

n(S∪C) = n( C - (S∩C) ) + n( S - (S∩C) ) + n(S∩C) = 6 + 4 + 15 = 25

Now, [tex]P(S)=\frac{21}{25}=0.84[/tex]

[tex]P(S\:or\:C,\:but\:not\:Both)=P((S\cup C)-(S\cap C))=\frac{10}{25}=0.40[/tex]

[tex]P(C)=\frac{19}{25}=0.76[/tex]

[tex]P(S\cup C)=\frac{25}{25}=1.00[/tex]

[tex]P(C\:,\:but\:not\:S)=P(C - (S\cap C))=\frac{4}{25}=0.16[/tex]

[tex]P(S\cap C)=\frac{15}{25}=0.60[/tex]

Therefore, Match the answers as above.

Final answer:

To determine the probability that the first ball has a smaller number than 4, we count the three balls with numbers 1, 2, and 3, resulting in a 3 out of 11 probability. The probability that the first ball was even is calculated by counting the even-numbered balls (2, 6, 8, 10, 12), which gives a probability of 5 out of 11.

Explanation:

The question asks about the probability that the first ball drawn from a box of 12 numbered balls is smaller or even, given that the second ball drawn has the number 4 on it and the drawing occurs without replacement. We know that once the second ball has been confirmed as the number 4, there are 11 remaining possibilities for the first ball.

To find the probability that the first ball had a smaller number than 4, we count the balls that are numbered less than 4. There are 3 such balls: 1, 2, and 3. Therefore, the probability is 3 out of 11 that the first ball had a smaller number.

To calculate the probability that the first ball had an even number, we consider only the even-numbered balls among the 11 remaining. These are 2, 6, 8, 10, and 12. So, there are 5 even-numbered balls, making the probability 5 out of 11 that the first ball was even.

Answer:

The equation is:

[tex]3(2) ^ t=1,536[/tex]

After [tex]t=9\ months[/tex]

Step-by-step explanation:

This situation can be represented by an exponential growth equation of the form

[tex]y = a (b) ^ {t}[/tex]

Where a is the initial amount

b is the growth rate

t is the time in months

In this case the initial amount is $ 3. Then [tex]a=3[/tex]

if she initially saves $3 and plans to double the amount she saves each month then

[tex]b=2[/tex]

The bike Jane wants is $1,536 at the local bike shop.

Then [tex]y=1,536[/tex]

The equation is:

[tex]3(2) ^ t=1,536[/tex]

Now we solve the equation for t

[tex]3(2) ^ t=1,536[/tex]

[tex](2) ^ t=\frac{1,536}{3}[/tex]

[tex](2) ^ t=512[/tex]

[tex]log_2(2) ^ t=log_2(512)[/tex]

[tex]t=log_2(512)[/tex]

[tex]t=9\ months[/tex]

Answer:

t=9 months. hope this helps

Answer:

  The rectangle is 7 km by 14 km. The 14 km dimension is parallel to the river.

Step-by-step explanation:

Let x represent the length of fence parallel to the river. The remaining fence is divided into two equal pieces for the ends of the enclosure. Then (28 -x)/2 will be the length of the side of the rectangle perpendicular to the river.

The total area of the enclosure is the product of length and width:

  Area = (x)(28-x)/2

This expression describes a parabola opening downward with zeros at x=0 and x=28. The vertex (maximum) is halfway between those zeros, so is at ...

  x = (0 +28)/2 = 14

Area is maximized when the dimension parallel to the river is 14 km and the ends of the enclosure are 7 km.

To maximize the area of the sanctuary, set up an equation with the length of the riverbank side. Differentiate and solve for x to find the dimensions of the rectangle.

To maximize the area of the sanctuary, we need to find the dimensions of the rectangle.

Let the length of the riverbank side be x km.

The remaining two sides of the rectangle will each be (28 - x/2) km, as the total fence length should be equal to x km along the riverbank and (28 - x/2) km for the other two sides.

The area of the rectangle is given by A = x * (28 - x/2). To maximize the area, we can differentiate A with respect to x, set it equal to 0, and solve for x.

Taking the derivative of A, we get dA/dx = 28 - 3x/2. Setting this equal to 0, we find 28 - 3x/2 = 0. Solving for x, we get x = 18.67 km.

Therefore, the dimensions of the rectangle to maximize the area of the sanctuary are approximately 18.67 km along the riverbank and (28 - 18.67/2) km for the other two sides.

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Answer with explanation:

Given the function f from R  to [tex](0,\infty)[/tex]

f: [tex]R\rightarrow(0,\infty)[/tex]

[tex]-f(x)=x^2[/tex]

To prove that  the function is objective from R to  [tex](0,\infty)[/tex]

Proof:

[tex]f:(0,\infty )\rightarrow(0,\infty)[/tex]

When we prove the function is bijective then we proves that function is one-one and onto.

First we prove that function is one-one

Let [tex]f(x_1)=f(x_2)[/tex]

[tex](x_1)^2=(x_2)^2[/tex]

Cancel power on both side then we get

[tex]x_1=x_2[/tex]

Hence, the function is one-one on domain [tex[(0,\infty)[/tex].

Now , we prove that function is onto function.

Let - f(x)=y

Then we get [tex]y=x^2[/tex]

[tex]x=\sqrt y[/tex]

The value of y is taken from [tex](0,\infty)[/tex]

Therefore, we can find pre image  for every value of y.

Hence, the function is onto function on domain [tex](0,\infty)[/tex]

Therefore, the given [tex]f:R\rightarrow(0.\infty)[/tex] is bijective function on [tex](0,\infty)[/tex] not on  whole domain  R .

Hence, proved.

Answer:

Step-by-step explanation:

Given

Find

Solution

a. Solve x plus y equals 81 for y.

  y equals 81 minus x

__

b. Substitute the result from part a into the equation P equals x squared y for the variable that is to be maximized.

  P equals x squared left parenthesis 81 minus x right parenthesis

__

c. Find the domain of the function P found in part b.

  left bracket 0 comma 81 right bracket

__

d. Find dP/dx. Solve the equation dP/dx = 0.

  P = 81x² -x³

  dP/dx = 162x -3x² = 3x(54 -x) = 0

The zero product rule tells us the solutions to this equation are x=0 and x=54, the values of x that make the factors be zero. x=0 is an extraneous solution for this problem so ...

  P is maximized at (x, y) = (54, 27).

The problem is about using the equations x + y = 81 and P=x^2y to find the maximum possible value of P. This involves solving for y, substituting the result into the P equation, determining the domain of P, and finding the derivative of P to solve for x.

We are given the system of equations: x + y = 81 and P = x^2*y and we're tasked with finding the optimum value for P given these constraints.

a. We solve for y in the equation x + y = 81, so y = 81 - x

b. We substitute the result from part a into the equation for P, P = x^2(81 - x)

c. The domain of function P is the set of all possible x values or [0, 81]. This is because x and y are non-negative and y is equal to 81 minus x, meaning the highest x can be is 81

d. With respect to maximization, step d usually involves calculating the derivative of P with respect to x, setting it equal to zero, and solving for x. If you apply the product rule and the chain rule, you would get dP/dx = 0 then solve for x

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Answer:

The range is 21

The interquartile range is 6

The variance is 33

The standard deviation (σ) is 5.74

Step-by-step explanation:

* Lets study the information to solve the problem

- The values of the data are 27 , 24 , 23 , 15 , 30 , 36 , 29 , 24

- They are eight values

* lets arrange them from small to big

∴ The values are 15 , 23 , 24 , 24 , 27 , 29 , 30 , 36

* Now lets solve the problem

# The range

- It is the difference between the largest and the smallest values

∵ The largest value is 36

∵ The smallest value is 15

∴ The range = 36 - 15 = 21

* The range is 21

# The interquartile range

- The steps to find the interquartile range is:

1- Arrange the values from the smallest to the largest

∴ The values are 15 , 23 , 24 , 24 , 27 , 29 , 30 , 36

2- Find the median

- The median is the middle value after arrange them

* If there are two values in the middle take their average

∵ The values are 8 then the 4th and the 5th are the values

∵ The 4th is 24 and the 5th is 27

∴ The median = [tex]\frac{24+27}{2}=\frac{51}{2}=25.5[/tex]

∴ The median is 25.5

3- Calculate the median of the lower quartile

- The lower quartile is the median of the first half data values

∵ There are 8 values

∴ The first half is the first four values

∴ The first half values are 15 , 23 , 24 , 24

∵ The middle values are 23 and 24

∴ The median of lower quartile = [tex]\frac{23+24}{2}=\frac{47}{2}= 23.5[/tex]

- Similar find the median of the upper quartile

- The upper quartile is the median of the second half data values

∵ There are 8 numbers

∴ The second half is the last four values

∴ The second half values are 27 , 29 , 30 , 36

∵ The middle values are 29 and 30

∴ The median of upper quartile = [tex]\frac{29+30}{2}=\frac{59}{2}=29.5[/tex]

4- The interquartile range (IQR) is the difference between the upper

    and the lower medians

∴ The interquartile range = 29.5 - 23.5 = 6

* The interquartile range is 6

# The variance

- The variance is the measure of how much values in a set of data are

  likely to differ from the mean value of the same data

- The steps to find the variance

1- Find the mean of the data

∵ The mean = sum of the data ÷ the number of the values

∵ The sum = 15 + 23 + 24 + 24 + 27 + 29 + 30 + 36 = 208

∵ The number of values is 8

∴ The mean = [tex]\frac{208}{8}=26[/tex]

∴ The mean is 26

2- Subtract the mean from each value and square the answer

∴ 15 - 26 = -11 ⇒ (-11)² = 121

∴ 23 - 26 = -3 ⇒ (-3)² = 9

∴ 24 - 26 = -2 ⇒ (-2)² = 4

∴ 24 - 26 = -2 ⇒ (-2)² = 4

∴ 27 - 26 = 1 ⇒ (1)² = 1

∴ 29 - 26 = 3 ⇒ (3)² = 9

∴ 30 - 26 = 4 ⇒ (4)² = 16

∴ 36 - 26 = 10 ⇒ (10)² = 100

3- Add all of these squared answer and divide the sum by the number

   of the values

∴ The sum = 121 + 9 + 4 + 4 + 1 + 9 + 16 + 100 = 264

∵ They are 8 values

∴ The variance (σ²) = [tex]\frac{264}{8}=33[/tex]

* The variance is 33

# The standard deviation

- It is the square root of the variance

∵ The variance = 33

∴ The standard deviation (σ) = √33 = 5.74

* The standard deviation (σ) is 5.74

The plane [tex]z=9[/tex] lies above the paraboloid [tex]z=x^2+y^2[/tex], so the volume of the bounded region [tex]R[/tex] is given by

[tex]\displaystyle\iiint_R\mathrm dV=\int_{-3}^3\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}\int_{x^2+y^2}^9\mathrm dz\,\mathrm dy\,\mathrm dx[/tex]

Convert to cylindrical coordinates, setting

[tex]\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=z\end{cases}\implies\mathrm dx\,\mathm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm d\theta\,\mathrm dz[/tex]

and the integral is equivalent to

[tex]\displaystyle\int_0^{2\pi}\int_0^3\int_{r^2}^9r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=2\pi\int_0^3(9r-r^3)\,\mathrm dr=\boxed{\frac{81\pi}2}[/tex]

The volume of the solid enclosed by the paraboloid z = x² + y² and the plane z = 9 is found by using double integrals in polar coordinates. The volume is calculated as 81π cubic units.

To find the volume of a solid enclosed by the paraboloid z = x² + y² and the plane z = 9, you have to use the method of double integrals in polar coordinates. The cone extends from z = 0 at its apex to z = 9 at the top, which is given by the plane. Hence, we can imagine this region as a bunch of thin disks or pancakes that lie above circles in the xy-plane and pile up to form the parcel of the parabolic solid under the plane z = 9.

In this case, we have to integrate over the region R, which is a disk of radius 3 (it's the projection on the xy-plane under the plane z = 9), with the height of a 'thin disk' as z = x² + y² = r² (in polar coordinates). Therefore, the volume V can be given as:

V = ∫∫R(z*r*dr*dθ) = ∫02π∫03(r²*r*dr*dθ) = 2π* [03 0.25r⁴] = 2π*(40.5-0) = 81π cubic units.

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Answer:yes

Step-by-step explanation:

Theoretically he could just open the doors leading to the seven cells to the opposing side of the bored, then move forward the seven more, ending in the corner cell diagonal from his orgiu position.

No, I don’t think so. If he passes through each column of cells, he gets to the end cell to the left or right from him instead of the one exactly diagonal from him.

Hope this helps!

Answer:

the smallest angle is x, and then the second angle is 3x and the third angle is 3x+50. I don't know what the smallest angle is though.

Step-by-step explanation:

first I don't see the smallest angle that's why I called it x. Next you said 3 times the smallest angle so 3 times x. Finally you said the third angle is 50 more than the second one so you add 50 to 3x. I can't tell you what the answer is though because you didn't tell me the smallest angles measure.

Answer: [tex]y(t)=Acos(t)+Bsin(t)[/tex]

Step-by-step explanation:

To find the solution of a given differential equation ay''+by'+cy=0, a≠0, you have to consider the quadratic polynomial ax²+bx+c=0, called the characteristic polynomial.  

Using the quadratic formula, this polynomial will always have one or two roots, for example r and s. The general solution of the differential equation is:

[tex]y(t)= Ae^{rt}+Be^{st}[/tex] , if the roots r and s are real numbers and r≠s.

[tex]y(t)= A e^{rt}+B*t*e^{rt}[/tex] , if r=s is real.

[tex]y(t)=Acos(\beta t)e^{\alpha t} +Bsin(\beta t)e^{\alpha t}[/tex] , if the roots r and s are complex numbers α+βi and  α−βi

.

In this case, the characteristic polynomial is:

[tex]x^{2} +1=0\\x^{2} =-1\\x1=i; x2=-i[/tex]

Since the roots are complex numbers, with α=0 and β=1, then the answer is: [tex]y(t)=Acos(t)+Bsin(t)[/tex]

Final answer:

To find the total number of different value meal packages possible at Ron's Subs, multiply the number of choices for drinks (4), sandwiches (3), and chips (3), resulting in 4 × 3 × 3 = 36 possible combinations.

Explanation:

To find the total number of different value meal packages possible, we calculate the product of the number of choices for each category. In this case, there are 4 types of drinks, 3 types of sandwiches, and 3 types of chips. Thus, the calculation is as follows:

To find the total number of combinations, we multiply the number of choices for each category:

4 (drinks) × 3 (sandwiches) × 3 (chips) = 36 different value meal packages.

Therefore, at Ron's Subs, there are 36 possible different value meal packages a customer can choose from.

Answer:

Let's say that the mile that the cyclist going west  is a.

And so the one heading North is (a+5)

a+(a+5)=25

a+a+5=25

2a=20

a=10

So the one heading West has traveled 10 mi.

ANSWER

36,-36

EXPLANATION

The given function is:

[tex]f(x) = {x}^{2} + 12x + 6[/tex]

To write this function in vertex form;

We need to add and subtract the square of half the coefficient of x.

The coefficient of x is 12.

Half of it is 6.

The square of 6 is 36.

Therefore we add and subtract 36.

Hence the zero pair is:

36, -36.

The correct answer is D.

Answer:

Last option: 36,-36​

Step-by-step explanation:

The vertex form of the function of a parabola is:

[tex]y=a(x-h)^2+k[/tex]

Where (h,k) is the vertex.

To write the given function in vertex form, we need to Complete the square.

Given the Standard form:

[tex]y=ax^2+bx+c[/tex]

We need to add and subtract [tex](\frac{b}{2})^2[/tex] on one side in order to complete the square.

Then, given [tex]y=x^2+12x+6[/tex], we know that:

[tex](\frac{12}{2})^2=6^2=36[/tex]

Then, completing the square, we get:

 [tex]y=x^2+12x+(36)+6-(36)[/tex]

[tex]y=(x+6)^2-30[/tex] (Vertex form)

Therefore, the answer is: 36,-36​

Answer:

Step-by-step explanation:

If i am not mistaken I believe it is 2.75 lemme know if that is right!

Answer:

Step-by-step explanation:

Using 2016 as a reference (t=0), the exponential equation for winnings can be written as ...

  w(t) = 1480000×(1480000/200)^(t/121)

where 1480000 is the winnings in the reference year, and the ratio 1480000/200 is the ratio of winnings increase over the 121 years from 1895 to 2016.

This can be approximated by ...

  w(t) ≈ 1,480,000×1.07640850764^t

In this form, we can see that the annual percentage increase is ...

  1.0764 -1 = 7.64%

__

Then the winner's check in 2043, 27 years after 2016, is predicted to be ...

  w(27) = $1,480,000×(1.0764...)^27 ≈ $10,805,478.41 ≈ $10,805,000

The percentage increase per year in the winner's prize money over the period is 6065.57%. The winner's prize money in 2043 would be approximately $15,190,712.55.

To calculate the percentage increase per year, we need to find the average annual growth rate over the time period. First, we calculate the total percentage increase by taking the difference between the final and initial values, divided by the initial value.

In this case, it is (($1,480,000 - $200) / $200) * 100 = 740,000%. Then, we divide this percentage by the number of years, which is 2016 - 1895 + 1 = 122. So the annual percentage increase is 740,000% / 122 = 6065.57%.

To calculate the winner's prize in 2043, we need to find the number of years from 2016 to 2043, which is 2043 - 2016 = 27.

Then, we use the compound interest formula to calculate the future value: $1,480,000 * (1 + (6065.57% / 100))^27 = $15,190,712.55. So the winner's prize in 2043 would be approximately $15,190,712.55.

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Answer:

x = -6 and y = 9

Step-by-step explanation:

It is given that,

0.12x - 0.07y = -1.35    -----(1)

0.4x + 0.8y = 4.8   -----(2)

To find the value of x and y

eq(1) * 100 ⇒

12x + 7y = -135   -----(3)

eq(2) /0.4 ⇒

x + 2y = 12   -----(4)

eq(4) * 12 ⇒

12x + 24y = 144   ---(5)

eq(5) - eq(3) ⇒

12x + 24y = 144   ---(5)

12x - 7y = -135  -----(3)

 0  + 31y = 279

y = 279/31 = 9

Substitute the value of y in eq(4)

x + 2y = 12   -----(4)

x + 2*9 = 12

x = 12 - 18 = -6

Therefore x = -6 and y = 9

Answer:

The total cost of resurface the path is [tex]\$600[/tex]

Step-by-step explanation:

step 1

Find the area of the path

The area of the path is equal to the area of the path plus the swimming pool minus the area of the swimming pool

[tex]A=(14+3+3)(30+3+3)-(14)(30)[/tex]

[tex]A=(20)(36)-(14)(30)[/tex]

[tex]A=300\ ft^{2}[/tex]

step 2

Find the cost of resurface the path

Multiply the area of the path by $2 per square foot

[tex]300*2=\$600[/tex]

the total cost of resurfacing the path is $600.

To calculate the total cost of resurfacing the path around the swimming pool, you first need to determine the area of the path. The swimming pool measures 14 feet by 30 feet, and the path is 3 feet wide. To find the area of the outer rectangle, which includes the pool and the path, you calculate the width and length including the path. This gives you a width of (14 + 2*3) feet and a length of (30 + 2*3) feet, as the path goes all the way around, adding twice the width of the path to each dimension.

The outer rectangle's dimensions are therefore 20 feet by 36 feet. The area of the outer rectangle is 20 feet * 36 feet = 720 square feet. The area of the pool itself is 14 feet * 30 feet = 420 square feet. To find the area of just the path, you subtract the area of the pool from the area of the outer rectangle: 720 square feet - 420 square feet = 300 square feet. The cost to resurface the path is $2 per square foot, so the total cost is 300 square feet * $2/square foot = $600.

In the [tex]x[/tex] direction we consider the [tex]m=2[/tex] subintervals [0, 1] and [1, 2] (each with length 1), while in the [tex]y[/tex] direction we consider the [tex]n=2[/tex] subintervals [0, 2] and [2, 4] (with length 2). Then the lower right corners of the cells in the partition of [tex]R[/tex] are (1, 0), (2, 0), (1, 2), (2, 2).

Let [tex]f(x,y)=7x+5y^2[/tex]. The volume of the solid is approximately

[tex]\displaystyle\iint_Rf(x,y)\,\mathrm dx\,\mathrm dy\approx f(1,0)\cdot1\cdot2+f(2,0)\cdot1\cdot2+f(1,2)\cdot1\cdot2+f(2,2)\cdot1\cdot2=\boxed{164}[/tex]

###

More generally, the lower-right-corner Riemann sum over [tex]m=\mu[/tex] and [tex]n=\nu[/tex] subintervals would be

[tex]\displaystyle\sum_{m=1}^\mu\sum_{n=1}^\nu\left(7\frac{2m}\mu+5\left(\frac{4n-4}\nu\right)^2\right)\frac{2-0}\mu\frac{4-0}\nu=\frac83\left(101+\frac{21}\mu+\frac{40}{\nu^2}-\frac{120}\nu\right)[/tex]

Then taking the limits as [tex]\mu\to\infty[/tex] and [tex]\nu\to\infty[/tex] leaves us with an exact volume of [tex]\dfrac{808}3[/tex].

The Riemann sum is used to provide an estimate of the volume of a solid under the function surface z = 7x + 5y² and above the rectangle R = [0, 2] × [0, 4]. The rectangle is divided into four equal parts, and the function's value at specific points, multiplied by the area of the base, provides the estimate.

This Mathematics problem requires estimating the volume of a solid under the surface z = 7x + 5y² and above the rectangle R = [0, 2]⨯[0, 4] using a Riemann sum. This method is commonly used in calculus to approximate the definite integral of a function.

Applying a Riemann sum with m=n=2 implies the rectangle is split into 4 equal rectangles for the estimation. Our sample delta x and delta y = rectangle's length/2, for this example, Δx = 2/2 = 1 and Δy = 4/2 = 2. The lower right corners points will be (1,2), (1,4), (2,2) and (2,4).

We then find volume estimates by taking the function's value at these sample points and multiplying it by the area of the base. This gives us: ((7*1 + 5*2²) + (7*1 + 5*4²) + (7*2 + 5*2²) + (7*2 + 5*4²))* (Δx*Δy). Simplifying the expression gives us the estimated volume using the Riemann sum.

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Step-by-step answer:

Given:

mean, mu = 200 m

standard deviation, sigma = 30 m

sample size, N = 5

Maximum deviation for no damage, D = 100 m

Solution:

Z-score for maximum deviation

= (D-mu)/sigma

= (100-200)/30

= -10/3

From normal distribution tables, the probability of right tail with

Z= - 10/3

is 0.9995709, which represents the probability that the parachute will open at 100m or more.

Thus, by the multiplication rule, the probability that all five parachutes will ALL open at 100m or more is the product of the individual probabilities, i.e.

P(all five safe) = 0.9995709^5 = 0.9978565

So there is an approximately 1-0.9978565 = 0.214% probability that at least one of the five parachutes will open below 100m

The probability that at least one out of five parachutes causes equipment damage, given that the parachute opening altitude is normally distributed with a mean of 200m and standard deviation of 30m, is approximately 0.2%.

The situation described is a question of probability related to the normal distribution. In this case, we are asked to find the probability of a parachute opening at less than 100m, which will cause damage. First, we need to standardize the value to a z-score. The z-score is calculated by subtracting the mean from the value of interest and dividing by the standard deviation. In this case, it will be (100-200)/30, which equals to about -3.33.

By looking at a z-table or using a statistical calculator we find that the probability of single parachute causing damage is approximately 0.0004. However, the question is interested in the probability of at least one out of five parachutes causing damage. This can be approached as 1 minus the probability of none of the five causing damage, which will be 1 - (1-0.0004)^5. Thus, the resulting probability of equipment damage to the payload of at least one of five independently dropped parachutes is approximately 0.002 or 0.2%.

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The mean value (expected value) of the given discrete variable x is,

⇒ 2.35.

The process of combining two or more numbers is called the Addition. The 4 main properties of addition are commutative, associative, distributive, and additive identity.

Now, For the mean value of the given discrete variable x, we need to multiply each value of x by its corresponding probability, and then add up these products.

Hence, the expected value of x can be calculated as follows:

Mean = (0 x 0.2) + (1 x 0.2) + (2 x 0.15) + (3 x 0.15) + (4 x 0.15) + (5 x 0.1) + (6 x 0.05)

Simplifying this expression, we get:

Mean = 0 + 0.2 + 0.3 + 0.45 + 0.6 + 0.5 + 0.3

Mean = 2.35

Therefore, the mean value (expected value) of the given discrete variable x is 2.35.

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Answer:

The equation is:

[tex]y = 103- 23d[/tex]

Step-by-step explanation:

The initial amount was 103 grams.

After one day of use the remaining amount of soap was 80 grams.

So the amount of time he spent in one day was:

[tex]103-80 = 23[/tex]

Each day margaret spends 23 grams of soap.

if d represents the number of days elapsed then, the amount of soap "y" that Margaret spends after days is:

[tex]y = 103- 23d[/tex]

Margaret uses 5.75 grams of soap each day. The equation that shows the amount of soap remaining after d days of use is: S = 103 - 5.75d.

Based on the information provided, we can find the rate of soap loss, measured in grams per day. Initially, Margaret's soap weighed 103 grams and 4 days later, it weighed 80 grams so we know that a total of 23 grams of soap was used over this 4-day period.

Therefore, Margaret is using (103-80) / 4 = 23 / 4 = 5.75 grams of soap each day. Given this daily usage rate, we can say that after d days, the amount of soap remaining can be calculated by subtracting the total soap used from the initial weight. So, our equation will be: S = 103 - 5.75d, where S represents the soap remaining and d represents the number of days.

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ANSWER

[tex]y = \cot(x - \frac{\pi}{3} ) + 2[/tex]

EXPLANATION

The cotangent function that is fully transformed is of the form

[tex]y =a \cot(bx + c) + d[/tex]

where 'a' is the amplitude.

[tex] \frac{\pi}{b} = \pi[/tex]

is the period.

This implies that b=1

The phase shift is

[tex] \frac{c}{b} = - \frac{\pi}{3} [/tex]

Substitute b=1 to get;

[tex]c = - \frac{\pi}{3} [/tex]

and d=2 is the vertical shift.

We choose a=1 to get the required function as

[tex]y = \cot(x - \frac{\pi}{3} ) + 2[/tex]

Answer:

  [tex]1000\pm 100\sqrt{55}[/tex]

Step-by-step explanation:

A TI-84 or "normal" calculator is designed to evaluate expressions numerically. It can tell you the numerical value of this expression is the set of values

  {1741.619849, 258.3801513}

but it cannot simplify the expression.

This expression can be simplified by evaluating the fraction and removing double factors from under the radical:

[tex]\dfrac{2000\pm\sqrt{2200000}}{2}=\dfrac{2000}{2}\pm\sqrt{\dfrac{2200000}{2^2}}=1000\pm\sqrt{550000}\\\\=1000\pm\sqrt{100^2\cdot 55}=1000\pm 100\sqrt{55}[/tex]

Check the picture below.  So the parabola looks more or less like so.

let's recall that the vertex is half-way between the focus point and the directrix, at "p" units away from both.

Let's notice that the focus point is below the directrix, that means the parabola is vertical, namely the squared variable is the "x", and it also means that it's opening downwards as  you see in the picture, namely that "p" is negative, in this case "p" is 1 unit, and thus is -1.

[tex]\bf \textit{parabola vertex form with focus point distance} \\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ \stackrel{\textit{we'll use this one}}{4p(y- k)=(x- h)^2} \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=-2\\ k=5\\ p=-1 \end{cases}\implies 4(-1)(y-5)=[x-(-2)]^2\implies -4(y-5)=(x+2)^2 \\\\\\ y-5=-\cfrac{1}{4}(x+2)^2\implies y=-\cfrac{1}{4}(x+2)^2+5[/tex]

Answer:

Width = 3 feet

Length = 7 feet

Step-by-step explanation:

x represents the width of doghouse so,

Width = x

Length is 4 times more than its width

Length = x+4

Area of doghouse = 21 square feet

We know

Area of rectangle = Length * Width

21 = (x+4)*x

21 = x^2+4x

=> x^2+4x-21=0

Solving the above quadratic equation by factorization to find the value of x

x^2+7x-3x-21=0

x(x+7)-3(x+7)=0

(x+7)(x-3)=0

x+7 =0 and x-3=0

x= -7 and x =3

Since the width of rectangle can never be negative so, x=3

Width =x = 3 feet

Length = x+4 = 3+4 = 7 feet

Answer:  (x + 7)(x - 3) = 0

               width (x) = 3

               length (x+4) = 7

Step-by-step explanation:

Area (A) = length (l) × width (x)

       21   =  (x + 4)     ×    (x)

       21   =  x² + 4x                   distributed (x) into (x + 4)

        0   =  x² + 4x - 21             subtracted 21 from both sides

        0  = (x + 7)(x - 3)               factored quadratic equation

 0 = x + 7   or    0 = x - 3          applied Zero Product Property

 x = -7        or     x = 3               solved each equation

x = -7 is not valid because lengths cannot be negative

so x = 3

and length ... x + 4 = (3) + 4 = 7

Answer:

[tex]\large\boxed{x=\dfrac{k-14}{5}}[/tex]

Step-by-step explanation:

[tex]5x=k-14\qquad\text{divide both sides by 5}\\\\\dfrac{\not5x}{\not5}=\dfrac{k-14}{5}\\\\x=\dfrac{k-14}{5}[/tex]

Answer:

[tex]5x = k - 14 \\ \frac{5x}{5} = \frac{k - 14}{5} \\ x = \frac{k - 14}{5} [/tex]

Recall that for [tex]|x|<1[/tex], we have

[tex]\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n[/tex]

Then for [tex]|-x^2|<1[/tex], or [tex]|x|<1[/tex], we have

[tex]\displaystyle\frac1{1+x^2}=\frac1{1-(-x^2)}=\sum_{n=0}^\infty(-x^2)^n=\sum_{n=0}^\infty(-1)^nx^{2n}[/tex]

Multiply this series by [tex]2x[/tex] to get the Taylor series for [tex]f(x)[/tex]:

[tex]f(x)=\dfrac{2x}{1+x^2}=\displaystyle2\sum_{n=0}^\infty(-1)^nx^{2n+1}[/tex]

Notice that

[tex]\dfrac{\mathrm d(\ln(1+x^2))}{\mathrm dx}=\dfrac{2x}{1+x^2}[/tex]

so to find the Taylor series for [tex]g(x)=\ln(1+x^2)[/tex], we integrate the Taylor series for [tex]f(x)[/tex]:

[tex]g(x)=\displaystyle\int f(x)\,\mathrm dx=C+2\sum_{n=0}^\infty\frac{(-1)^nx^{2n+2}}{2n+2}[/tex]

Since [tex]g(0)=\ln(1+0^2)=\ln1=0[/tex], it follows that [tex]C=0[/tex] and

[tex]g(x)=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n+2}}{n+1}[/tex]

which converges for [tex]|x|<1[/tex] as well.

Following are the calculation to the Taylor Series:

Geometric series:

[tex]\to \Sigma_{\infty}^{n=0} \ a+ar+ar^2+.....+[/tex] which converges to [tex]\frac{a}{1-r} \ \ for\ \ |r| < 1[/tex].

Remembering that:

[tex]\to \frac{2x}{1+x^2}=\frac{2x}{1-(-x^2)}[/tex]  Taking  [tex]a=2x \ \ \ \ \ \ \ \ \ r=-x^2\\\\[/tex]

Using the Taylor series:

[tex]\to \frac{2x}{1+x^2}= \Sigma_{\infity}^{n=0} \ 2x \times (-x^2)^{n} = \Sigma_{\infity}^{n=0} 2x \times (-1)^{n} \times (x^{2n})[/tex]

[tex]\to \frac{2x}{1+x^2}= \Sigma_{\infty}^{n=0} (-1)^{n} \times 2 \times x^{2n+1} = 2x -2x^3+2x^5+............+[/tex]

In the given scenario we will converge the [tex]|x^2| < 1 |x| < 1[/tex].  Now, realize:

[tex]\to \frac{2x}{1+x^2} \ dx = \In (1+x^2) \\\\[/tex]

Integrating the series for [tex]\frac{2x}{1+x^2}[/tex]  :

[tex]\to \In( 1+x^2)=\int \Sigma_{\infty}^{n=0} \ (-1)^{n} \times 2 \times x^{2n+1}\ dx\\\\[/tex]

                  [tex]=\Sigma_{\infity}^{n=0} \frac{(-1)^{n} \times 2 \times x^{2n+2}}{ 2n+2}\\\\=\Sigma_{\infity}^{n=0} \frac{(-1)^{n} \times x^{2n+2}}{ n+1}\\\\= x^2 -\frac{x^4}{2}+\frac{x^6}{3}+............+[/tex]

Since integrating a number has no effect on its radius of converge, this series similarly converges for [tex]\bold{|x| < 1}[/tex].

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