What is the volume of the cone with radius 4 ft and height 10 ft? Round to the nearest cubic foot.

A) 126 ft3


B) 200 ft3


C) 251 ft3


D) 168 ft3

Answer:

the desired equation is y = 4x + 1

Step-by-step explanation:

As we move to the right from (-9, -35) to (2, 9), x increases by 11 and y increases by 44.  Thus, the slope of the line in question is

m = rise / run = 44/11 = 4.

Using the slope-intercept form of the equation of a straight line, we substitute 4 for m, 2 for x and 9 for y, obtaining:

y = mx + b →  9 = 4(2) + b.  Thus, b = 1, and the desired equation is

y = 4x + 1

By using the slope-intercept form of a line and the given anchor points, we find that the equation of the line is y= 4x - 1.

The subject of this question is to find the equation of a trend line using two anchor points (-9,-35) and (2,9). We can calculate the equation of a line using the slope-intercept form y = mx + b, where m represents the slope and b represents the y-intercept.

First, calculate the slope (m) which is (y2-y1)/(x2-x1) = (9 - (-35))/(2 - (-9)) = 44/11 = 4.

Then, with the slope (m = 4) and one point (2,9), plug in these values into the slope-intercept form to solve for the y-intercept (b). 9 = 4*2 + b. Solving for b gives -1.

So, the equation of the trend line is y= 4x - 1.

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Answer:

12.62

Step-by-step explanation:

you divide 3 by 8, then add the quotient of that to 3, then subtract that from 16

Answer:

Givens

First, we need to the number of pages dedicated to advertisements.

Let's transform the mixed number into a fraction

[tex]3\frac{3}{8}=\frac{27}{8}[/tex]

Now, let's multiply this fraction with the number of pages

[tex]\frac{27}{8} \times 16= 54[/tex]

That is, there are 54 pages dedicated to advertisements.

Pages without advertisements are 5/8, which is

[tex]\frac{5}{8} \times 16=5(2)=10[/tex]

Answer: Second Option

"the summation of 880 times one fourth to the i minus 1 power, from i equals 1 to infinity. ; the sum is 1,173"

Step-by-step explanation:

We know that infinite geometrical series have the following form:

[tex]\sum_{i=1}^{\infty}a_1(r)^{n-1}[/tex]

Where [tex]a_1[/tex] is the first term of the sequence and "r" is common ratio

In this case

[tex]a_1 = 880\\\\r=\frac{1}{4}[/tex]

So the series is:

[tex]\sum_{i=1}^{\infty}880(\frac{1}{4})^{n-1}[/tex]

By definition if we have a geometric series of the form

[tex]\sum_{i=1}^{\infty}a_1(r)^{n-1}[/tex]

Then the series converges to  [tex]\frac{a_1}{1-r}[/tex]   if [tex]0<|r|<1[/tex]

In this case [tex]r = \frac{1}{4}[/tex] and [tex]a_1=880[/tex]  then the series converges to [tex]\frac{880}{1-\frac{1}{4}} = 1,173.3[/tex]

Finally the answer is the second option

Answer:

A, g(x) = 7x + 5

Step-by-step explanation:

applying these translations to the parent function f(x) = x, we would get the following equation:

g(x) = 7x + 5

a vertical compression is written before the parent function (in this case f(x)=x), and a shift up is written next to the function. both of these are without parentheses

the answer would be A, g(x) = 7x + 5

Answer:

[tex]7\sqrt[11]{x^{5}}[/tex]

Step-by-step explanation:

we know that

[tex]a^{\frac{n}{m}}=\sqrt[m]{a^{n}}[/tex]

In this problem we have

[tex]7x^{\frac{5}{11}}[/tex]

therefore

[tex]7x^{\frac{5}{11}}=7\sqrt[11]{x^{5}}[/tex]

Given the speeds of the dog and cat, the dog will catch the cat in approximately 33.33 seconds by covering the 200-meter distance at a relative speed of 6 m/s.

Problem: A dog and a cat are 200 meters apart. The dog runs at 30 m/s, and the cat runs at 24 m/s. How soon will the dog catch the cat?

Calculate the relative speed at which the dog is gaining on the cat: 30 m/s - 24 m/s = 6 m/s.

Divide the initial distance (200 meters) by the relative speed (6 m/s) to find the time it takes for the dog to catch the cat: 200 m / 6 m/s = 33.33 seconds.

Answer:

option A

Step-by-step explanation:

Step 1

X

[tex]x=\left[\begin{array}{ccc}b&a\\4&a\end{array}\right][/tex]

Step 2

2Y

[tex]\left[\begin{array}{ccc}2c&2d\\2a&2b\end{array}\right][/tex]

Step 3

X - 2Y = Z

[tex]2Y=\left[\begin{array}{ccc}b&a\\4&a\end{array}\right]-\left[\begin{array}{ccc}2c&2d\\2a&2b\end{array}\right]=\left[\begin{array}{ccc}a&c\\16&b\end{array}\right][/tex]

Step 4

Four equations are formed

Equation 1

b - 2c = a

Equation 2

a - 2d = c

Equation 3

4 - 2a = 16

-2a = 16 - 4

-2a = 12

Equation 4

a -2b = b

-6 - 2b = b

-6 = b + 2b

-6 = 3b

Plug values of a and b in equation 1 and 2

b - 2c = a

-2 -2c = -6

-2c = -6 + 2

-2c = -4

c = -4/-2

a - 2d = c

-6 -2d = 2

-2d = 2+6

-2d = 8

 d = 8/-2

Answer: last option.

Step-by-step explanation:

To apply the Distributive property, remember that:

[tex]c(a-b)=ca-cb[/tex]

Then, applying this, you get:

[tex]-4(x - 2) - 3x = 2(5x - 7) + 6\\\\(-4)(x)+(-4)(-2)-3x=(2)(5x)+(2)(-7)+6\\\\-4x+8-3x=10x-14+6[/tex]

Combine like terms means that you need to add the like terms.

Therefore, you get:

[tex]-7x+8=10x-8[/tex]

You can observe that the expression obtained matches with the expression provided in the last option.

Answer:

[tex]x^2+y^2-8x-4y+11=0[/tex]

Step-by-step explanation:

We want to find the equation of the circle: [tex](x-4)^2+(y-2)^2=9[/tex] in general form.

We need to expand the parenthesis to obtain: [tex]x^2-8x+16+y^2-4y+4=9[/tex]

This implies that:

[tex]x^2+y^2-8x-4y+20=9[/tex]

We add -9 to both sides of the equattion to get:

[tex]x^2+y^2-8x-4y+20-9=0[/tex]

Simplify the constant terms to get:

[tex]x^2+y^2-8x-4y+11=0[/tex]

The general form of the equation is x^2 + y^2 - 8x -4y - 11 = 0

The equation is given as:

(x-4)^2+(y-2)^2=9

Evaluate the exponents

x^2 - 8x + 16 + y^2 -4y + 4 = 9

Collect like terms

x^2 - 8x +  y^2 -4y - 9 + 16 + 4 = 0

Evaluate the like terms

x^2 - 8x +  y^2 -4y - 11 = 0

Rewrite as:

x^2 + y^2 - 8x -4y - 11 = 0

Hence, the general form of the equation is x^2 + y^2 - 8x -4y - 11 = 0

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Answer:

d is the answer i believe

Step-by-step explanation:

A triangle is 180°. So you can do:

3.2n + 6.4n + 2.4n = 180   Simplify

12n = 180

n = 15   Now that you know the value of n, you can plug it into each individual angle/equation

∠X = 3.2n    plug in 15 for n

∠X = 3.2(15)

∠X = 48°

∠Y = 6.4(15)

∠Y = 96°

∠Z = 2.4(15)

∠Z = 36°

With

[tex]\vec r(t)=4t\,\vec\imath+6t\,\vec\jmath-t^2\,\vec k[/tex]

we have

[tex]\mathrm d\vec r=(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt[/tex]

The vector field evaluated over this parameterization is

[tex]\vec f(x,y,z)=\vec f(x(t),y(t),z(t))=4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k[/tex]

so the line integral is

[tex]\displaystyle\int_{-1}^1(4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k)\cdot(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt[/tex]

[tex]=\displaystyle\int_{-1}^1(16t+6t^2-12t^2)\,\mathrm dt=-4[/tex]

To evaluate the line integral c f · dr, substitute the values of r(t) into f(x, y, z) to get a new vector function f(t). Find the derivative of r(t) using the chain rule. Take the dot product of f(t) and r'(t) and integrate the result with respect to t over the given bounds.

To evaluate the line integral c f · dr, we need to find the dot product of the vector function f and the derivative of r(t). Since c is given by r(t) and the bounds are -1 ≤ t ≤ 1, we can substitute the values of r(t) into f(x, y, z) and compute the dot product.

Compute the integral to find the final answer.

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Answer:

A. at (8,7) the maximum value is 98

Step-by-step explanation:

First draw the region, that is bounded by all inequalities. This is the triangle with vertices (6,1), (2,5) and (8,7).

Now you can see where the line f(x,y)=7x+6y intersect this region. The maximum value will be at endpoints of this region:

Thus, the maximum value of the function is 98 at the point (8,7).

hope it helps you!!!!!!!!!!!!!!

Answer:

The negative solution is between -3 and -2

The positive solution is between 11 and 13

Step-by-step explanation:

we have

[tex]0=x^{2} -10x-27[/tex]

The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to

[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]

in this problem we have

[tex]x^{2} -10x-27=0[/tex]  

so

[tex]a=1\\b=-10\\c=-27[/tex]

substitute in the formula

[tex]x=\frac{10(+/-)\sqrt{-10^{2}-4(1)(-27)}} {2(1)}[/tex]

[tex]x=\frac{10(+/-)\sqrt{208}} {2}[/tex]

[tex]x=\frac{10(+)\sqrt{208}} {2}=12.21[/tex]

[tex]x=\frac{10(-)\sqrt{208}} {2}=-2.21[/tex]

therefore

The negative solution is between -3 and -2

The positive solution is between 11 and 13

Answer:

Negative solution is between -3 and -2

Positive solution is between 12 and 13

Step-by-step explanation:

The statement that best describes the domain and range of p(x) and q(x) is:

             p(x) and q(x) have the same domain and the same range.

We are given a function p(x) as:

[tex]p(x)=6-x[/tex]

AS the function is a polynomial function.

Hence it is defined everywhere for all the real values.

Hence, the domain of the function p(x) is: All  Real numbers.

and the range of the function p(x) is: All the real numbers.

and the function q(x) is given by:

[tex]q(x)=6x[/tex]

which is also a polynomial function.

Hence, it also has the same domain and range.

Domain and range are specific sets for each function. For given case, p(x) and q(x) have the same domain and range.

Domain is the set of values for which the given function is defined.

Range is the set of all values which the given function can output.

The domain and range of given functions are:

For any real number value of x, p(x) just takes 6-x(negates the input and add 6 to it), thus, its always defined, and thus, its domain is all real numbers.

Since p = 6-x is possible to go negatively infinite and positively infinite and always continuous, thus, its range is all real numbers(all numbers are possible as its output)

We can prove the above statement. Let some real number T is not in the range of p(x). But we have T = 6-x => x = 6-T which is a real number, thus, for input 6-T, there is output T. Thus, its a contradiction, and thus, all real numbers are in range of p(x).

Thus,

Its scaling all numbers. All numbers can be multiplied by 6 and produce a valid result. Thus, its domain is all real numbers.

Suppose that we've T as a real number. Then we can get this as output if we put input as x = T/6 (since then 6x = 6(T/6) = T)

Thus, all real numbers are in its output set, thus, its range is all real numbers.

The Range of q(x): [tex]x \in \mathbb R[/tex] and Domain of q(x): [tex]x \in \mathbb R[/tex]

Hence, for given case, p(x) and q(x) have the same domain and range.

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Answer:

63.2702623...

Step-by-step explanation:

The ratio between FE and DE is the tangent of the angle EDF.[tex]tan 49 = \frac x {55}[/tex] or [tex]x= 55*tan 49[/tex]. With a calculator, you get 63.2702623... Cut where needed

55x tan49 =63.270

63.270.

Answer:

[tex]\frac{1}{6}[/tex]

Step-by-step explanation:

In probability theory, "AND" means multiplication and "OR" means "addition".

We can find the 2 probabilities separately and multiply them (as there is "AND")

So,

Probability number rolled greater than 4 = number of numbers that are greater than 4/total number of numbers

There are 2 numbers greater than 4 in a die (5 & 6) and total 6 numbers, so

P(number greater than 4) = 2/6

Now,

Probability that tails come up in coin toss = 1/2 (there are 1 tail and 1 head in a coin)

Hence,

P(number greater than 4 and tails in coin) = 2/6 * 1/2 = 1/6

To find the probability of rolling a number greater than 4 on a die and getting tails on a coin toss, you multiply the individual probabilities: (1/3) for the die roll and (1/2) for the coin toss, resulting in a combined probability of 1/6.

The question is asking to find the probability of a specific combined event involving the roll of a number cube (a six-sided die) and the flip of a coin. To solve this problem, we need to calculate the probability that the number cube shows a number greater than 4 (which can be either a 5 or 6) and that the coin toss results in tails.

First, we find the probability of rolling a number greater than 4 on a six-sided die. There are 2 favorable outcomes (5 or 6) out of 6 possible outcomes, so the probability of this event is 2/6, which simplifies to 1/3.

Next, we calculate the probability of getting tails on a coin flip. Since a coin has two sides, and only one side is tails, the probability is 1/2.

To find the combined probability of both events happening together, we multiply the probabilities of the individual events:

Combined Probability = Probability (Number > 4) × Probability (Tails) = (1/3) × (1/2)

Therefore, the combined probability is:

(1/3) × (1/2) = 1/6

Answer:

x = 8

Step-by-step explanation:

Since the triangle is right use Pythagoras' theorem to solve for x

The square on the hypotenuse of a right triangle is equal to the sum of the squares on the other 2 sides, thus

x² + 15² = 17²

x² + 225 = 289 ( subtract 225 from both sides )

x² = 64 ( take the square root of both sides )

x = [tex]\sqrt{64}[/tex] = 8

Answer:

ABCD is an isosceles trapezoid.

Step-by-step explanation:

The line AB will be horizontal (parallel to the x axis), because the y values are both 3. It is 9--2 = 11 units long.

CD is also parallel to the  x axis because the y values of C and D are both 6. Itis 5 - 2 = 3 units long.

The x values of the four points are all different so ABCD is a trapezoid.

Let's check the lengths of the line segments AC and BD:

AC = √((5--2)^2 + (6-3)^2 = √58.

BD =  √((9-2)^2 + (3-6)^2 = √58.

They are equal in length so:

ABCD is an isosceles trapezoid.

Answer:

40

Step-by-step explanation:

-5 x -8 = 40

40, because both numbers are negative, the answer is positive. So you would just multiply like normal.

For this case we must find the area of the figure composed of a triangle and a rectangle.

Triangle area:

[tex]A_ {t} = \frac {b * h} {2}[/tex]

Where b is the base and h is the height.

Area of the rectangle:

[tex]A_ {r} = a * b[/tex]

Where a and b are the sides.

The base of the triangle measures:

[tex]13-5 = 8[/tex]

We find the height by trigonometry:

[tex]tg (45) = \frac {h} {b}\\1 = \frac {h} {b}\\b = h[/tex]

So:

[tex]A_ {t} = \frac {8 * 8} {2}\\A_ {t} = 32 \ ft ^ 2[/tex]

On the other hand:

[tex]A_ {r} = 5 * 8\\A_ {r} = 40 \ ft ^ 2[/tex]

Thus, the total are the sum:

[tex](32 + 40) ft ^ 2 = 72 \ ft ^ 2[/tex]

Answer:

Option A

Option is A~ the answer is A

Hi again.

Answer

= option d, 225π mm^2

Circumference = 2πr

30π = 2πr

30 = 2r

30 / 2 = r

15 = r

Area = π[tex]r^{2}[/tex]

[tex]15^{2}[/tex]π

225π

The area of the circle in terms of the π will be 225π mm²

The area of the circle is defined as the space occupied by the circle in the three-dimensional plane. The circle is the locus of the point equidistant from its centre.

It is given in the question that:-

Circumference = 2πr

30π = 2πr

30 = 2r

30 / 2 = r

15 = r

Area = πr²

Area =π(15)²

Area = 225π  mm²

Hence the area of the circle will be 225π  mm²

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Answer:

The number of people in the group must divide 40. 8 is the only selection that divides 40.

Considering the factors of 40 for even distribution of the slips of paper, 8 people can be fairly included in the group as 8 is a factor of 40. The correct answer is option b) 8.

To determine how many people could be in the group when forty slips of paper numbered 1 through 40 are distributed, and a random number generator selects a single number from this range, we need to consider the factors of 40. Each person must get at least one slip of paper, and the distribution needs to be equal to maintain fairness. The factors of 40 are 1, 2, 4, 5, 8, 10, 20, and 40. However, the only possible numbers of people in the group, given by the options, are 6, 8, 12, and 15. Out of these, 8 is a factor of 40, meaning the slips of paper can be distributed evenly among 8 people.

Thus, the correct answer to the student's question is option b) 8.

Answer:

2 2/3 cups of sugar

Step-by-step explanation:

If your ratio is 3 flour to 4 sugar, and you use 2 cups of flour, set up your proportion as follows:

[tex]\frac{3}{4}=\frac{2}{x}[/tex]

Cross multiply to get 3x = 8 and x = 8/3 which is 2 and 2/3 cups of sugar.

The cups of sugar used is 3/2 corresponding to 2 cups of flour.

If we are using the ratio of 3 flour to 4 sugar.

We have 2 cups of flour.

Let us consider the amount of sugar be 'y'.

By writing the proportion /as follows:

[tex]\frac{3}{4} = \;\frac{2}{y}[/tex]

4y =6

y =3/2

Thus, the amount of sugar used is 3/2.

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Answer:

The minimum number of boxes needed is [tex]4[/tex] and the cost is [tex]\$10.40[/tex]

Step-by-step explanation:

step 1

Find the cost of one box

by proportion

[tex]\frac{5}{13}=\frac{1}{x}\\ \\x=13/5\\ \\x=\$2.60[/tex]

step 2

Find the number of boxes for 48 ounces of pasta

by proportion

[tex]\frac{1}{12}=\frac{x}{48}\\ \\x=48/12\\ \\x=4\ boxes[/tex]

step 3

Find the cost of 4 boxes

we know that

The cost of one box is [tex]\$2.60[/tex]

so

The cost of 4 boxes is

[tex]\$2.60(4)=\$10.40[/tex]

Final answer:

To determine the cost for 48 ounces of pasta, one must calculate the cost per box and multiply by the number of boxes required to reach 48 ounces. The cost for the needed 4 boxes is $10.40.

Explanation:

The cost of one box of pasta is calculated by dividing the total cost of the pasta by the number of boxes. Since the total cost of 5 boxes of pasta is $13, to find the cost per box, we divide $13 by 5, which is $2.60 per box. To find the cost of the minimum number of boxes needed to total 48 ounces of pasta, we then determine how many boxes are needed. Since each box contains 12 ounces of pasta, we need 48 ounces / 12 ounces per box = 4 boxes of pasta. Finally, we multiply the number of boxes by the cost per box, which is 4 boxes × $2.60 per box = $10.40.

The formula of the surface area of a cube is 6 x s²

→ s = 9

→ s² = 9²

→ s² = 81

→ 6 x 81 = 486

So, the surface area of the cube is 486 units².

1. Volume is Length x width x height.

Volume = 6 x 4 x 8 = 192 ft^2

Answer is D.

2. Divide the volume by the height to get the area of the base.

Area of base = 312 / 12 = 26 in^2

Answer is D.

3. A 1/2 x 8 x 6 = 24 x 12 = 288 cm^3

   B.  (12 +6)/2 x 5 = 45 x 14 = 630 m^3

4. See attached picture.

[tex]\bf -\cfrac{5}{1-cos(-x)}\implies -\cfrac{5}{\underset{\textit{symmetry identity}}{1-cos(x)}}\impliedby \begin{array}{llll} \textit{let's multiply top/bottom}\\ \textit{by the conjugate 1+cos(x)} \end{array} \\\\\\ \cfrac{-5}{1-cos(x)}\cdot \cfrac{1+cos(x)}{1+cos(x)}\implies \cfrac{-5(1+cos(x))}{\underset{\textit{difference of squares}}{[1-cos(x)][1+cos(x)]}} \\\\\\[/tex]

[tex]\bf \cfrac{-5[1+cos(x)]}{1^2-cos^2(x)}\implies \cfrac{-5-5cos(x)}{\underset{\textit{pythagorean identity}}{1-cos^2(x)}}\implies \cfrac{-5-5cos(x)}{sin^2(x)} \\\\\\ \cfrac{-5}{sin^2(x)}-\cfrac{5cos(x)}{sin^2(x)}\implies -5\cdot \cfrac{1}{sin^2(x)}-5\cdot \cfrac{1}{sin(x)}\cdot \cfrac{cos(x)}{sin(x)} \\\\\\ -5\cdot csc^2(x)-5\cdot csc(x)\cdot cot(x)\implies \boxed{-5csc^2(x)-5csc(x)cot(x)}[/tex]