If $450 is invested at 6% compounded A (annually), B (quarterly), C (monthly), what is the amount after 7 years? How much interest is earned?

Answer:

$17,277.07

Step-by-step explanation:

Present value of annuity is the present worth of cash flow that is to be received in the future, if future value is known, rate of interest is r and time is n then PV of annuity is

PV of annuity = [tex]\frac{P[1-(1+r)^{-n}]}{r}[/tex]

                      = [tex]\frac{3000[1-(1+0.10)^{-9}]}{0.10}[/tex]

                      = [tex]\frac{3000[1-(1.10)^{-9}]}{0.10}[/tex]

                      = [tex]\frac{3000[1-0.4240976184]}{0.10}[/tex]

                      = [tex]\frac{3000(0.5759023816)}{0.10}[/tex]

                      = [tex]\frac{1,727.7071448}{0.10}[/tex]

                      = 17,277.071448 ≈ $17,277.07

This question involves operations on sets to identify specific members based on conditions. Part a) resolves to D. {2,4}, while part b) finds the solution to be B. {2,4,6,7,8,9,10,11}, highlighting the application of intersection, union, and complement operations in set theory.

To solve these problems, we need to understand the operations on sets such as intersection (A∩B), union (A∪B), and the complement of a set (Bc). For part a), we identify set A as {2,4,6,8,10}, B as {1,3,5}, and C as {1,2,3,4,5}. A∩(B∪C) means we're looking for the intersection of A with the union of B and C. Since B∪C = {1,2,3,4,5}, intersecting this with A gives us D. {2,4} as the answer.

For part b), (A∩Bc)∪(B∩C)c means we're looking at elements in A but not in B, combined with elements not in both B and C. Since Bc = {6,7,8,9,10,11} and (B∩C)c = {6,7,8,9,10,11}, union these two gives us answer B. {2,4,6,7,8,9,10,11}, by including A∩Bc = {2,4,6,8,10} and excluding duplicates when union with (B∩C)c.

a.

[tex]f(x,y)=e^{-(x-a)^2-(y-b)^2}\implies\begin{cases}f_x=-2(x-a)e^{-(x-a)^2-(y-b)^2}\\f_y=-2(y-b)e^{-(x-a)^2-(y-b)^2}\end{cases}[/tex]

Critical points occur where [tex]f_x=f_y=0[/tex]. The exponential factor is always positive, so we have

[tex]\begin{cases}-2(x-a)=0\\-2(y-b)=0\end{cases}\implies(x,y)=\boxed{(a,b)}[/tex]

b. As the previous answer established, the critical point occurs at (-3, 8) if [tex]\boxed{a=-3}[/tex] and [tex]\boxed{b=8}[/tex].

c. Check the determinant of the Hessian matrix of [tex]f(x,y)[/tex]:

[tex]\mathbf H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}[/tex]

The second-order derivatives are

[tex]f_{xx}=(-2+4(x-a)^2)e^{-(x-a)^2-(y-b)^2}[/tex]

[tex]f_{xy}=4(x-a)(y-b)e^{-(x-a)^2-(y-b)^2}[/tex]

[tex]f_{yx}=4(x-a)(y-b)e^{-(x-a)^2-(y-b)^2}[/tex]

[tex]f_{yy}=(-2+4(y-b)^2)e^{-(x-a)^2-(y-b)^2}[/tex]

so that the determinant of the Hessian is

[tex]\det\mathbf H(x,y)=f_{xx}f_{yy}-{f_{xy}}^2=\left((4(x-a)^2-2)(4(y-b)^2-2)-16(x-a)^2(y-b)^2\right)e^{-2(x-a)^2-2(y-b)^2}[/tex]

[tex]\det\mathbf H(x,y)=(16(x-a)^2(y-b)^2-8(x-a)^2-8(y-b)^2)+4)e^{-2(x-a)^2-2(y-b)^2}[/tex]

The sign of the determinant is unchanged by the exponential term so we can ignore it. For [tex]a=x=-3[/tex] and [tex]b=y=8[/tex], the remaining factor in the determinant has a value of 4, which is positive. At this point we also have

[tex]f_{xx}(-3,8;a=-3,b=8)=-2[/tex]

which is negative, and this indicates that (-3, 8) is a local maximum.

The critical point of the given function is (a, b). For it to be at (-3, 8), a and b should be -3 and 8 respectively. This point (-3, 8) is a local maximum for the function.

The critical point of the given equation f(x, y) = e−(x − a)² − (y − b)² are the coordinates (a, b).

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Answer:

15.7 years

Step-by-step explanation:

we know that

The deforestation is a exponential function of the form

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

where

y ----> the number of trees still remaining in the forest

x ----> the number of years

a is the initial value (a=500,000 threes)

b is the base

b=100%-4.7%=95.3%=95.3/100=0.953

substitute

[tex]y=500,000(0.953)^{x}[/tex]

The linear equation of planting threes in the region is equal to

[tex]y=15,000x[/tex]

using a graphing tool

Solve the system of equations

The intersection point is (15.7,235,110)

see the attached figure

therefore

For x=15.7 years

The number of trees they have planted will be equal to the number of trees still remaining in the forest

Answer: There is only one solution, and it is viable.

Step-by-step explanation:

Answer:

1) x = 17,  line a parallel to line b

2) x = 18,  line a parallel to line b

Step-by-step explanation:

If line a parallel to line b then (10x - 40) + 50 = 180

Solve for x

10x - 40 + 50 = 180

Combine like terms

10x + 10 = 180

10x = 170

  x = 17

x = 17,  line a parallel to line b

-------------------------------------------------

If line a parallel to line b then 5x - 16 = 74

Solve for x

5x - 16 = 74

5x = 90

  x = 18

x = 18,  line a parallel to line b

The value of x which makes line a parallel to line b can be found by equating the slopes of the two lines and solving for x.

In mathematics, two lines a and b are parallel if and only if their slopes are equal. When we are given the equations of the lines and are asked to find the value of x that make the lines parallel, we start by setting the slopes of the two lines equal to each other. Let's assume now that line a is represented by y = mx + b1 and line b by y = nx + b2. In order for line a to be parallel to line b, m must be equal to n. Therefore, you solve for x from the equation m=n.

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Answer:

take 47.99 x .20 = 9.598

$9.60 off

then take 47.99 - 9.60 = $ 38.39

take 38.39 x .06 = 2.3034

$ 2.30 (tax)

add 38.39 + 2.30 = $40.69 or $40.70 is the final purchase price

(the two amounts depends on your choice answer or how it is rounded)

Step-by-step explanation:

Answer:

Option C is correct

Step-by-step explanation:

We need to factorize the expression:

[tex]x^2+10x+24[/tex]

For factorization we need to break the middle term such that the product is equal to 24x^2 and the sum is equal to 10x

We know that 6*4 = 24 and 6+4 =10

So, solving

[tex]=x^2+6x+4x+24[/tex]

Taking common

[tex]=x(x+6)+4(x+6)[/tex]

[tex]=(x+4)(x+6)[/tex]

So, the factors of

[tex]x^2+10x+24[/tex]

are

[tex](x+4)(x+6)[/tex]

Hence Option C is correct

The correct answer is actually D. The person who had answered before me had accurate math, but they were confused. The answer  

(x + 6)(x + 4) is on D, not C.

(a+3)(a-2)

Multiply the two brackets together

a^2-2a+3a+3*-2

a^2+a-6

Answer is a^2+a-6

ANSWER

[tex]{a}^{2} + a - 6[/tex]

EXPLANATION

The given expression is

[tex](a + 3)(a - 2)[/tex]

We expand using the distributive property to obtain:

[tex]a(a - 2) + 3(a - 2)[/tex]

We multiply out the parenthesis to get:

[tex] {a}^{2} - 2a + 3a - 6[/tex]

Let us now simplify by combining the middle terms to obtain;

[tex]{a}^{2} + a - 6[/tex]

The work is equal to the line integral of [tex]\vec F[/tex] over each line segment.

Parameterize the paths

The work done by [tex]\vec F[/tex] over each segment (call them [tex]C_1,\ldots,C_4[/tex]) is

[tex]\displaystyle\int_{C_1}\vec F\cdot\mathrm d\vec r_1=\int_0^2\vec0\cdot\vec\imath\,\mathrm dt=0[/tex]

[tex]\displaystyle\int_{C_2}\vec F\cdot\mathrm d\vec r_2=\int_0^1(t^2\,\vec\imath+24t\,\vec\jmath+32t^2\,\vec k)\cdot(4\,\vec\jmath+\vec k)\,\mathrm dt=\int_0^1(96t+32t^2)\,\mathrm dt=\frac{176}3[/tex]

[tex]\displaystyle\int_{C_3}\vec F\cdot\mathrm d\vec r_3=\int_0^2(\vec\imath+(24-12t)\,\vec\jmath+32\,\vec k)\cdot(-\vec\imath)\,\mathrm dr=-\int_0^2\mathrm dt=-2[/tex]

[tex]\displaystyle\int_{C_4}\vec F\cdot\mathrm d\vec r_4=\int_0^1((1-t)^2\,\vec\imath+2(4-4t)^2\,\vec k)\cdot(-4\,\vec\jmath-\vec k)\,\mathrm dt=-2\int_0^1(4-4t)^2\,\mathrm dt=-\frac{32}3[/tex]

Then the total work done by [tex]\vec F[/tex] over the particle's path is 46.

The total work done by the force field as the particle follows this path is 4/3 units of work.

To find the total work done by the force field as the particle moves along the path, we'll calculate the line integrals for each segment and then sum them up:

For the first segment (from the origin to (2, 0, 0)):

Parameterization: r1(t) = (2t, 0, 0), t from 0 to 1.

Work on this segment: ∫[0,1] F(r1(t)) · r1'(t) dt

F(r1(t)) = (0, 0, 0), as z = 0.

r1'(t) = (2, 0, 0).

∫[0,1] F(r1(t)) · r1'(t) dt = ∫[0,1] (0) dt = 0.

For the second segment (from (2, 0, 0) to (2, 4, 1)):

Parameterization: r2(t) = (2, 4t, t), t from 0 to 1.

Work on this segment: ∫[0,1] F(r2(t)) · r2'(t) dt

F(r2(t)) = (t^2, 0, 8t^2).

r2'(t) = (0, 4, 1).

∫[0,1] F(r2(t)) · r2'(t) dt = ∫[0,1] (0 + 0 + 8t^2) dt = ∫[0,1] 8t^2 dt = [8t^3/3] from 0 to 1 = (8/3).

For the third segment (from (2, 4, 1) to (0, 4, 1)):

Parameterization: r3(t) = (2 - 2t, 4, 1), t from 0 to 1.

Work on this segment: ∫[0,1] F(r3(t)) · r3'(t) dt

F(r3(t)) = (1, 0, 8).

r3'(t) = (-2, 0, 0).

∫[0,1] F(r3(t)) · r3'(t) dt = ∫[0,1] (-2) dt = [-2t] from 0 to 1 = -2.

For the fourth segment (from (0, 4, 1) back to the origin):

Parameterization: r4(t) = (0, 4 - 4t, 1), t from 0 to 1.

Work on this segment: ∫[0,1] F(r4(t)) · r4'(t) dt

F(r4(t)) = (1, 0, 32 - 8t).

r4'(t) = (0, -4, 0).

∫[0,1] F(r4(t)) · r4'(t) dt = ∫[0,1] (0 + 0 + 0) dt = 0.

Now, sum up the work done on each segment:

Total work = 0 + (8/3) - 2 + 0 = 4/3.

So, the total work done by the force field as the particle follows this path is 4/3 units of work.

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In this question (https://brainly.com/question/12792658) I derived the Taylor series for [tex]\mathrm{sinc}\,x[/tex] about [tex]x=0[/tex]:

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

Then the Taylor series for

[tex]f(x)=\displaystyle\int_0^x\mathrm{sinc}\,t\,\mathrm dt[/tex]

is obtained by integrating the series above:

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

We have [tex]f(0)=0[/tex], so [tex]C=0[/tex] and so

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

which converges by the ratio test if the following limit is less than 1:

[tex]\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}x^{2n+3}}{(2n+3)^2(2n+2)!}}{\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}}\right|=|x^2|\lim_{n\to\infty}\frac{(2n+1)^2(2n)!}{(2n+3)^2(2n+2)!}[/tex]

Like in the linked problem, the limit is 0 so the series for [tex]f(x)[/tex] converges everywhere.

The Taylor series for the function f(x) = ∫ sinc(t)dt based at 0 is derived from the Taylor series of sinc(x) by integrating it term by term, given in summation notation as ∑ (-1)ⁿ * xⁿ⁺¹ / (n+1)! for n=0 to n=∞. The series converges for all real numbers (-∞, ∞).

In order to find the Taylor series for the function f(x) = ∫ sinc(t)dt based at 0, one can use the Taylor series for sinc(x) and integrate term by term. We know the Taylor series for sinc(x) is x - x³/3! + x⁵/5! - ..., so the Taylor series for f(x) can be written as x²/2 - x⁴/4*3! + x⁶/6*5! - ... . In summation notation, this is ∑ (-1)ⁿ * xⁿ⁺¹ / (n+1)! for n=0 to n=∞.

The Taylor series for any function converges to the function itself within a certain interval called the radius of convergence. For the Taylor series of sinc(x), due to the nature of sine being bounded between -1 and 1, the series will converge for all real numbers (-∞, ∞).

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Answer:

Step-by-step explanation:

A ratio is defined as a comparison of two amounts by division.It is of the form of [tex]\frac{p}{q}[/tex] where p and q are the quantities.

A. 1580 people/square mile in division form

1580[tex]\frac{people}{\text{square mile}}[/tex]

This satisfies ratio definition as this compares two quantities People and Square mile and is of the form   [tex]\frac{p}{q}[/tex]

Where,

p: People, q:  Square mile

Example,

[tex]\frac{1580\times people}{1\times squaremile}[/tex]

expresses that per 1 square mile there are 1580 people. Thus rate satisfies the ratio definition.

B. 360 kilowatt-hour/4 months

in division form [tex]\frac{\text{360 kilowatt-hour}}{\text{4months}}[/tex]

This satisfies ratio definition as this compares two quantities kilowatt-hour and months and is of the form   [tex]\frac{p}{q}[/tex]

where, p: kilowatt-hour and q: people

Example,

[tex]\frac{\360\times kilowatt-hour}{4\times months}[/tex]

expresses that per 4 months there are 360 kilowatt-hour. Thus, rate satisfies the ratio definition.

C. 450 people/year

In division form 450[tex]\frac{people}{year}[/tex]

This satisfies ratio definition as this compares two quantities people and year and is of the form   [tex]\frac{p}{q}[/tex]

where, p: people and q: year

Example,

[tex]\frac{450\times people}{1\times\text{year}}[/tex]

expresses that per year there are 450 people. Thus, rate satisfies the ratio definition.

D. 355 calories/6 ounces

In division form [tex]\frac{355clories}{6ounces}[/tex]

This satisfies ratio definition as this compares two quantities calories and ounces and is of the form   [tex]\frac{p}{q}[/tex]

where p: calories and q: ounces

Example,

[tex]\frac{355\times calories}{6\times\text{ounces}}[/tex]

expresses that per 6 ounces there are 355 calories. Thus, rate satisfies the ratio definition.

Answer:

Yes they do.

And yes they do follow a normal distribution.

Percentages are close to 68-95-99.7%, we can declare that yes, the 68-95-99.7% rule is roughly followed and yes data appear to follow a normal distribution.

It's the probability curve of a continuous distribution that's most likely symmetric around the mean. On the Z curve, at Z=0, the chance is 50-50. A bell-shaped curve is another name for it.

We have a data of final exam scores of 20 Introductory.

a) Range of 1 standard deviation:

(77.7 – 8.44, 77.7 + 8.44)                [69.3, 86.1]

Range of 2 standard deviation:

(77.7 – 2(8.44), 77.7 + 2(8.44))            [60.8, 94.6]

Range of 3 standard deviation:

(77.7 – 3(8.44), 77.7 + 3(8.44))           [52.4, 103.0]

Number of data points lie within 1 standard deviation = 14

Percent of data points lie within 1 SD = (14/20)×100 = 70%

Number of data points lie within 2 SD = 19

Percent of data points lie within 1 SD = (19/20)×100 = 95%

Number of data points lie within 3 SD = 20

Percent of data points lie within 1 SD = (20/20)×100 = 100%

Because these percentages are close to 68-95-99.7%, we can declare that yes, the 68-95-99.7% rule is roughly followed.

b)

Because the histogram in the graph is symmetric, and the normal probability plot reveals that the points are very close to a straight line, the data appears to follow a normal distribution.

Thus, percentages are close to 68-95-99.7%, we can declare that yes, the 68-95-99.7% rule is roughly followed and yes data appear to follow a normal distribution.

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Answer:

The wedge cut from the first octant ⟹ z ≥ 0 and y ≥ 0 ⟹ 12−3y^2 ≥ 0 ⟹ 0 ≤ y ≤ 2  

0 ≤ y ≤ 2 and x = 2-y ⟹ 0 ≤ x ≤ 2  

V = ∫∫∫ dzdydx  

dz has changed from zero to 12−3y^2  

dy has changed from zero to 2-x  

dx has changed from zero to 2  

V = ∫∫∫ dzdydx = ∫∫ (12−3y^2) dydx = ∫ 12(2-x)-(2-x)^3 dx =  

24(2)-6(2)^2+(2-2)^4/4 -(2-0)^4/4 = 20

Step-by-step explanation:

It can be deduced that the volume of the wedge cut from the first octant will be 20.

From the information, the wedge cut from the first octant will be z ≥ 0 and y ≥ 0 = 12−3y² ≥ 0 = 0 ≤ y ≤ 2  

Also, it can be deduced that 0 ≤ y ≤ 2 and x = 2-y ⟹ 0 ≤ x ≤ 2. Therefore, V = ∫dzdydx. In this case,

dz = 12−3y  

dy = 2-x  

dx = 2  

V = ∫dzdydx = ∫(12−3y²) dydx = ∫12(2-x)-(2-x)³dx

= [24(2) - 6(2)² + (2-2)⁴/4 -(2-0)⁴]/4

= 20

In conclusion, the volume of the wedge cut from the first octant will be 20.

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Answer:

should be 53 if im right

Final answer:

The probability of choosing a green ball from the three boxes, given their individual selection probabilities and color distributions, is calculated using the law of total probability. The overall probability of selecting a green ball is found to be 29/80, or roughly 36.25%.

Explanation:

The question asks for the probability of choosing a green ball from three different boxes, given their individual probabilities of being chosen and the distribution of red and green balls in each box. To solve this, we employ the law of total probability which combines the probability of each event (selecting a box) with the conditional probability of finding a green ball within that selected box.

Box 1: Probability of green ball = 5 green balls / (5 red + 5 green) = 1/2

Box 2: Probability of green ball = 3 green balls / (7 red + 3 green) = 3/10

Box 3: Probability of green ball = 4 green balls / (6 red + 4 green) = 2/5

The overall probability is calculated as: P(Green) = P(Box 1) * P(Green|Box 1) + P(Box 2) * P(Green|Box 2) + P(Box 3) * P(Green|Box 3) = (1/4) * (1/2) + (1/2) * (3/10) + (1/4) * (2/5) = 1/8 + 3/20 + 1/10 = 29/80.

Therefore, the probability that the ball chosen is green is 29/80 or approximately 36.25%.

Intersecting chords theorem:

[tex]7\cdot7=10x[/tex]

i.e. when two chords intersect, the products of the resulting line segments' lengths are equal. Then

[tex]10x=49\implies x=\dfrac{49}{10}=\boxed{4.9}[/tex]

Answer:

4.9

Step-by-step explanation:

got it right

Answer:

  66.2

Step-by-step explanation:

We know that the middle 68% of the distribution is between -1 and +1 standard deviations from the mean. -1 standard deviation is a score of ...

  69.9 - 3.7 = 66.2

_____

Comment on the question

The way the question is worded, "A" can be any value less than 68.1, down to -∞. The question does does not require the A-B range to be symmetrical.

Answer: Hence, Option 'B' is correct.

Step-by-step explanation:

Since we have given that

Number of births = 1400

Number of birth of girls = 1378

Number of birth of boys is given by

[tex]1400-1378\\\\=22[/tex]

so, the number of girls is significantly higher than the number of boys.

So, the number of births of girls is significantly high.

Hence, Option 'B' is correct.

B. The number of girls is significantly high.

When evaluating whether the number of girl births in a sample is significantly high or low, we can reference the expected natural ratio of girls to boys, which is typically 100:105. Given that in the scenario 1378 out of 1400 births were girls, this significantly deviates from the expected natural ratio. For comparison, an article in Newsweek states that the natural ratio is 100:105, and in China, it is 100:114, which equals 46.7 percent girls. If we consider a sample where out of 150 births, there are 60 girls (or 40%), this is lower than the expected percentage based on China's statistics but not implausible. However, in the case of the scenario with 1378 girls out of 1400 births, the proportion of girls is approximately 98.43%, which seems very unlikely given the natural ratio, suggesting an unusual or non-random process may be involved.

Therefore, based on subjective judgment and without applying more precise statistical tests, the number of girls being 1378 out of 1400 births is significantly high compared to natural birth rates or the stated birth rate in China. This leads us to select the correct answer: B. The number of girls is significantly high.

Answer:

quantitative

Step-by-step explanation:

Answer: There are 22 whole cheese sandwiches that can be prepared.

Step-by-step explanation:

Since we have given that

Number of slices of bread = 44

Number of slices of cheese = 75

According to question, a cheese sandwich consists of 2 slices of bread and 3 slices of cheese.

So, we need to find the number of whole cheese sandwiches that can be prepared.

Number of sandwich containing only slice of bread is given by

[tex]\dfrac{44}{2}=22[/tex]

Number of sandwich containing only slice of cheese is given by

[tex]\dfrac{75}{3}=25[/tex]

As we know that each sandwich should contain both slice of bread and slice of cheese.

So, Least of (22, 25) = 22

Hence, there are 22 whole cheese sandwiches that can be prepared.

First confirm that [tex]y_1=xe^{-x}[/tex] is a solution to the ODE,

[tex]y''+2y'+y=0[/tex]

We have

[tex]{y_1}'=e^{-x}-xe^{-x}=(1-x)e^{-x}[/tex]

[tex]{y_1}''=-e^{-x}-(1-x)e^{-x}=(-2+x)e^{-x}[/tex]

Substituting into the ODE gives

[tex](-2+x)e^{-x}+2(1-x)e^{-x}+xe^{-x}=0[/tex]

Suppose [tex]y_2(x)=v(x)y_1(x)[/tex] is another solution to this ODE. Then

[tex]{y_2}'=v'y_1+v{y_1}'[/tex]

[tex]{y_2}''=v''y_1+2v'{y_1}'+v{y_1}''[/tex]

and substituting these into the ODE yields

[tex](v''y_1+2v'{y_1}'+v{y_1}'')+2(v'y_1+v{y_1}')+vy_1=0[/tex]

[tex]xe^{-x}v''+2e^{-x}v'=0[/tex]

[tex]xv''+2v'=0[/tex]

Let [tex]w(x)=v'(x)[/tex]. Then the remaining ODE is linear in [tex]w[/tex]:

[tex]xw'+2w=0[/tex]

Multiply both sides by the integrating factor, [tex]x[/tex], and condense the left hand side as a derivative of a product:

[tex]x^2w'+2xw=(x^2w)'=0[/tex]

Integrate both sides with respect to [tex]x[/tex] and solve for [tex]w[/tex]:

[tex]x^2w=C_1\implies w=C_1x^{-2}[/tex]

Back-substitute and integrate both sides with respect to [tex]x[/tex] to solve for [tex]v[/tex]:

[tex]v'=C_1x^{-2}\implies v=-C_1x^{-1}+C_2[/tex]

Back-substitute again to solve for [tex]y_2[/tex]:

[tex]\dfrac{y_2}{y_1}=C_2-\dfrac{C_1}x[/tex]

[tex]\implies y_2=C_2xe^{-x}-C_1e^{-x}[/tex]

[tex]y_1[/tex] already captures the solution [tex]xe^{-x}[/tex], so the remaining one is

[tex]\boxed{y_2=e^{-x}}[/tex]

A differential equation shows the relationship between functions and their derivatives.

The equation of [tex]y_2(x)[/tex] is: [tex]y_2 = -e^{-x}[/tex]

The given parameters are:

[tex]y_2 = y_1(x) \int\frac{ e^{(\int -P(x)\ dx)} }{ y_1^2(x) }dx[/tex]

[tex]y" + 2y' + y = 0[/tex]

[tex]y_1 = xe^{-x}[/tex]

The general equation is:

[tex]y" + P(x) y' + Q(x)y = 0[/tex]

Compare the above equation to [tex]y" + 2y' + y = 0[/tex]

[tex]P(x) = 2[/tex]

Integrate:

[tex]\int\limits^x_0 P(x') dx'= \int\limits^x_0 2 dx'[/tex]

[tex]\int\limits^x_0 P(x') dx'= 2x|\limits^x_0[/tex]

[tex]\int\limits^x_0 P(x') dx'= 2[x - 0][/tex]

[tex]\int\limits^x_0 P(x') dx'= 2[x ][/tex]

[tex]\int\limits^x_0 P(x') dx'= 2x[/tex]

We have:

[tex]y_2 = y_1(x) \int\frac{ e^{(\int -P(x)\ dx)} }{ y_1^2(x) }dx[/tex]

The above equation becomes:

[tex]y_2 = y_1(x) \int\frac{e^{(\int -2 dx)} }{ y_1^2(x) }dx[/tex]

Substitute [tex]y_1 = xe^{-x}[/tex]

[tex]y_2 = xe^{-x} \int\frac{ e^{(\int -2 dx)} }{ (xe^{-x})^2 }dx[/tex]

Integrate

[tex]y_2 = xe^{-x} \int\frac{ e^{-2x}}{ (xe^{-x})^2 }dx[/tex]

Evaluate the exponents

[tex]y_2 = xe^{-x} \int\frac{ e^{-2x}}{ x^2e^{-2x} }dx[/tex]

Cancel out common factors

[tex]y_2 = xe^{-x} \int\frac{1}{ x^2 }dx[/tex]

Rewrite as:

[tex]y_2 = xe^{-x} \int x^{-2}dx[/tex]

Integrate

[tex]y_2 = xe^{-x} \times -\frac{1}{x} + c[/tex]

[tex]y_2 = -e^{-x} + c[/tex]

Set c to 0.

[tex]y_2 = -e^{-x} + 0[/tex]

[tex]y_2 = -e^{-x}[/tex]

Hence, the equation of [tex]y_2(x)[/tex] is:

[tex]y_2 = -e^{-x}[/tex]

Read more about differential equations at:

https://brainly.com/question/14620493

[tex]\sin(3x+13)=\cos(4x)\\\sin(3x+13)=\cos(90-4x)\\3x+13=90-4x\\7x=77\\x=11[/tex]

Answer:

1. Y

2. N

3. N

4. N

Step-by-step explanation:

Let's use the second equation, since it seems to be easier to use.

To check if an ordered pair is a solution, plug it in to the equation.

1. [tex]-10+18=8[/tex] --> Y

2. [tex]25-12=13[/tex] --> N

3. [tex]0-9=-9[/tex] --> N

4. [tex]35-27=8[/tex] --> N

Edit : The 4th equation doesn't work for the first equation, whereas the first one still does.

Answer:

(-2,-6)

Step-by-step explanation:

-9x +2y = 6

5x - 3y = 8

1) Make one of the coefficients the same - y.

-9x +2y = 6 * 3

5x - 3y = 8 * 3

-27x +6y = 18

10x - 6y = 16

2) Add the new equations.

(-27x +6y = 18) + (10x - 6y = 16)

(-27x +6y) + (10x - 6y) = 18 + 16

-17x = 34

3) Divide to find the value of x

-17x = 34

x = 34/-17

x = -2

4) Substitute x into either equation to find the value of y.

-9(-2) +2y = 6

18 +2y = 6

2y = -12

y = -12/2

y = -6

5(-2) - 3y = 8

-10 - 3y = 8

-3y = 18

y = 18/-3

y = -6

Your answer is (-2,-6).

Try this solution:

the rule: if f1>f2, then E1>E2, where f1;f2 - frequency, E1;E2 - energy of light.

The formula is L=c/f, where L - the wavelength, c - 3*10⁸ m/s, f - frequency.

Frequency for the wavelength 519 nm. is:

[tex]f=\frac{c}{L}=\frac{3*10^8}{519*10^{-9}}=\frac{3*10^17}{519}=578034682080924=5.78*10^{14}( \frac{1}{sec})[/tex]

Answer: the energy of light of wavelength 519 nm.

Answer:

Option D: [tex]7.59*10^{2}[/tex]

Step-by-step explanation:

we know that

The volume of Saturn is about [tex]8.27*10^{14}\ km^{3}[/tex]

The volume of Earth is about [tex]1.09*10^{12}\ km^{3}[/tex]

Dividing Saturn's volume by Earth's volume

[tex]\frac{8.27*10^{14}}{1.09*10^{12}} =(\frac{8.27}{1.09})*10^{14-12} =7.59*10^{2}[/tex]

The correct answer is option D. The quotient in scientific notation is [tex]75.9 \times 10^2[/tex]

To find the number of Earths that can fit inside Saturn, we divide the volume of Saturn by the volume of Earth. Given the volumes:

Saturn's volume [tex]= 8.27 \times 10^{26}\ km^3[/tex]

Earth's volume [tex]= 1.09 \times 10^{24}\ km^3[/tex]

The calculation is as follows:

Number of Earths in Saturn = Saturn's volume / Earth's volume

[tex]=\frac{ (8.27 \times 10^{26})} {(1.09 \times 10^{24})}[/tex]

To simplify this, we divide the coefficients and subtract the exponents of 10:

[tex]= \frac{8.27}{ 1.09} \times \frac{10^{26}} { 10^{24}}[/tex]

[tex]= 7.59 \times 10^2[/tex]

[tex]\bf 12~~,~~\stackrel{12+3}{15}~~,~~\stackrel{15+3}{18}~~,~~\stackrel{18+3}{21}~\hspace{10em}\stackrel{\textit{common difference}}{d=3} \\\\[-0.35em] ~\dotfill\\\\ n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} a_n=n^{th}\ term\\ n=\textit{term position}\\ a_1=\textit{first term}\\ d=\textit{common difference}\\ \cline{1-1} a_1=12\\ d=3 \end{cases} \\\\\\ a_n=12+(n-1)3\implies a_n=12+3n-3\implies a_n=3n+9[/tex]

Answer:

tₙ = 3(3 + n)

Step-by-step explanation:

Points to remember

nth term of an AP

tₙ = a + (n - 1)d

Where a - first term of AP

d - Common difference of AP

To find the nth term  

The given series is,

12,15,18,21 .....

Here a = 12 and d = 15 - 12 = 3

tₙ = a + (n - 1)d

  = 12 + (n - 1)3

  =12 + 3n - 3

  = 9 + 3n

  = 3(3 + n)

Therefore tₙ = 3(3 + n)

Answer:

[tex]f'''(x)=\frac{3}{x^{2}}[/tex]

Step-by-step explanation:

We are given with Second-order derivative of function f(x).

[tex]f''(x)=9-\frac{3}{x}[/tex]

We need to find Third-order derivative of the function f(x).

[tex]f''(x)=9-\frac{3}{x}=9-3x^{-1}[/tex]

We know that,

f'''(x) = (f''(x))'

So,

[tex]f'''(x)=\frac{\mathrm{d}\,f''(x)}{\mathrm{d} x}[/tex]

[tex]f'''(x)=\frac{\mathrm{d}\,(9-3x^{-1})}{\mathrm{d} x}[/tex]

[tex]f'''(x)=\frac{\mathrm{d}\,9}{\mathrm{d} x}-\frac{\mathrm{d}\,3x^{-1}}{\mathrm{d} x}[/tex]

[tex]f'''(x)=0-3\frac{\mathrm{d}\,x^{-1}}{\mathrm{d} x}[/tex]

[tex]f'''(x)=-3(-1)x^{-1-1}[/tex]

[tex]f'''(x)=3x^{-2}[/tex]

[tex]f'''(x)=\frac{3}{x^{2}}[/tex]

Therefore, [tex]f'''(x)=\frac{3}{x^{2}}[/tex]

Answer:

  12.683%

Step-by-step explanation:

The effective annual rate is given by ...

  (1 +r/n)^n -1

where r is the nominal annual rate, and n is the number of compoundings per year. Filling in the given numbers, we have ...

  effective rate = (1 +0.12/12)^12 -1 ≈ 0.12683 = 12.683%

Answer:

$1314.37

Step-by-step explanation:

We have to calculate final value i.e. balance to earn $100,000 annually from interest.

= [tex]\frac{100,000}{0.08}[/tex] = $1,250,000

Now, N = n × y  = 12 × 25 = 300

         I  = 8% =  APR = 0.08

        PV = 0  = PMT = 0

        FV = 1,250,000 = A

[tex]A=\frac{PMT\times [(1+\frac{apr}{n})^{ny}-1]}{\frac{apr}{n}}[/tex]

[tex]PMT=\frac{A\times (\frac{APR}{n})}{[(1+\frac{APR}{n})^{ny}-1]}[/tex]

[tex]PMT=\frac{1,250,000\times (\frac{0.08}{12})}{[(1+\frac{0.08}{12})^{12\times 25}-1]}[/tex]

[tex]PMT=\frac{1,250,000\times (0.006667)}{[(1+\frac{0.08}{12})^{12\times 25}-1]}[/tex]

[tex]PMT=\frac{1,250,000\times (0.006667)}{[(1+0.006667)^{300}-1]}[/tex]

[tex]PMT=\frac{\frac{25000}{3}}{[1.006667^{300}-1]}[/tex]

[tex]PMT=\frac{\frac{25000}{3}}{6.340176}[/tex]

Monthly payment (PMT) = $1314.369409 ≈ $1314.37

$1314.37 is required monthly payment in order to $100,000 interest.

Answer:(-6,312), (6,312)

Step by Step explanation:

10x^2-y=48

y=-48+10x^2

2(-48+10x^2)=16x^2+48

Solve the equation for x.

x=-6

x=6

y=-48+10(-6)^2

y=-48+10×6^2

y=312

y=312