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2 years ago

What is the slope of a line perpendicular to the line whose equation is

Mathematics

2 answers:

gizmo_the_mogwai [7]2 years ago

5 0

Your slope is 2/5 and your y intercept is (0,-18)

UNO [17]2 years ago

3 0

Slope is 2/5 and y intercept should be (0,-18)

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Help!! 20points! thank you need by tomorrow.

Yakvenalex [24]

Answer:

1. f(x,y) = (x, -y) 2. f(x,y) = (-x,y).

Step-by-step explanation:

I hope that helps

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3 years ago

The formula for centripetal acceleration, a, is given below, where v is the velocity of the object and r is the object's distanc

julsineya [31]

Answer:

Step-by-step explanation:

For this case we have the following equation:

a = v ^ 2 / r

From here, we must clear the value of speed.

We have then:

v ^ 2 = a * r

v = root (a * r)

Answer:

the formula for the velocity is:

v = root (a * r)

3 0

2 years ago

Read 2 more answers

If you answer all 4 i will give you brainliest please answer ASAP

marshall27 [118]

Answer:

9. The graph will shift down 2 units

10. The graph will be moved right by 4 units

11. The graph will be vertically stretched by 2 units

Step-by-step explanation:

PLEASE MARK AS BRAINLIEST! HOPE THIS HELPS! :)

6 0

2 years ago

Three-quarters of a number is 15. Work out four-fifths of the number.

azamat

Answer:

Step-by-step explanation:

let the number be x , then

$\frac{3}{4}$ x = 15 ( multiply both sides by 4 to clear the fraction )

3x = 60 ( divide both sides by 3 )

x = 20 , then

$\frac{4}{5}$ × 20 = 4 × 4 = 16

3 0

2 years ago

Read 2 more answers

What is the Laplace Transform of 7t^3 using the definition (and not the shortcut method)

Leokris [45]

Answer:

Step-by-step explanation:

By definition of Laplace transform we have

L{f(t)} = $L{{f(t)}}=\int_{0}^{\infty }e^{-st}f(t)dt\\\\Given\\f(t)=7t^{3}\\\\\therefore L[7t^{3}]=\int_{0}^{\infty }e^{-st}7t^{3}dt\\\\$

Now to solve the integral on the right hand side we shall use Integration by parts Taking $7t^{3}$ as first function thus we have

$\int_{0}^{\infty }e^{-st}7t^{3}dt=7\int_{0}^{\infty }e^{-st}t^{3}dt\\\\= [t^3\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(3t^2)\int e^{-st}dt]dt\\\\=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\$

Again repeating the same procedure we get

$=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt= \frac{3}{s}[t^2\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t^2)\int e^{-st}dt]dt\\\\=\frac{3}{s}[0-\int_{0}^{\infty }\frac{2t^{1}}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{2}}[\int_{0}^{\infty }te^{-st}dt]\\\\$

Again repeating the same procedure we get

$\frac{3\times 2}{s^2}[\int_{0}^{\infty }te^{-st}dt]= \frac{3\times 2}{s^{2}}[t\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t)\int e^{-st}dt]dt\\\\=\frac{3\times 2}{s^2}[0-\int_{0}^{\infty }\frac{1}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{3}}[\int_{0}^{\infty }e^{-st}dt]\\\\$

Now solving this integral we have

$\int_{0}^{\infty }e^{-st}dt=\frac{1}{-s}[\frac{1}{e^\infty }-\frac{1}{1}]\\\\\int_{0}^{\infty }e^{-st}dt=\frac{1}{s}$

Thus we have

$L[7t^{3}]=\frac{7\times 3\times 2}{s^4}$

where s is any complex parameter

5 0

2 years ago