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2 years ago

Select the equation of the line that is equivalent to the equation: y-1=4(x-3).

Mathematics

1 answer:

Morgarella [4.7K]2 years ago

3 0

The answer is C
First multiple 4 into bracket then put +1 in both side so it will be y = 4x-11

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How to solve quadratic inequalities.

Digiron [165]

Find all the zeros of the polynomial, and arrange the zeros in increasing order. ...

Plot those numbers on the number line as open or closed points based upon the original inequality symbol.

Choose a test value in each interval to see if the interval satisfies the inequality or not.

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1 year ago

Write and solve an inequality that represents the values of x, in feet.

SVEN [57.7K]

Answer:

x < 16 ft

Step-by-step explanation:

Perimeter = x

If it is less than 16, the equation is:

x < 16

-Chetan K

3 0

2 years ago

What is 30+11x-12-6x

Sladkaya [172]

11x-6x=5x
30-12=18
5x+18

8 0

3 years ago

Read 2 more answers

Find the probability of not rolling prime numbers on both dice

gladu [14]

Here’s the prime numbers on one dice: 2 3 5, so 3/6. If you have two die then your answer is 6/12. You have a 50 percent chance of not rolling prime numbers on both.

6 0

2 years ago

Consider the initial value problem yâ€²+2y=4t,y(0)=8.

Xelga [282]

Answer:

Please read the complete procedure below:

Step-by-step explanation:

You have the following initial value problem:

$y'+2y=4t\\\\y(0)=8$

a) The algebraic equation obtain by using the Laplace transform is:

$L[y']+2L[y]=4L[t]\\\\L[y']=sY(s)-y(0)\ \ \ \ (1)\\\\L[t]=\frac{1}{s^2}\ \ \ \ \ (2)\\\\$

next, you replace (1) and (2):

$sY(s)-y(0)+2Y(s)=\frac{4}{s^2}\\\\sY(s)+2Y(s)-8=\frac{4}{s^2}$ (this is the algebraic equation)

$sY(s)+2Y(s)-8=\frac{4}{s^2}\\\\Y(s)[s+2]=\frac{4}{s^2}+8\\\\Y(s)=\frac{4+8s^2}{s^2(s+2)}$ (this is the solution for Y(s))

$y(t)=L^{-1}Y(s)=L^{-1}[\frac{4}{s^2(s+2)}+\frac{8}{s+2}]\\\\=L^{-1}[\frac{4}{s^2(s+2)}]+L^{-1}[\frac{8}{s+2}]\\\\=L^{-1}[\frac{4}{s^2(s+2)}]+8e^{-2t}$

To find the inverse Laplace transform of the first term you use partial fractions:

$\frac{4}{s^2(s+2)}=\frac{-s+2}{s^2}+\frac{1}{s+2}\\\\=(\frac{-1}{s}+\frac{2}{s^2})+\frac{1}{s+2}$

Thus, you have:

$y(t)=L^{-1}[\frac{4}{s^2(s+2)}]+8e^{-2t}\\\\y(t)=L^{-1}[\frac{-1}{s}+\frac{2}{s^2}]+L^{-1}[\frac{1}{s+2}]+8e^{-2t}\\\\y(t)=-1+2t+e^{-2t}+8e^{-2t}=-1+2t+9e^{-2t}$

(this is the solution to the differential equation)

5 0

3 years ago