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3 years ago

May someone explain to me how to do this?

Mathematics

1 answer:

lara31 [8.8K]3 years ago

4 0

Answer:

All you gotta do is go to "desmos" and type in all of the 4 solutions. A, B, C, and D. and see which answer matches the graph

Step-by-step explanation:

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What is the length of one side of a square that has an area of 196 inches

labwork [276]

I think it would be 14

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3 years ago

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What are the x and y- intercepts for the line 4(y+3)=−2(x+6)?<br><br> Please give an actual answer.

kow [346]

Hello!

X and Y intercepts are found when graphing an equation. You know the X and Y intercepts by when a point crosses the X axis, and for the Y intercept, when a point crosses the Y axis.

First off we must simplify the equation.

4y + 12 = -2x - 12

Here is the photo of the graph attached.

By looking at the graph we can see the answer is...

X intercept = (-12,0)
Y intercept = (0,-6)

6 0

4 years ago

-4x+9(10x-1) <br> please help me this is distributive property

kotykmax [81]

Let me walk you through this real quick:

-4x + 9 (10x - 1)
-4x + 90x - 9 (open up the brackets)
90x - 4x - 9
= 86x - 9

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3 years ago

Tell which property the statement illustrates 5•p=p•5

solong [7]

Answer:

Yuhh the answers c dawhggg trus me i just did the same thing lawl!

Step-by-step explanation:

7 0

3 years ago

Use the laplace transform to solve the given initial-value problem. y' 5y = e4t, y(0) = 2

Basile [38]

The Laplace transform of the given initial-value problem

$y' 5y = e^{4t}, y(0) = 2$ is mathematically given as

$y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}$

<h3>What is the Laplace transform of the given initial-value problem? y' 5y = e4t, y(0) = 2?</h3>

Generally, the equation for the problem is mathematically given as

$&\text { Sol:- } \quad y^{\prime}+s y=e^{4 t}, y(0)=2 \\\\&\text { Taking Laplace transform of (1) } \\\\&\quad L\left[y^{\prime}+5 y\right]=\left[\left[e^{4 t}\right]\right. \\\\&\Rightarrow \quad L\left[y^{\prime}\right]+5 L[y]=\frac{1}{s-4} \\\\&\Rightarrow \quad s y(s)-y(0)+5 y(s)=\frac{1}{s-4} \\\\&\Rightarrow \quad(s+5) y(s)=\frac{1}{s-4}+2 \\\\&\Rightarrow \quad y(s)=\frac{1}{s+5}\left[\frac{1}{s-4}+2\right]=\frac{2 s-7}{(s+5)(s-4)}\end{aligned}$

$\begin{aligned}&\text { Let } \frac{2 s-7}{(s+5)(s-4)}=\frac{a_{0}}{s-4}+\frac{a_{1}}{s+5} \\&\Rightarrow 2 s-7=a_{0}(s+s)+a_{1}(s-4)\end{aligned}$

$put $s=-s \Rightarrow a_{1}=\frac{17}{9}$$

$\begin{aligned}\text { put } s &=4 \Rightarrow a_{0}=\frac{1}{9} \\\Rightarrow \quad y(s) &=\frac{1}{9(s-4)}+\frac{17}{9(s+s)}\end{aligned}$

In conclusion, Taking inverse Laplace tranoform

$L^{-1}[y(s)]=\frac{1}{9} L^{-1}\left[\frac{1}{s-4}\right]+\frac{17}{9} L^{-1}\left[\frac{1}{s+5}\right]$ \\\\$

$y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}$