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solmaris [256]
3 years ago
7

Segment DF bisects angle EDG. Enter the value of x. The diagram is not to scale.​

Mathematics
1 answer:
maksim [4K]3 years ago
7 0

Answer:

\displaystyle 6 = x

Step-by-step explanation:

Since DF is a SEGMENT BISECTOR, you set 15x and 8x + 42 equal to each other:

8x + 42 = 15x

- 8x - 8x

___________

\displaystyle \frac{42}{7} = \frac{7x}{7} \\ \\ 6 = x

I am joyous to assist you anytime.

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8 0
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Egal loggers got away with 4 acres of trees at a reserve in central Mexico. That number represents a 97 percent decline from two
geniusboy [140]

Answer: with 133.33 acres of trees

Step-by-step explanation:

4 acres of trees, is a 97% decrease of the previous amount.

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6 0
3 years ago
I need help with this math problem
QveST [7]

Step-by-step explanation:

Use the following ratio:

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7 0
3 years ago
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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}

Read more about Laplace tranoform

brainly.com/question/14487937

#SPJ4

6 0
2 years ago
ASAP 20 POINTS
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4 0
4 years ago
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