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3 years ago

How to solve 10 (g+5)=2(g+9)

Mathematics

1 answer:

Ghella [55]3 years ago

4 0

first multiply the outside numbers by both the numbers in the ()

10xg=10g 10x5=50 2xg=2g 2x9=18

then combine like terms

10g+50=2g+18

8g=32

then divide

g=4

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Solve for x please <br><br> 1/6x=10

ElenaW [278]

Answer:

x=60

Step-by-step explanation:

1/6x=10

multiply both sides by 6 to get x

x=60

6 0

3 years ago

Read 2 more answers

Consider the matrix shown below:

velikii [3]

Answer:

$A)\ \ \ \ \left[\begin{array}{ccc}8&-5\\-3&2\\\end{array}\right]$

Step-by-step explanation:

Given the matrix: $\left[\begin{array}{ccc}2&5\\3&8\\\end{array}\right]$ , it's inverse is calculated using the formula:

$\left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right]^{-1}=\frac{1}{det\left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right] }\left[\begin{array}{ccc}d&-b\\-c&a\\\end{array}\right]$

#Therefore, we calculate as;

$\frac{1}{det\left[\begin{array}{ccc}2&5\\3&8\\\end{array}\right] }\left[\begin{array}{ccc}8&-5\\-3&2\\\end{array}\right] \\\\\\\\\#det\left[\begin{array}{ccc}2&5\\3&8\\\end{array}\right] =1\\\\\\\\=\frac{1}{1}\left[\begin{array}{ccc}8&-5\\-3&2\\\end{array}\right] \\\\\\\\=\left[\begin{array}{ccc}8&-5\\-3&2\\\end{array}\right]$

Hence, the inverse of the matrix is $\left[\begin{array}{ccc}8&-5\\-3&2\\\end{array}\right]$

8 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

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

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