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3 years ago

16/5 x 36/12 x 25/18

Mathematics

1 answer:

solmaris [256]3 years ago

6 0

Answer

$\frac{16}{5}$ * $\frac{36}{12}$ * $\frac{25}{18}$ = $\frac{4}{1} *\frac{2}{3} *\frac{5}{1}=\frac{40}{3} =13\frac{1}{3}$

Step-by-step explanation:

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3 years ago

There are 10 mystery books and 7 nonfiction book and 5 science books and 2 westerns what is the ratio of mystery books to the to

Artyom0805 [142]

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3 years ago

5. -[-2(x-4) - |1-3|] = -2x18

irina [24]

Solving the expression $-[-2(x-4) - |1-3|] = -2x18$ we get value of x: $x=\frac{3}{19}$

Step-by-step explanation:

We need to solve the equation $-[-2(x-4) - |1-3|] = -2x18$ and find value of x

Solving:

$-[-2(x-4) - |1-3|] = -2x18$

Multiply -2 with terms inside the bracket

$-[-2x+8 - |1-3|] = -2x18$

Solving |1-3| we get |-2|=2

$-[-2x+8 - 2] = -2x18$

$-[-2x+6] = -2x18$

Multiply (-) sign with terms inside the bracket

$+2x-6 = -2x18$

Multiply 2x with 18

$2x-6 = -36x$

Add 36x on both sdides

$2x-6+36x =0$

$38x-6 =0$

Add 6 on both sides:

$38x =6$

Divide both sides by 38

$x=\frac{6}{38}$

Simplifying:

$x=\frac{3}{19}$

So, Solving the expression $-[-2(x-4) - |1-3|] = -2x18$ we get value of x: $x=\frac{3}{19}$

Keywords: Solving Equations

Learn more about Solving Equations at:

#learnwithBrainly

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4 years ago

Two students use different methods to solve this multiplication problem: 1/2 • - 2 2/5

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Answer:

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Step-by-step explanation:

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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

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

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