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3 years ago

Can someone explain how to solve this

Mathematics

1 answer:

DochEvi [55]3 years ago

3 0

The answer is D. Here’s the explanation:
Take out the constants: 20/4 x a^(-4)b(-3)c^(2) / a^(2)b^(-6)c^(5)
Simplify the 20/4 to 5.
Use the Quotient Rule which is x^a / x^b = x^(a-b)
5a^(-4-2)b(-3+6)c(2-5)
Simplify (-4-2) to -6.
Simplify (-3+6) to 3.
Simplify (2-5) to -3.
You’re left with 5a^(-6)b^(3)c^(-3)
Use negative power rule which is x^(-a) = 1/x^(a)
5 x 1/a^(6)b^(3) x 1/c^(3)
Final Answer: 5b^(3)/a^(6)c^(3)

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Find the value of x . 12.5,-10,-7.5,x ; The mean is 11.5

Tems11 [23]

Answer:

Step-by-step explanation:

The mean is obtained by adding up the values in the data set and then divide by the number of values that you added:

$\bar X=\sum{X_i}\\$

Given our mean as 11.5 and n=4:

$\bar X=\frac{1}{n}\sum{X_i}\\\\\\\\\therefore\sum{X_i}\\=11.5\times 4=46$

#Knowing our sum as 65, we can obtain x by subtracting sum of known values from 65:

$x=46-(12.5+-10+-7.5)\\=51$

Hence, the value of x is 51

8 0

4 years ago

Solve the initial value problem 2ty" + 10ty' + 8y = 0, for t > 0, y(1) = 1, y'(1) = 0.

Eva8 [605]

I think you meant to write

$2t^2y''+10ty'+8y=0$

which is an ODE of Cauchy-Euler type. Let $y=t^m$ . Then

$y'=mt^{m-1}$

$y''=m(m-1)t^{m-2}$

Substituting $y$ and its derivatives into the ODE gives

$2m(m-1)t^m+10mt^m+8t^m=0$

Divide through by $t^m$ , which we can do because $t\neq0$ :

$2m(m-1)+10m+8=2m^2+8m+8=2(m+2)^2=0\implies m=-2$

Since this root has multiplicity 2, we get the characteristic solution

$y_c=C_1t^{-2}+C_2t^{-2}\ln t$

If you're not sure where the logarithm comes from, scroll to the bottom for a bit more in-depth explanation.

With the given initial values, we find

$y(1)=1\implies1=C_1$

$y'(1)=0\implies0=-2C_1+C_2\implies C_2=2$

so that the particular solution is

$\boxed{y(t)=t^{-2}+2t^{-2}\ln t}$

# # #

Under the hood, we're actually substituting $t=e^u$ , so that $u=\ln t$ . When we do this, we need to account for the derivative of $y$ wrt the new variable $u$ . By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{\mathrm dy}{\mathrm du}\dfrac{\mathrm du}{\mathrm dt}=\dfrac1t\dfrac{\mathrm dy}{\mathrm du}$

Since $\frac{\mathrm dy}{\mathrm dt}$ is a function of $t$ , we can treat $\frac{\mathrm dy}{\mathrm du}$ in the same way, so denote this by $f(t)$ . By the quotient rule,

$\dfrac{\mathrm d^2y}{\mathrm dt^2}=\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac ft\right]=\dfrac{t\frac{\mathrm df}{\mathrm dt}-f}{t^2}$

and by the chain rule,

$\dfrac{\mathrm df}{\mathrm dt}=\dfrac{\mathrm df}{\mathrm du}\dfrac{\mathrm du}{\mathrm dt}=\dfrac1t\dfrac{\mathrm df}{\mathrm du}$

where

$\dfrac{\mathrm df}{\mathrm du}=\dfrac{\mathrm d}{\mathrm du}\left[\dfrac{\mathrm dy}{\mathrm du}\right]=\dfrac{\mathrm d^2y}{\mathrm du^2}$

so that

$\dfrac{\mathrm d^2y}{\mathrm dt^2}=\dfrac{\frac{\mathrm d^2y}{\mathrm du^2}-\frac{\mathrm dy}{\mathrm du}}{t^2}=\dfrac1{t^2}\left(\dfrac{\mathrm d^2y}{\mathrm du^2}-\dfrac{\mathrm dy}{\mathrm du}\right)$

Plug all this into the original ODE to get a new one that is linear in $u$ with constant coefficients:

$2t^2\left(\dfrac{\frac{\mathrm d^2y}{\mathrm du^2}-\frac{\mathrm d y}{\mathrm du}}{t^2}\right)+10t\left(\dfrac{\frac{\mathrm dy}{\mathrm du}}t\right)+8y=0$