Ask question

Ask question

All categories

Misha Larkins [42]

2 years ago

15

Express the following fractions as the sum or difference of two fractions (x^2+y^2)/x^4

1 answer:

zlopas [31]2 years ago

4 0

Answer:

$\frac{1}{x^2}$ + $\frac{y^2}{x^4}$

Step-by-step explanation:

$\frac{x^2 + y^2}{x^4}$ = $\frac{x^2}{x^4}$ + $\frac{y^2}{x^4}$ = $\frac{1}{x^2}$ + $\frac{y^2}{x^4}$

You might be interested in

Solve for b. -9/3 + b = -1 b = -2 please help i suck at math ;---------;

Digiron [165]

Answer:%28a-b%29%2Fb=c

a-b=bc

a=bc%2Bb

a=b%28c%2B1%29

a%2F%28c%2B1%29=b

b=a%2F%28c%2B1%29

So the answer is b=a%2F%28c%2B1%29

New! FREE algebra solver that shows work:

3x-x+2=4

plz mark me the braniest

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

A waffle house offers 19 different types of syrups. If a customer can choose any number of syrups to have with their waffle, how

bogdanovich [222]

Answer:

Step-by-step explanation:

they all just have 19 or less

3 0

3 years ago

Equivalent fraction for ½.

labwork [276]

Answer:

2/4

Step-by-step explanation:

half/whole for any number really

5 0

2 years ago

Problem 4: Solve the initial value problem

pishuonlain [190]

Separate the variables:

$y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx$

Separate the left side into partial fractions. We want coefficients a and b such that

$\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}$

$\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}$

$\implies 1 = a(y-2)+b(y+1)$

$\implies 1 = (a+b)y - 2a+b$

$\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13$

So we have

$\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx$

Integrating both sides yields

$\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx$

$\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C$

$\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C$

$\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C$

$\dfrac{y-2}{y+1} = e^{3x + C}$

$\dfrac{y-2}{y+1} = Ce^{3x}$

With the initial condition y(0) = 1, we find

$\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12$

so that the particular solution is

$\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}$

It's not too hard to solve explicitly for y; notice that

$\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}$

Then

$1 - \dfrac3{y+1} = -\dfrac12 e^{3x}$

$\dfrac3{y+1} = 1 + \dfrac12 e^{3x}$

$\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}$

$y+1 = \dfrac6{2+e^{3x}}$

$y = \dfrac6{2+e^{3x}} - 1$

$\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}$

7 0

2 years ago

kwesi is putting on sunscreen. he uses 3 ml to cover 45 cm squared of his skin. he wants to know how many milliliters of sunscre

Anastasy [175]

Kwesi needs 16 ml of sunscreen.

6 0

3 years ago

Read 2 more answers