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3 years ago

Need help don’t know how to do this

Mathematics

1 answer:

DerKrebs [107]3 years ago

4 0

I am not sure but I think the answer is 8

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12 – 2x= -2(y – x)<br> -2x = -2(y – x) – 12<br> x= (y - x) + 6

Vilka [71]

Answer:

x=o and y= -6

Step-by-step explanation:

12-2x = - 2(y-x)

or,12-2x = -2y + 2x

or, 12 = - 2y + 4x

or, 12/2= 2x - y

or, 2x - y = 6

• 2x -6= y.....eqn_1

-2x=-2y+2x -12

or,4x +2y= -12

or,4x + 2(2x-6)= -12

or, 4x+4x-12=-12

or, 8x= 0

• x= o

putting the value of x in eqn...3

o=(y-0)+6

or, o= y + 6

•y = -6

5 0

2 years ago

Identify the zeros of the quadratic function.

Amanda [17]

Answer:

y = -2 and y = -5 thus: d)

Step-by-step explanation:

5 0

2 years ago

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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

We paid $8 for 2 hamburgers, which is a rate of $_<br> per hamburger

ipn [44]

ANSWER: $4 per hamburger

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2 years ago

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The advertised price of a cell phone was $99.99 after a $50 mail-in rebate. What is the price of the cell phone before the rebat

Nataly_w [17]

Answer:

149.99

Step-by-step explanation:add 50

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2 years ago

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