Ask question

Ask question

All categories

stepladder [879]

3 years ago

8

6. On Saturday, you bowl at Mar Vista Bowl, where renting shoes

1 answer:

olga_2 [115]3 years ago

6 0

Answer:

3.50 +3.25 = 6.75

Sory If wrong :)

I hope This Help'S

You might be interested in

What The results of this equations.?

Otrada [13]

A.9×6x - ( 9×2)=
64x-18 = 0
64x=18
x=18/64 =1/3 =0.3
B. 64x+18=0
64x= -18
x =-18/64
x= -1/3
x = -0.3

6 0

3 years ago

What is the prime factorization of 616?

Leni [432]

A would be the correct answer:)

Hope this helps!

3 0

3 years ago

Read 2 more answers

Which of the following are steps to use when formulating all that apply equation? Check an A. Write each fact as a variable expr

kramer

Answer:

a

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

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

A line segment has endpoints at (3, 2) and (2, –3). Which reflection will produce an image with endpoints at (3, –2) and (2, 3)?

Alex73 [517]

It is a reflection over the x-axis

4 0

3 years ago

Read 2 more answers