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3 years ago

Find the area of the triangle below.

Mathematics

2 answers:

nordsb [41]3 years ago

7 0

Answer:

3.6 cm

Step-by-step explanation:

Base times height divide by 2.

Nikitich [7]3 years ago

6 0

I think the answer is 3.6. I got the answer because I used the triangle formula to find the area. Which is base times height.

-Dhruva;)

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Find the volume of a rectangular prism with a length of 9 cm, a height of 13 cm, and a width of 7 cm.

Sati [7]

Answer:

819 cm³

Step-by-step explanation:

V=L·H·W

V=9·13·7

V=819 cm³

4 0

3 years ago

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iVinArrow [24]

Answer:

3.92%

Step-by-step explanation:

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3 years ago

Enter numbers to evaluate -5 -8

Slav-nsk [51]

Answer:

-13

Step-by-step explanation: hope this help

8 0

3 years ago

Sean is stacking cans of beans of the market. His shelf is 10 inches tall, 10 inches deep, and 50 inches wide. The cans are 3 in

N76 [4]

Basically you get the volume first:

10 times 10 times 50

Which is 5,000

And we're given the cans volume already:

9.42

So what we do now is divided the area by the cans for our answer:

5,000/9.42

Which is 530.785563

So Sean can stack 530 cans of beans in his shelf.

(when rounding is is 531 cans)

6 0

3 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