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2 years ago

Please help number 3 through 4

Mathematics

1 answer:

Maurinko [17]2 years ago

3 0

as per graph 3 other points (s, h) can be

( 20, 50)

(40,40)

(60, 30)

no solution wouldn't change. graph will still habe the same point, just representional view of graph would change.

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Your friend deposits $1900 in an account that earns 6% compounded

iVinArrow [24]

Answer:

$2,128

Step-by-step explanation:

6% + 6% = 12%

12% x 1900 = 228

1900 + 228 = 2128

her accoun balance will be $2128 after 2 years

mark as brainlest :)

6 0

2 years ago

Read 2 more answers

A rectangle is 25 in. long and 13 in. wide.<br><br><br><br> What is the area of the rectangle?

sineoko [7]

area = I x w

so 25 x 13 = 325 square inches

4 0

3 years ago

Read 2 more answers

A vertical right circular cylindrical tank has height h=8 feet high and diameter d=6 feet. It is full of kerosene weighing 50 po

Mariana [72]

Answer:

The work done to pump all of the kerosene from the tank to an outlet is $W=45238.9\: J$

Step-by-step explanation:

The work is defined by:

$W=\int dFdx$ (1)

The force here will be the product between the volume and the kerosene weighing, so we have :

$dF=\pi R^{2}dy*50$

This force will be in-lbs.

Where R is the radius (3 feet)

Then using (1), we have:

$W=\int \pi R^{2}dy*50(8-y)$

Here 8-y is a distance at some point of the tank. Now, to get the work done from the base to the top of the tank we will need to take integral from 0 to 8 feet.

$W=\int_{0}^{8} \pi 3^{2}dy*50(8-y)$

$W=450\pi \int_{0}^{8}dy(8-y)$

$W=450\pi(\int_{0}^{8} 8dy-\int_{0}^{8} ydy)$

$W=450\pi(8y|_{0}^{8} -\frac{y^{2}}{2}|_{0}^{8})$

$W=450\pi(8*8 -\frac{8^{2}}{2})$

$W=450\pi(64 -\frac{64}{2})$

Therefore, the work done to pump all of the kerosene from the tank to an outlet is $W=45238.9\: J$

I hope it helps you!

6 0

2 years ago

Write an equivalent expression.

djyliett [7]

8.5 the answer is and replace the areas

4 0

2 years ago

<img src="https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B80%7D%20" id="TexFormula1" title=" \sqrt[3]{80} " alt=" \sqrt[3]{80} " ali

Marianna [84]

Start by decomposing the number inside the root into primes

Then group the terms into cubes if possible

$\begin{gathered} 80=2\cdot2\cdot2\cdot2\cdot5 \\ 80=2^3\cdot2\cdot5 \\ 80=10\cdot2^3 \end{gathered}$

rewrite the root

$\sqrt[3]{80}=\sqrt[3]{10\cdot2^3}$

then cancel the terms that are cubes and bring them out of the root

$\sqrt[3]{80}=2\sqrt[3]{10}$

7 0

11 months ago

Please help number 3 through 4​

Please help number 3 through 4