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3 years ago

7

I'll pick brainiest. Please help me!! Answer as best as you can. THANK YOU!!

1 answer:

rodikova [14]3 years ago

5 0

The volume of the cube is 54.87 but rounded to the nearest tenth is 54.9

The exact volume of the cylinder is 1770.28 but rounded is 1770.29 and the approximate volume of the cylinder is 1769.36

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What is 90 added to 85

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90+85=175
Thats the answer

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6 months ago

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Help Please!!!!!<br><br> Evaluate the expression<br> when n = 4<br> 2n-7

Montano1993 [528]

2•4= 8 8-7= 1
The answer is 1

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3 years ago

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Can somebody please help???

beks73 [17]

Darnel needs to spend three days making jam if he wants to jar 15 liters.

Step-by-step explanation:

If you are to assume the relationship is proportional then if he makes 10 liters in 2 days you divide 10 by 2 and you get 5. This means he makes 5 liters a day. Then you add 5 to 10 and you get 15 which means you are adding another day(5) So he needs to work an extra day which would give you three dyas total. Hope this helped!

4 0

2 years ago

A rectangular swimming pool is twice as long as it is wide. A small concrete walkway surrounds the pool. The walkway is a 2 feet

Ksivusya [100]

Answer:

The width and the length of the pool are 12 ft and 24 ft respectively.

Step-by-step explanation:

The length (L) of the rectangular swimming pool is twice its wide (W):

$L_{1} = 2W_{1}$

Also, the area of the walkway of 2 feet wide is 448:

$W_{2} = 2 ft$

$A_{T} = W_{2}*L_{2} = 448 ft^{2}$

Where 1 is for the swimming pool (lower rectangle) and 2 is for the walkway more the pool (bigger rectangle).

The total area is related to the pool area and the walkway area as follows:

$A_{T} = A_{1} + A_{w}$ (1)

The area of the pool is given by:

$A_{1} = L_{1}*W_{1}$

$A_{1} = (2W_{1})*W_{1} = 2W_{1}^{2}$ (2)

And the area of the walkway is:

$A_{w} = 2(L_{2}*2 + W_{1}*2) = 4L_{2} + 4W_{1}$ (3)

Where the length of the bigger rectangle is related to the lower rectangle as follows:

$L_{2} = 4 + L_{1} = 4 + 2W_{1}$ (4)

By entering equations (4), (3), and (2) into equation (1) we have:

$A_{T} = A_{1} + A_{w}$

$A_{T} = 2W_{1}^{2} + 4L_{2} + 4W_{1}$

$448 = 2W_{1}^{2} + 4(4 + 2W_{1}) + 4W_{1}$

$224 = W_{1}^{2} + 8 + 4W_{1} + 2W_{1}$

$224 = W_{1}^{2} + 8 + 6W_{1}$

By solving the above quadratic equation we have:

W₁ = 12 ft

Hence, the width of the pool is 12 feet, and the length is:

$L_{1} = 2W_{1} = 2*12 ft = 24 ft$

Therefore, the width and the length of the pool are 12 ft and 24 ft respectively.

I hope it helps you!

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2 years ago

Graphing linear equations y+5=-2(x-2)

levacccp [35]

First put the equation in y=mx+b form then plot the b and count off the slope.

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3 years ago