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3 years ago

5

Help yalllll did put links What is the volume of the object? Use 3.14 as

1 answer:

scoundrel [369]3 years ago

4 0

Answer:

V= 3,617.28 cm^3

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0=-20.5x+16.35x-0.1245

Pachacha [2.7K]

The value of x in the given equation $-20.5x+16.35x -0.1245 = 0\\$ is 0.03

<u>Step-by-step explanation:</u>

The given equation is that $-20.5x+16.35x -0.1245 = 0\\$

<u>The steps to be followed to solve the equation are :</u>

Add the like terms together to reduce the equation in a simplified form.

Here, there are two x terms and they must be reduced to a single term.

For this, add the both terms together so that the equation is simplified into one x term and a constant term.

⇒ $(-20.5x+16.35x) - 0.1245 = 0$

⇒ $4.15x-0.1245 = 0$

To eliminate the constant term on the left side of the equation, add 0.1245 on both sides.

⇒ $4.15x -0.1245+0.1245 = 0.1245$

⇒ $4.15x = 0.1245$

Now, the equation is further simplified by dividing 4.15 on both sides,

⇒ $4.15x / 4.15 = 0.1245/4.15$

⇒ $x = 0.03$

Therefore, the value of x is 0.03

7 0

3 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!

8 0

3 years ago

Please I need help on this ASAP

Virty [35]

Answer:

True, False, True, False, False

Step-by-step explanation:

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3 years ago

The volley ball team at west view high school is comparing T -shirts where they can purchase their practice shirts. The graph re

valkas [14]

Idk I need hekpfbfjsnbfbs snan

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3 years ago

A frame is 3 in wide and 9 in tall. if it is reduced to a height of 3 in, then how wide will it be?

Nimfa-mama [501]

It would be 1 inch wide.

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3 years ago

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