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nalin [4]
3 years ago
13

Choose the equation of the line through (4, –5) and perpendicular to 6x + 12y = 13.

Mathematics
1 answer:
saw5 [17]3 years ago
5 0

Answer:

Kotz I don’t know the answer because I’m to lazy to do it - but I need to talk to you again... it’s me Ellis - you don’t have to rejoin the server but message me 503 333 0177.

Step-by-step explanation:

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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
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