Mathematics, 02.02.2021 01:40 hollandhogenson

A new ice skating rink will be approximately rectangular in shape and will have an area of 19,720 square feet. Using an equation for the perimeter P, of the skating rink in terms of its width W, what are the dimensions of the skating rink if the perimeter is 634 feet? The dimensions of the rink are ? feet by ? feet.

Answers: 3

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Answer from: alex24gonzales

The dimension of the rectangular rink is 85ft by 232ft

Perimeter and Areas of triangle

Given the following parameters

Area = 19,720 square feet.Perimeter = 634 feet

The formula for finding the area is:

A = lw

lw = 19,720

l = 19,720/w

If the perimeter is 634 feet

2(l+w) = 634

2(19,720/w + w) = 634

39,440/w + 2w = 634

39,440 + 2w^2 = 634w

2w^2 - 634w + 39,440 = 0

On factorizing, w=85 or w=232

Since l = 19,720/w

l = 19,720/232

l = 85

Hence the dimension of the rectangular rink is 85ft by 232ft

Learn more on area of triangle here: link

Answer from: glowbaby123

The dimension is: 232ft by 85ft

Step-by-step explanation:

Given

Shape: Rectangle

A = 19720 --- Area

P = 634 --- Perimeter

Required

Determine the dimension

Area is calculated as:

A = L * W

This gives

19720 = L * W

Perimeter is calculated as:

P = 2(L + W)

This gives

634 = 2(L + W)

Divide both sides by 2

317 = L + W

Make L the subject

L = 317 -W

Substitute 317 - W for L in 19720 = L * W

19720 = (317 - W)*W

Open brackets

19720 = 317W - W^2

Rewrite as:

W^2 - 317W + 19720 = 0

Expand

W^2 -232W - 85W + 19720 = 0

Factorize:

W(W -232) - 85(W - 232)= 0

(W - 85) (W - 232)= 0

Split

$W = 85\ or\ W = 232$

Recall that:

L = 317 -W

$L= 317 - 85\ or\ L = 317 - 232$

$L= 232\ or\ L = 85$

Hence, the dimension is: 232ft by 85ft

Answer from: Quest

They sought new allies to germany tried to get mexico to join and put pressure on the us. the us joined the allies because of the lucitania

Answer from: Quest

hello!

to begin with, we're given that on earth, a rock weighs 20 pounds, and while on jupiter, that same rock weighs 42 pounds. now, we're trying to find the weight of a rock that weighs 10.5 pounds on earth, on jupiter.

to solve this, first, set up a proportion, cross-multiply, and solve it algebraically.

setting up the proportion is the hardest and the most confusing part of the problem, but it's simple.

if on earth, the rock weighs 20 lbs, and on jupiter it weighs 42 lbs, our first part of our proportion is 20/42 lbs. the second part has to be equal to each other. this means that 10.5 has to be on the numerator of the fraction, and x, the value we're trying to find, has to be on the denominator. remember that the rock weighs 10.5 lbs on earth, and we're trying to find the weight of it on jupiter (which is x).

20/42 = 10.5/x

next, you would cross-multiply.

20x = 42(10.5)

20x = 441

now, solve it algebraically by using division.

x = 22.05 lbs

this can be rounded to 22.1 lbs.

therefore, the weight of the rock on jupiter is 22.1 lbs, the fourth choice.

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