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In-s [12.5K]
1 year ago
14

Quadrilateral ABCD is similiar to quadrilateral EFGH. The lengths of the three longest sides in quadrilateral ABCD are 24 feet,

16 feet, and 12 feet long. If the two shortest sides of quadrilateral EFGH are 9 feet long and 18 feet long, how long is the 4th side on quadrilateral ABCD?

Mathematics
2 answers:
masha68 [24]1 year ago
5 0

Well the sides of these 2 quadrilaterals are proportional to each other , so you would have to find the scale factor of the 2 then see which sides have already proportional pairs , then multiply *or / to find the last one .


Answer: D. 11.7 ft

OlgaM077 [116]1 year ago
5 0

check the picture below.

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Answer:

$4.081

Step-by-step explanation:

We have been given that Melissa owns a credit card with a 8.9% APR. The balance after her last billing cycle was $550.      

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\text{The finance charge}=\$550\times 0.00742

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1 year ago
Given: PRST square
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Answer:

7a²/16

Step-by-step explanation:

Area of the triangle PTS

½ × a × a

a²/2

Length of PS:

sqrt(a² + a²)

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Length of MS:

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Step-by-step explanation:

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1 year ago
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The area of the rectangle is 64.8 square centimeters. What is the perimeter of the rectangle? One group of ten tenths and one gr
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Answer:

(a) Perimeter = 32.2\ cm

(b) 100

Step-by-step explanation:

Solving (a):

Given

Shape: Rectangle

Area = 64.8

Required

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Area is calculated as:

Area = L * W

Where

L = Length and W = Width

Substitute 64.8 for Area

64.8 = L * W

Make L the subject:

L = \frac{64.8}{W}

Perimeter is calculated as:

P = 2 * (L + W)

Substitute 64.8/W for L

P = 2 * (\frac{64.8}{W} + W)

P = \frac{129.6}{W} + 2W

To solve further, we take the derivative of P and set it to 0, afterwards.

dP = -\frac{129.6}{W^2} + 2

Set to 0

0 = -\frac{129.6}{W^2} + 2

Collect Like Terms

\frac{129.6}{W^2} = 2

Cross Multiply:

2W^2= 129.6

Divide through by 2

W^2 = 64.8

Take square roots

W = \sqrt{64.8

W = 8.05

Recall that:

L = \frac{64.8}{W}

So:

L= \frac{64.8}{8.05}\\

L= 8.05

The perimeter is:

Perimeter = 2 * (8.05 + 8.05)

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Perimeter = 32.2\ cm

Solving (b):

Given

((1 Group of 10 tenths) and (1 group of 8 tenths))/(6 groups of 3 tenths)

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So, the expression becomes:

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This gives:

((1 Group of \frac{10}{10}) and (1 group of \frac{8}{10}))/(6 groups of \frac{3}{10})

Group means product, so the expression becomes:

\frac{(1 * \frac{10}{10} \ and\ 1 * \frac{8}{10})}{6 * \frac{3}{10}}

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\frac{(1 * \frac{10}{10} +  1 * \frac{8}{10})}{6 * \frac{3}{10}}

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\frac{(1 +  0.80)}{1.80}

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Mrs Atkins is going to choose two students from her class to take part in a competition.
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Given:

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