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Ilia_Sergeevich [38]
3 years ago
14

Given that WA = 5x – 8 and WC = 3x + 2, find WB. A. WB = 5 B. WB = 8 C. WB = 10 D. WB = 17

Mathematics
2 answers:
rodikova [14]3 years ago
7 0
The rest of the question is the attached figure.
============================================
Δ AYW a right triangle at Y ⇒⇒⇒ ∴ WA² = AY² + YW²
And AY = YB ⇒⇒⇒ ∴ WA² = YB² + YW²   → (1)
Δ BYW a right triangle at Y ⇒⇒⇒ ∴ WB² = BY² + YW²  → (2)
From (1) , (2)  ⇒⇒⇒ ∴ WA = WB  →→ (3)
Δ CXW a right triangle at Y ⇒⇒⇒ ∴ WC² = CX² + XW²
And CX = XB ⇒⇒⇒ ∴ WC² = XB² + XW²   → (4)
Δ BXW a right triangle at Y ⇒⇒⇒ ∴ WB² = XB² + XW²  → (5)
From (4) , (5)  ⇒⇒⇒ ∴ WC = WB  →→ (6)
From (3) , (6)
WA = WB = WC
given ⇒⇒⇒ WA = 5x – 8 and WC = 3x + 2
∴ <span> 5x – 8 = 3x + 2</span>
Solve for x ⇒⇒⇒ ∴ x = 5
∴ WB = WA = WC = 3*5 + 2 = 17

The correct answer is option D. WB = 17






Dimas [21]3 years ago
4 0

Answer: 17

Step-by-step explanation:

5x - 8)^2 = h^2

(3x + 2)^2 = l^2

5x - 8 = 3x + 2

5x - 3x = 8 + 2

2x = 10

x = 5

WB = 5x - 8

WB = 5(5) - 8

WB = 17

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C is correct.


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5 + 1.5x = total cost

Step-by-step explanation:

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Round to the nearest tenth if necessary.
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=(−2−9)2+(7−12)2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√

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How do I set up this problem?
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Hey there!

We know that perimeter is all the sides added up.

70= 2w+2L <== this is the basic equation

We know that the rectangles length is three times as long as it is wide.

So, you know that is is three times the length of the width. So, put 3 times w in as length and times that by two since there are two length sides, and keep width the same. 2w because there are two width sides. Then, solve.

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We can check this by plugging it in to our first equation and seeing if it works. First, we need to multiply 8.75 by 3 to get the value for length.

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So, the length is 26.25 cm and the width is 8.75 cm.

I hope this helps!

~kaikers

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