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Kaylis [27]
2 years ago
15

The perimeter of a rectangle is 42 inches. If the length of the rectangle is 14 inches, which equation could be used to find the

width, x? A. 2(x + 14) = 42 B. x + 2(14) = 42 C. 14(x + 2) = 42 D. x + 14 = 42
Mathematics
2 answers:
Cerrena [4.2K]2 years ago
8 0

Answer:

Option A is correct

equation 2(x+14)=42 could be used.

Step-by-step explanation:

Given: The length of a rectangle (l) is 14 inches and perimeter of rectangle is 42 inches.

Perimeter(P)of a rectangle in inches is given by:

P= (2l+w) ; where l  is the length of the rectangle and w is the width of the rectangle.

Substitute the value of  P=42 inches , w =x inches and l = 14 inches in the above formula we get;

42 = 2 \cdot (14+x)

Divide by 2 on both sides  we get;

21 = 14+x

On simplify, we get;

x = 21-14 = 7 inches

therefore, the width of the rectangle is, x = 7 inches

Check :

Substitute the value of x =7 in option A.

2(x+14) =42

2(7+14)=42

2\cdot 21 =42

42 = 42      

Hence, the only equation that could be used to find the width, x is,  2(x + 14) = 42




Sever21 [200]2 years ago
7 0
A. 2(x+14) = 42

(2x = 14
  x = 7

2(7) + 2(14) = 14 + 28 = 42)

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Third, now we have 2 values of x. Find the value of y for each of x
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Step-by-step explanation:

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7 0
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Given that WA = 5x – 8 and WC = 3x + 2, find WB. A. WB = 5 B. WB = 8 C. WB = 10 D. WB = 17
rodikova [14]
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






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