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son4ous [18]
3 years ago
6

A quilted block users eight parallelogram shaped pieces of clothe each with a base of 3/3/4 inches. How much fabric is needed to

make the parallelogram pieces for 24 blocks? Write in square feet. ( Hint:144in second power =1ft second power
Mathematics
1 answer:
borishaifa [10]3 years ago
7 0

Answer:

The answer to the question is

37.5 ft² of fabric is needed to make the parallelogram pieces for 24 blocks.

Step-by-step explanation:

To answer the question, we note that  a parallelogram is a four sided polygon having two opposite parallel sides.

Therefore in this question, we select our parallelogram to be a rectangle

Number of pieces of parallelogram in a quilted block = 8

Number of parallelogram in each quadrant of the quilted block = 2

Therefore the height of the rectangle = 2 × the base = 2×3\frac{3}{4} = 7.5 inches

Therefore the area of each parallelogram = base × height

= \frac{15}{4}*\frac{15}{2} = \frac{225}{8} in²

since each quilted block uses 8 parallelogram, then the amount of fabric for each quilted block is

\frac{225}{8} *8 = 225 in^2

Hence 24 blocks wil require 24×225 in² = 5400 in²

However, since 144 in² = 1 ft² we have

5400 in² = 5400 in²× 1/144  ft²/in² = 37.5 ft².

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AD←→ is tangent to circle M at point D. The measure of ∠DMQ is 58º.
Marat540 [252]

Answer:

32°

Step-by-step explanation:

Given:

∠DMQ = 58º

In this circle, the radius is DM. Since AD is tangent to the circle M, at point D, and the angle between a tangent and a radius is 90°

Therefore, ∠MDQ = 90°

The total angle in a triangle is 180°. Since we have the values of ∠MDQ and ∠DMQ, ∠DQM will be calculated as:

180 = ∠DMQ + ∠MDQ + ∠DQM

Solving for ∠DQM, we have:

∠DQM = 180 - ∠DMQ - ∠MDQ

∠DQM = 180 - 90 - 58

∠DQM = 32°

The measure of ∠DQM is 32°

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3 years ago
Look at the table on texts assigned to students. A 4-column table with 3 rows titled texts assigned to students. The first colum
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Answer is C) 0.1 on edge.

3 0
3 years ago
Read 2 more answers
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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