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Alex_Xolod [135]
3 years ago
11

Margot is sewing a ribbon on a seam along the perimeter of a square pillow. The side length of the pillow is 2x2+1 inches. She p

lans to make a similar pillow, including the ribbon, whose side length is 4x−7 inches. What expression can be used for the length of ribbon that she needs for both pillows, and what is the length if x = 3.5?
2x2+4x−6; 22.0 inches
2x2+4x−6; 32.5 inches
4(2x2+4x−6;) 88.0 inches
4(2x2+4x−6;) 130.0 inches
Mathematics
2 answers:
polet [3.4K]3 years ago
8 0
Infinity Many Solutions

Write above. Thank me tomorrow when it’s right



- Harvard university professor
Ierofanga [76]3 years ago
7 0

Answer:

4(2x2+4x−6;) 130.0 inches

Step-by-step explanation:

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What must be true of PQ?
yulyashka [42]
Interesting question
Usually when you look at something like that construction, you think that AB has been bisected by PQ and that the two segments are perpendicular. They are perpendicular but nowhere is that stated. So the answer is C because all the other answers are wrong. 

PQ is congruent AB is not correct. As long as the arcs are equal and meet above and below AB there is no proof of congruency. In your mind widen the compass legs so that they are wider than AB and redraw the arcs. You get a larger PQ, but it has all the original properties of PQ except size.

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The two lines are not parallel. They are perpendicular. That can be proven. They meet at right angles to each other (also provable). 


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I believe the answer is 275 miles. Hope this helps!
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Sam could make 18 cups
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Find the ordered pair for the equation: y = 2x + 10 where x = 3
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Use Hooke's Law to determine the work done by the variable force in the spring problem. A force of 450 newtons stretches a sprin
vlada-n [284]

Answer:

The work done is 202.50Nm

Step-by-step explanation:

Given

F =450N

x_1 = 30cm

x_2 = 60cm

Required

The work done

First, we calculate the spring constant (k)

F = kx_1

450N = k *30cm

k = \frac{450N}{30cm}

k =15N/cm

So:

F = kx_1

F(x) = 15x

The work done using Hooke's law is:

W =\int\limits^a_b {F(x)} \, dx

This gives:

W =\int\limits^{60}_{30} {15x} \, dx

Rewrite as:

W =15\int\limits^{60}_{30} {x} \, dx

Integrate

W =15 \frac{x^2}{2}|\limits^{60}_{30}

This gives:

W =15 *\frac{60^2 - 30^2}{2}

W =15 *\frac{2700}{2}

W =15 *1350

W =20250N-cm

Convert to Nm

W =\frac{20250Nm}{100}

W =202.50Nm

7 0
3 years ago
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