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Igoryamba
3 years ago
12

Nicole made costumes for the annual school play. She used a total length of 1468.8 in of blue fabric. The fabric came in 3 rolls

that each had the same length. How much fabric was in each roll? Write your answer in yards.
Mathematics
1 answer:
Charra [1.4K]3 years ago
5 0
1468.8/3 = 489.3 in
convert answer into yards (divide by 36)
there was 13.6 yards of fabric in each roll.
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Joshua used the Commutative Property. 
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The diameter of the bottom of a lamp shade is twice the diameter of the top of the lamp shade. The top of the lamp shade has a 9
GarryVolchara [31]

Answer:

see explanation

Step-by-step explanation:

The circumference (C) is calculated as

C = πd ← d is the diameter

diameter = 2 × 9 = 18 in ( twice the top diameter ), thus

C = 18π in ← exact value ≈ 56.55 in ( to 2 dec. places )

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3 years ago
A piece of wire 30 m long is cut into two pieces. One piece is bent into a square and the other is bent into a circle.
antiseptic1488 [7]

Answer:

a) 0 m

b) 16.8 m

Step-by-step explanation:

A piece of wire, 30 m long, is cut in two sections: a and b. Then, the relation between a and b is:

a+b=30\\\\b=30-a

The section "a" is used to make a square and the section "b" is used to make a circle.

The section "a" will be the perimeter of the square, so the square side will be:

l=a/4

Then, the area of the square is:

A_s=l^2=(a/4)^2=a^2/16

The section "b" will be the perimeter of the circle. Then, the radius of the circle will be:

2\pi r=b=30-a\\\\r=\dfrac{30-a}{2\pi}

The area of the circle will be:

A_c=\pi r^2=\pi\left(\dfrac{30-a}{2\pi}\right)^2=\pi\left(\dfrac{900-60a+a^2}{4\pi^2}\right)=\dfrac{900-60a+a^2}{4\pi}

The total area enclosed in this two figures is:

A=A_s+A_c=\dfrac{a^2}{16}+\dfrac{900-60a+a^2}{4\pi}=\left(\dfrac{1}{16}+\dfrac{1}{4\pi}\right)a^2-\dfrac{60a}{4\pi}+\dfrac{900}{4\pi}

To calculate the extreme values of the total area, we derive and equal to 0:

\left(\dfrac{1}{16}+\dfrac{1}{4\pi}\right)a^2-\dfrac{60a}{4\pi}+\dfrac{900}{4\pi}\\\\\\\dfrac{dA}{da}=\left(\dfrac{1}{16}+\dfrac{1}{4\pi}\right)(2a)-\dfrac{60}{4\pi}+0=0\\\\\\\left(\dfrac{1}{8}+\dfrac{1}{2\pi}\right)a=\dfrac{15}{\pi}\\\\\\\dfrac{\pi+4}{8\pi}\cdot a=\dfrac{15}{\pi}\\\\\\\dfrac{\pi+4}{8}\cdot a=15\\\\\\a=15\cdot \dfrac{8}{\pi+4}\approx 16.8

We obtain one value for the extreme value, that is a=16.8.

We can derive again and calculate the value of the second derivative at a=16.8 in order to know if the extreme value is a minimum (the second derivative has a positive value) or is a maximum (the second derivative has a negative value):

\dfrac{d^2A}{da^2}=\left(\dfrac{1}{16}+\dfrac{1}{4\pi}\right)(2)-0=\dfrac{1}{8}+\dfrac{1}{2\pi}>0

As the second derivative is positive at a=16.8, this value is a minimum.

In order to find the maximum area, we analyze the function. It is a parabola, which decreases until a=16.8, and then increases.

Then, the maximum value has to be at a=0 or a=30, that are the extremes of the range of valid solutions.

When a=0 (and therefore, b=30), all the wire is used for the circle, so the total area is a circle, which surface is:

A=\pi r^2=\pi\left( \dfrac{30}{2\pi}\right)^2=\dfrac{900}{4\pi}\approx71.62

When a=30, all the wire is used for the square, so the total area is:

A=a^2/16=30^2/16=900/16=56.25

The maximum value happens for a=0.

3 0
3 years ago
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