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kodGreya [7K]
3 years ago
12

Given: 315/1575. Write the prime fractorization of the numerator and the denominator, then write the fraction in lowest terms.

Mathematics
1 answer:
Ne4ueva [31]3 years ago
3 0
The prime factorization of the numerator, 315, is 5 x 3 x 3 x 7. That of the denominator, 1575 is 5 x 5 x 3 x 3 x 7. To convert the fraction to its lowest form, cancel all the factors that appear to both the numerator and denominator. We will be left with 1/5. 
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If r = 3.2, what is the area of the shaded region? Round your answer to the nearest hundredth. Use your calculator's value of π.
MrRa [10]

Answer: The shaded area is 99.55 units squared.

Step-by-step explanation:

r = 3.2

Now, we can see that the sides of the square are equal to two times the diameter of the circles (or four times the radius of the circles), so the length of the sides of the square is:

L = 2*(2*3.2) = 12.8

The area of the square is A1 = L^2 = 12.8*12.8 = 163.84 units squared.

the shaded semicircle has a diameter of 4 times r (so the radius is 2 times r), and the area is equal to half the area of a circle:

A2 = (1/2)*pi*(2r)^2 = (1/2)*3.14*(6.4)^2 = 64.31 units squared.

And now we must subtract the area of the four smaller circles inside the square, the area of each one is:

A3 = pi*r^2 = 3.14*(3.2)^2 = 32.15 units squared.

Then the shaded area is:

A = A1 + A2 - 4*A3 = 163.84 + 64.31 - 4* 32.15 = 99.55 units squared.

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I have no idea what I am doing. Here is one of my problems,
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A team of ten cyclists wants to choose leaders. in how many ways can the team choose one captain and then two assistants?
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3 years ago
Lizzie has 30 coins that total $4.80. All of her coins are dimes, D, and quarters, Q. Algebraically determine how many of each t
Gnom [1K]

Lizzie has 18 dimes and 12 quarters

<em><u>Solution:</u></em>

Let "d" be the number of dimes

Let "q" be the number of quarters

We know that,

value of 1 dime = $ 0.10

value of 1 quarter = $ 0.25

Given that LIzzie has 30 coins

number of dimes + number of quarters = 30

d + q = 30 ---- eqn 1

Also given that the coins total $ 4.80

number of dimes x  value of 1 dime + number of quarters x  value of 1 quarter = 4.80

d \times 0.10 + q \times 0.25 = 4.80

0.1d + 0.25q = 4.8 ------ eqn 2

Let us solve eqn 1 and eqn 2

From eqn 1,

d = 30 - q ---- eqn 3

Substitute eqn 3 in eqn 2

0.1(30 - q) + 0.25q = 4.8

3 - 0.1q + 0.25q = 4.8

0.15q = 1.8

<h3>q = 12</h3>

From eqn 3,

d = 30 - 12

<h3>d = 18</h3>

Thus she has 18 dimes and 12 quarters

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