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VashaNatasha [74]

4 years ago

5

Dakota read 130 pages over the weekend. Dajuan read 165 pages over the weekend. Which prime number do the prime factorization of

130 and 165 have in common?

1 answer:

Amanda [17]4 years ago

6 0

In order to find the common prime factors of 130 and 165, let us first find prime factors of 130 and 165 individually and then we can compare.

Prime factors of 130 can be found as:

130 is a composite number because it has factors other than 1 and itself.

It is not a prime number. The 8 factors of 130 are 1, 2, 5, 10, 13, 26, 65, and 130. The factor pairs of 130 are 1 x 130, 2 x 65, 5 x 26, and 10 x 13.

The proper factors of 130 are 1, 2, 5, 10, 13, 26, and 65.

The prime factors of 130 are 2, 5, and 13.

Prime factors of 165 can be found as:

165 is a composite number because it has factors other than 1 and itself. It is not a prime number.

The 8 factors of 165 are 1, 3, 5, 11, 15, 33, 55, and 165. The factor pairs of 165 are 1 x 165, 3 x 55, 5 x 33, and 11 x 15.

The proper factors of 165 are 1, 3, 5, 11, 15, 33, and 55.

The prime factors of 165 are 3, 5, and 11.

Answer: On comparing the two, we can see that 5 is the common prime factor of both 130 and 165.

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Step-by-step explanation:

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

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r is the rate of interest in decimal

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n is the number of times interest is compounded per year

so

Part A) Annual

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=1$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{1})^{1*5}$

$A=\$47,400(1.07)^{5}$

$A=\$66,480.95$

Part B) Semiannual

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=2$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{2})^{2*5}$

$A=\$47,400(1.035)^{10}$

$A=\$66,862.38$

Part C) Monthly

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=12$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{12})^{12*5}$

$A=\$47,400(1.0058)^{60}$

$A=\$67,195.44$

Part D) Daily

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=365$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{365})^{365*5}$

$A=\$47,400(1.0002)^{1,825}$

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