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erica [24]
3 years ago
7

List all factors of the number 52. SHOW ALL WORK!!!

Mathematics
1 answer:
Nutka1998 [239]3 years ago
7 0

Answer:

Factors of number 52

Factors of 52: 1, 2, 4, 13, 26 and 52.

Negative Factors of 52: -1, -2, -4, -13, -26 and -52.

Prime Factors of 52: 2, 13.

Prime Factorization of 52: 2 × 2 × 13 = 22 × 13.

Sum of Factors of 52: 98.

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A salesperson works 40 hours per week at a job where he has two options for being paid. Option A is an hourly wage of ​$28. Opti
irina [24]

Answer:

$9500

Step-by-step explanation:

Given, Option A: Hourly wage is $19 and the salesperson works 40 hours per week. So, he will earn in a week= 19 X 40 = $760

Now, according to option b, he will get 8% commission on weekly sales.

Let. x = the amount of weekly sales.

To earn the same amount of option A, he will have to equal the 8% of x to $760

So, 8x/100= 760

Or, 8x= 76000

Or, x= 76000/8= 9,500

the salesman needs to make a weekly sales of $9,500 to earn the same amount with two options.

8 0
3 years ago
Read 2 more answers
Show all work to factor x^4 − 17x^2 + 16 completely.
Dmitrij [34]

Answer:

x^{4}-17x^{2} +16 = (x -1)(x + 1)(x - 4)(x + 4)

Step-by-step explanation:

At first, let us find the first two factors of x^{4}-17x^{2} +16

∵ The sign of the last term is positive

∴ The middle signs of the two factors are the same

∵ The sign of the middle term is negative

∴ The middle signs of the two factors are negative

∵ x^{4} = x² × x² ⇒ first terms of the two factors

∵ 16 = -1 × -16 ⇒ second terms of the two factors

∵ x²(-1) + x²(-16) = -x² + -16x² = -17x² ⇒ the value of the middle term

∴ (x² - 1) and (x² - 16) are the factors of x^{4}-17x^{2} +16

Now let us factorize each factor

→ The factors of the binomial a² - b² (difference of two squares) are

   (a - b) and (a + b)

∵ x² - 1 is the difference of two squares

∴ Its factors are (x - 1) and (x + 1)

∵ x² - 16 is the difference of two squares

∴ Its factors are (x - 4) and (x + 4)

∵ (x -1), (x + 1), (x - 4), and (x + 4) are the factors of (x² - 1) and (x² - 16)

∵ (x² - 1) and (x² - 16) are the factors of x^{4}-17x^{2} +16

∴ (x -1), (x + 1), (x - 4), and (x + 4) are the factors of x^{4}-17x^{2} +16

∴  x^{4}-17x^{2} +16 = (x -1)(x + 1)(x - 4)(x + 4)

8 0
2 years ago
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Answer:

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