Can you please help me

Mathematics

1 answer:

skad [1K]3 years ago

4 0

Answer:

explanations:

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Simplify the expression:<br><br> 7x + 3y – 2 + 6x – 1 + y2

Hitman42 [59]

Answer: 13x+3y-3+y^2

Step-by-step explanation:

(7x+6x)+3y+(-2-1)+y^2

13x+3y-3+y^2

7 0

3 years ago

n is a positive integer with exactly two different divisors greater than 1, how many positive factors does n^2 have? A. 4 B. 5 C

Evgesh-ka [11]

Answer:

C. 5

Step-by-step explanation:

Remember that for any integer $n$ , the integers $1 \text{and} n$ are both divisors (or factors) of $n$ . First, we will prove that n is a square, and then we will compute the factors of the n².

In this case, the integer n has exactly two different divisors greater than 1. It's impossible that $n=1$ , since 1 doesn't have positive factors greater than 1. Then $n>1$ , therefore $n$ itself is one of the required divisors. Denote by $a$ the other divisor greater than 1, and note that to satisfy the condition on the divisors, $a$ .

Because a divides n, there exists some integer $k$ such that $n=ak$ . We must have that $k>1$ , if not, then $k\leq 1$ , which implies that $ak=n\leq a$ , which contradicts the part above.

Now, $k>1$ and, by definition of divisibility, k divides n. Then k must be equal either to n or a, since we can't have three different divisors of this kind. If $k=n$ then $n=an$ and by cancellation, $1=a$ which is a contradiction. Therefore $k=a$ and $n=a^2$ .

We have that $n=(a^2)^2=a^4$ . We can write n as $n=a^3a=a^2a^2=a^4 \cdot 1$ . From the first equation, a divides n and a³ divides n. From the second equation, a² divides, and from the last one, 1 divides n and a⁴=n divides n. Thus n has exactly 5 positive factors.