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Licemer1 [7]
2 years ago
9

PARTIES Louise is having a party.

Mathematics
1 answer:
just olya [345]2 years ago
3 0

Answer:

32 guests

Step-by-step explanation:

96 divided by 3

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Your bank account hos a bolance of -$17. You deposit $70. What is your new balance? Show work:​
Norma-Jean [14]

Answer:

53

Step-by-step explanation:

70-17= 53

3 0
2 years ago
Read 2 more answers
Solve z ÷ (–7) = –1.<br> A. 7 B. -7 C. 1/7 D. 6
aalyn [17]

Answer:

A. 7

Step-by-step explanation:

z ÷ (–7) = –1.

multiply each side by -7

z ÷ (–7)*-7 = –1.*-7

z = 7

6 0
3 years ago
Read 2 more answers
If 5% of the total sale is 125 dollars what is the value of the ticket sale
RideAnS [48]
So the question is
:
5% of total sale is 125

'of' means mutiply
'is' means equals

percent means parts out of 100
5%=5/100=1/20
so now we have
1/20 times total sale=125
multiply both sides by 20 to clear fraction
20/20 times total sale=2500
1 times total sale=2500
total sale=2500
answer is total sale=$2500<span />
4 0
3 years ago
Hich ordered pair is the solution to the system of equations? {x=(1)/(2)y+5} {2x+3y=-14}
faltersainse [42]

Answer:

(2, - 6 )

Step-by-step explanation:

Given the 2 equations

x = \frac{1}{2} y + 5 → (1)

2x + 3y = - 14 → (2)

Substitute x = \frac{1}{2} y + 5 into (2)

2( \frac{1}{2} y + 5 ) + 3y = - 14 ← distribute and simplify left side

y + 10 + 3y = - 14

4y + 10 = - 14 ( subtract 10 from both sides )

4y = - 24 ( divide both sides by 4 )

y = - 6

Substitute y = - 6 into (1) for corresponding value of x

x = (0.5 × - 6 ) + 5 = - 3 + 5 = 2

Solution is (2, - 6 )

8 0
3 years ago
Read 2 more answers
If 2^n + 1 is an odd prime for some integer n, prove that n is a power of 2. (H
vovikov84 [41]

Step-by-step explanation:

We will prove by contradiction. Assume that 2^n + 1 is an odd prime but n is not a power of 2. Then, there exists an odd prime number p such that p\mid n. Then, for some integer k\geq 1,

n=p\times k.

Therefore

  1. 2^n + 1=2^{p\times k} + 1=(2^{k})^p + 1^p.

Here we will use the formula for the sum of odd powers, which states that, for a,b\in \mathbb{R} and an odd positive number n,

a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^2-...+b^{n-1})

Applying this formula in 1) we obtain that

2^n + 1=2^{p\times k} + 1=(2^{k})^p + 1^p=(2^k+1)(2^{k(p-1)}-2^{k(p-2)}+...-2^{k}+1).

Then, as 2^k+1>1 we have that 2^n+1 is not a prime number, which is a contradiction.

In conclusion, if 2^n+1 is an odd prime, then n must be a power of 2.

4 0
3 years ago
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