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Sauron [17]
3 years ago
6

12c - 4 = 14c - 10 c = ? !!!

Mathematics
1 answer:
Sergio [31]3 years ago
3 0
The answer to this question is c=3

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Katyanochek1 [597]
1 and 11/12 hours I think
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What’s the answer to this? I need to do this for extra credit and I have no idea what this is
sergiy2304 [10]

9514 1404 393

Answer:

  20

Step-by-step explanation:

As with any evaluation problem, take it step by step according to the order of operations.

The first thing you need to do here is compute a#b.

The given definition can be simplified a bit for evaluation purposes:

  a#b = a²b -ab² = ab(a -b)

Then for a=3 and b=-2, you have ...

  (3)#(-2) = (3)(-2)(3 -(-2)) = -6(5) = -30

Now, you are in a position to evaluate the expression you're asked for.

  \dfrac{(a\#b)^2}{15-(a\#b)}=\dfrac{(-30)^2}{15-(-30)}=\dfrac{900}{45}=\boxed{20}

4 0
3 years ago
What is best definition of combining like terms
Bond [772]
They are terms<span> that contain the same variables raised to the same power. Only the numerical coefficients are different. In an expression, only </span>like terms<span> can be combined. We combine </span>like terms<span> to shorten and simplify algebraic expressions, so we can work with them more easily.</span>
8 0
3 years ago
A regular octagon rotates 360° about its center. How many times does the image of the octagon coincide with the preimage during
Anuta_ua [19.1K]
This regular polygon, having eight sides and 45º angle in similar parts, when rotating 360º, will find a similar image and a similar angle of 45º, therefore dividing the angle of 360º by 45º will be the Number of coincidences<span> in image.

</span><span>Solving, thus, we have:
</span>\frac{360}{45} = \boxed{\boxed{8\:coincidences\:in\:image}}\end{array}}\qquad\quad\checkmark<span>



</span>
4 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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