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kotegsom [21]
4 years ago
14

Please help, which expression is equivalent to....

Mathematics
1 answer:
nadya68 [22]4 years ago
3 0

Answer:

a

Step-by-step explanation:

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Simplify plzzzzzz 6(12m+12b-2b)
Slav-nsk [51]

How to get answer by Mimiwhatsup:

6(12m+12b−2b)

=(6)(12m+12b+−2b)

=(6)(12m)+(6)(12b)+(6)(−2b)

=72m+72b−12b

=60b+72m

6 0
4 years ago
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Solve for b.<br><br> 8b−1=24b+4
Allushta [10]
8b - 1 = 24b + 4
8b - 1 + 1 = 24b + 4 + 1
8b = 24b + 5
8b - 24b = + 5
-16 = + 5
-5/16

Answer: b = -5/16

3 0
3 years ago
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Please show your work so i can see how to do it. The problem on the top btw is the one i want you to answer.
olga2289 [7]

Answer:

baby, I don't know

Step-by-step explanation:

6 0
3 years ago
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Please hurry ;^;
kompoz [17]

Answer:

28x+8y+2

Step-by-step explanation:

the answer is 28x+8y+2 because a math calculator said so (m a t h w a y)

7 0
3 years ago
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<img src="https://tex.z-dn.net/?f=%20%5Clarge%7B%5Cbold%20%5Cred%7B%20%5Csum%20%5Climits_%7B8%7D%5E%7B4%7D%20%7Bx%7D%5E%7B2%7D%2
kiruha [24]

Answer:

No solution

Step-by-step explanation:

We have

$\sum_{x=8}^{4}x^2 + 9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)$

For the sum it is not correct to assume

$\sum_{x=8}^{4}x^2= 8^2 + 7^2+6^2+5^2+4^2 = 64+49+36+25+16 = 190$

Note that for

$\sum_{x=a}^b f(x)$

it is assumed a\leq x \leq b and in your case \nexists x\in\mathbb{Z}: a\leq x\leq b for a>b

In fact, considering a set S we have

$\sum_{x=a}^b (S \cup \varnothing) = \sum_{x=a}^b S + \sum_{x=a}^b \varnothing $ that satisfy S = S \cup \varnothing

This means that, by definition \sum_{x=a}^b \varnothing = 0

Therefore,

$\sum_{x=8}^{4}x^2 = 0$

because the sum is empty.

For

9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)

we have other problems. Actually, this case is really bad.

Note that \cos^2(\infty) has no value. In fact, if we consider for the case

$\lim_{x \to \infty} \cos^2(x)$, the cosine function oscillates between [-1, 1] , and therefore it is undefined. Thus, we cannot evaluate

9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)

and then

$\sum_{x=8}^{4}x^2 + 9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)$

has no solution

7 0
3 years ago
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