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Fudgin [204]
3 years ago
9

HELP ME WITH THIS QUESTION PLEASE

Mathematics
1 answer:
Shkiper50 [21]3 years ago
8 0

Answer:

Step-by-step explanation:

1. 7x + 4

2. 3x - 2

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A cube with a volume of 75 cm³ is dilated by a factor of 5.
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ANSWER

The volume of the dilated cube is 9375cm³.


EXPLANATION

The volume of the original cube is 75 cm³.



The scale factor for the dilation is 5.


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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

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