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2 years ago

All simplified answers for 6x+1-(3x-1)

Mathematics

1 answer:

sertanlavr [38]2 years ago

7 0

Answer:

Step-by-step explanation:

Simplify ——————

3x - 1

Equation at the end of step 1 :

(6x - 1)

y - ———————— = 0

3x - 1

Step 2 :

Rewriting the whole as an Equivalent Fraction :

2.1 Subtracting a fraction from a whole

Rewrite the whole as a fraction using (3x-1) as the denominator :

y y • (3x - 1)

y = — = ————————————

1 (3x - 1)

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Step-by-step explanation:

We have

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

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Note that for

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

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

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because the sum is empty.

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

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

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has no solution

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