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4 years ago

2. Help me find (f/g)(x)

Mathematics

2 answers:

tiny-mole [99]4 years ago

6 0

2x(2x + 3)
That would be -----------------------
2x^2 + 13x + 15

We should factor the denominator and then reduce the original fraction:

2x(2x + 3) 2x x
----------------------- = ------------- = ---------
(2x + 3)(2)(x+5) 2(x+5) x+5

This is true only for x not equal to -5, not equal to -3/2.

ElenaW [278]4 years ago

5 0

Answer:

-5

Step-by-step explanation:

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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

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tino4ka555 [31]

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APEX

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3 years ago

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Sedbober [7]

Answer:

4.9

Step-by-step explanation:

$70 times 7% = 4.9 (You can round it if you want)

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strojnjashka [21]

area of quadrilateral (rhombus)=4square units

:

Solution given:

diagonal 1:AC=2 unit

diagonal 2:BD=4 unit

it looks like a rhombus

area of quadrilateral (rhombus)

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