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IRISSAK [1]
3 years ago
14

Plz answer the following No scams,links or random amswers or ill report​

Mathematics
1 answer:
zvonat [6]3 years ago
3 0

Answer:

f(2) = 12

f(x) = 7, x = -3, 1

Step-by-step explanation:

<u>a)</u>

plug in x as 2

f(x) = 2^2 + 2(2) + 4

f(x) = 4 + 4 + 4

f(x) = 12

<u>b)</u>

replace f(x) with 7

7 = x^2 + 2x + 4

x^2 + 2x - 3 (move 7 to other side)

Factor

ac: -3x^2

b: 2x

split b into 3x, -x

(x^2 -x) + (3x - 3)

    ↓             ↓

  x(x-1)  +  3(x-1)

Factor: (x-1)(x+3) = 0

Solve using Zero Product Property:

x - 1 = 0, x + 3 = 0

x = 1, x = -3

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SOLUTION:

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2 years ago
Which of the following is an irrational number?
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7 0
3 years ago
<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
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Answer:

No solution

Step-by-step explanation:

We have

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

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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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