Given: $f (x) = 2 + \sqrt{- 3 x + 1}$ $f(x) = 2+ sqrt(-3x+1)$ , $q (x) = \frac{5 x}{x + 2}$ $q(x) = (5x)/(x+2)$ , and $m (x) = ∣ ∣ 7 - x^{2} ∣ ∣ + x$ $m(x) = abs(7 - x^2) + x$ , how do you find q(10)?

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Given: $f (x) = 2 + \sqrt{- 3 x + 1}$ $f(x) = 2+ sqrt(-3x+1)$ , $q (x) = \frac{5 x}{x + 2}$ $q(x) = (5x)/(x+2)$ , and $m (x) = ∣ ∣ 7 - x^{2} ∣ ∣ + x$ $m(x) = abs(7 - x^2) + x$ , how do you find q(10)?

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Answer 1 · 2018-05-16T03:45:54

q(10) = about 6.17

Explanation:

Since you have a given of q(x) = $\frac{5 x}{x + 2}$ $(5x)/(x+2)$ , the answer can be found by substituting 10 for x in the equation.

This would then be rewritten as $\frac{5 \cdot 10}{10 + 2}$ $(5*10)/(10+2)$
The top and bottom would be simplified next: $\frac{50}{12}$ $50/12$
Then, in dividing the two numbers, the final answer is about 6.17, when rounded to the nearest hundredth's spot.

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Given: $f (x) = 2 + \sqrt{- 3 x + 1}$ $f(x) = 2+ sqrt(-3x+1)$ , $q (x) = \frac{5 x}{x + 2}$ $q(x) = (5x)/(x+2)$ , and $m (x) = ∣ ∣ 7 - x^{2} ∣ ∣ + x$ $m(x) = abs(7 - x^2) + x$ , how do you find q(10)?

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Explanation:

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Given: f(x)=2+√−3x+1f(x) = 2+ sqrt(-3x+1), q(x)=5xx+2q(x) = (5x)/(x+2), and m(x)=∣∣7−x2∣∣+xm(x) = abs(7 - x^2) + x, how do you find q(10)?

1 Answer

Explanation:

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Impact of this question

Given: $f (x) = 2 + \sqrt{- 3 x + 1}$ $f(x) = 2+ sqrt(-3x+1)$ , $q (x) = \frac{5 x}{x + 2}$ $q(x) = (5x)/(x+2)$ , and $m (x) = ∣ ∣ 7 - x^{2} ∣ ∣ + x$ $m(x) = abs(7 - x^2) + x$ , how do you find q(10)?