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Crazy boy [7]
3 years ago
7

An you help me now with these |x + 2| + 16 = 1 i need help :( |2x + 6| − 4 = 20 nd help solving

Mathematics
1 answer:
krok68 [10]3 years ago
3 0
|x + 2| + 16 = 1
<u>           - 16  - 16</u>
        |x + 2| = -15
         x + 2 = -15    U    x + 2 = 15
         <u>    - 2     - 2</u>          <u>    - 2    - 2</u>
               x = -17                 x = 13

|2x + 6| - 4 = 20
<u>            + 4   + 4</u>
     |2x + 6| = 24
      2x + 6 = 24    U    2x + 6 = -24
      <u>     - 6    - 6</u>           <u>     - 6      - 6</u>
            <u>2x</u> = <u>18</u>                 <u>2x</u> = <u>-30</u>
             2      2                   2       2
              x = 9                     x = -15
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Step-by-step explanation:

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Can someone help me with problem please? It’s special right triangle: decimal answers (Round to the nearest tenth)
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If line a is perpendicular to line b and the slope of line a is 7 whats the slope of line b
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Answer:

If a line is perpendicular to another line, that means that the slope is completely opposite that of the original line.  The first thing that we do to the slope is we negate the number which means that if we have a slope of -7 our slope because 7 in this step.  In our case our slope is 7 so in this step it becomes -7.

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Consider the equation below. f(x) = 2x3 + 3x2 − 336x (a) Find the interval on which f is increasing. (Enter your answer in inter
Vaselesa [24]

We have been given a function f(x)=2x^3+3x^2-336x. We are asked to find the interval on which function is increasing and decreasing.

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f'(x)=2(3)x^{2}+3(2)x^1-336

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So x=-8,7 are critical points and these will divide our function in 3 intervals (-\infty,-8)U(-8,7)U(7,\infty).

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Since f'(1), therefore, function is decreasing on interval (-8,7).

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Therefore, (-8,1856) is a local maximum and (7,-1519) is a local minimum.

(c) To find inflection points, we need to check where 2nd derivative is equal to 0.

Let us find 2nd derivative.

f''(x)=6(2)x^{1}+6

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Therefore, x=-\frac{1}{2} is an inflection point of given function.

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