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stiv31 [10]
3 years ago
5

How do you do this question?

Mathematics
1 answer:
s344n2d4d5 [400]3 years ago
8 0

Answer:

A) x ≤ -2 and 0 ≤ x ≤ 3

Step-by-step explanation:

g(x) is decreasing when g'(x) is negative.

Use second fundamental theorem of calculus to find g'(x).

g(x) = ∫₋₁ˣ (t³ − t² − 6t) / √(t² + 7) dt

g'(x) = (x³ − x² − 6x) / √(x² + 7) (1)

To find when g'(x) is negative, first find where it is 0.

0 = (x³ − x² − 6x) / √(x² + 7)

0 = x³ − x² − 6x

0 = x (x² − x − 6)

0 = x (x − 3) (x + 2)

x = -2, 0, or 3

Check the intervals before and after each zero.

x < -2, g'(x) < 0

-2 < x < 0, g'(x) > 0

0 < x < 3, g'(x) < 0

3 < x, g'(x) > 0

g(x) is decreasing on the intervals x ≤ -2 and 0 ≤ x ≤ 3.

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Write the equation of the line that passes through (3, 4) and (2, −1) in slope-intercept form. (2 points) a y = 3x − 7 b y = 3x
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Answer: y = 5x − 11

Step-by-step explanation:

The equation of a straight line can be represented in the slope-intercept form, y = mx + c

Where c = intercept

Slope, m =change in value of y on the vertical axis / change in value of x on the horizontal axis represent

change in the value of y = y2 - y1

Change in value of x = x2 -x1

y2 = final value of y

y 1 = initial value of y

x2 = final value of x

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The line passes through (3,4) and (2, -1),

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To determine the y intercept, we would substitute x = 3, y = 4 and m= 5 into

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4 = 5 × 3 + c

4 = 15 + c

c = 4 - 15 = - 11

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y = 5x - 11

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