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3 years ago

PLEASE HELP ASAP 15 BRAINLY POINTS

Mathematics

2 answers:

lana66690 [7]3 years ago

5 0

Answers

1) 1

2) 11

3) 4

4) D) Yes, the constant of variation is 3

Explanation

The slope of a line is given by

Slope = (change in y)/(change in x)

= (0 - -2)/(-4 - -6)

= 2/2

= 1

Slope = Δy/Δx

= (8 - -3)/(2 - 1)

= 11/1

= 11

Y varies directly as X. This is written as,

Y ∞ X

Put and equal sign then introduce the constant of proportionality, K.

Y = KX

K = Y/X

K = 6/18 = 1/3

∴ Y = (1/3)X

Y = (1/3) × 12

Y = 4

32Y = 96X

Dividing both sides by 32,

Y = 96/32 X

Y = 3X

3 been the constant, Y varies directly as X.

So, the equation 32y = 96x is a direct variation.

Constant of variation = 3

joja [24]3 years ago

3 0

the answers are 1) 1 2) 11 3) 4 4) D) 3

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(a) f is increasing at $(-\infty,-2) \cup (2,\infty)$ .

(b) f is decreasing at $(-2,2)$ .

(c) f is concave up at $(2, \infty)$

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Explanations:

(a) f is increasing when the derivative is positive. So, we find values of x such that the derivative is positive. Note that

$f'(x) = 15x^2 - 60
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So,

$
f'(x) \ \textgreater \ 0
\\
\\ \Leftrightarrow 15x^2 - 60 \ \textgreater \ 0
\\
\\ \Leftrightarrow 15(x - 2)(x + 2) \ \textgreater \ 0
\\
\\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textgreater \ 0} \text{ (1)}$

The zeroes of (x - 2)(x + 2) are 2 and -2. So we can obtain sign of (x - 2)(x + 2) by considering the following possible values of x:

-->> x < -2
-->> -2 < x < 2
--->> x > 2

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If -2 < x < 2, then x + 2 is positive but x - 2 is negative. So, (x - 2)(x + 2) < 0.
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So, (x - 2)(x + 2) is positive when x < -2 or x > 2. Since

$f'(x) \ \textgreater \ 0 \Leftrightarrow (x - 2)(x + 2) \ \textgreater \ 0$

Thus, f'(x) > 0 only when x < -2 or x > 2. Hence f is increasing at $(-\infty,-2) \cup (2,\infty)$ .

(b) f is decreasing only when the derivative of f is negative. Since

$f'(x) = 15x^2 - 60$

Using the similar computation in (a),

$f'(x) \ \textless \ \ 0 \\ \\ \Leftrightarrow 15x^2 - 60 \ \textless \ 0 \\ \\ \Leftrightarrow 15(x - 2)(x + 2) \ \ \textless \ 0 \\ \\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textless \ 0} \text{ (2)}$

Based on the computation in (a), (x - 2)(x + 2) < 0 only when -2 < x < 2.

Thus, f'(x) < 0 if and only if -2 < x < 2. Hence f is decreasing at (-2, 2)

(c) f is concave up if and only if the second derivative of f is positive. Note that

$f''(x) = 30x - 60$

Since,

$f''(x) \ \textgreater \ 0
\\
\\ \Leftrightarrow 30x - 60 \ \textgreater \ 0
\\
\\ \Leftrightarrow 30(x - 2) \ \textgreater \ 0
\\
\\ \Leftrightarrow x - 2 \ \textgreater \ 0
\\
\\ \Leftrightarrow \boxed{x \ \textgreater \ 2}$

Therefore, f is concave up at $(2, \infty)$ .

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$f''(x) = 30x - 60$

Using the similar computation in (c),

$f''(x) \ \textless \ 0 
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\\ \\ \Leftrightarrow 30(x - 2) \ \textless \ 0 
\\ \\ \Leftrightarrow x - 2 \ \textless \ 0 
\\ \\ \Leftrightarrow \boxed{x \ \textless \ 2}$

Therefore, f is concave down at $(-\infty, 2)$ .

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