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Nikolay [14]
3 years ago
13

On the interval (r, 7), at which x-value is the average rate of change 56?

Mathematics
1 answer:
denpristay [2]3 years ago
7 0
I need point 4,7 so lemme get this pls and thx
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7p+5q+4p-2q what is the result of combining the like terms
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Answer:

11p+7q

Step-by-step explanation:

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Choose the graph below that meets all three of the following requirements: Limit as x approaches 1 minusf(x) = 3 Limit as x appr
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Graph D (last one)

<u><em>Step-by-step explanation:</em></u>

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PLEASE ANSWER THIS AND IF YOU DO YOU GET <br> 1. FOLLOW<br> 2. BRAINLIEST
spin [16.1K]

Answer:

1. y= 2x + 5

2. y= 2x -16

3. - 7/5 x + 18/5

3 0
3 years ago
use the definition of countinuity to find the value of k so that the function is continuous for all real numbers
liubo4ka [24]

First of all, recall the definition of absolute value:

|x| = \begin{cases}x&\text{if }x\ge0\\-x&\text{if }x

So if <em>x</em> < 4, then <em>x</em> - 4 < 0, so |<em>x</em> - 4| = -(<em>x</em> - 4), and the first case in <em>h(x)</em> reduces to

\dfrac{|x-4|}{x-4}=\dfrac{-(x-4)}{x-4} = -1

Next, in order for <em>h(x)</em> to be continuous at <em>x</em> = 4, the limits from either side of <em>x</em> = 4 must be equal and have the same value as <em>h(x)</em> at <em>x</em> = 4. From the given definition of <em>h(x)</em>, we have

h(4) = 5k-4\cdot4 = 5k-16

Compute the one-sided limits:

• From the left:

\displaystyle \lim_{x\to4^-}h(x) = \lim_{x\to4}\frac{|x-4|}{x-4} = \lim_{x\to4}(-1) = -1

• From the right:

\displaystyle \lim_{x\to4^+}h(x) = \lim_{x\to4}(5k-4x) = 5k-16

If the limits are to be equal, then

-1 = 5<em>k</em> - 16

Solve for <em>k</em> :

-1 = 5<em>k</em> - 16

15 = 5<em>k</em>

<em>k</em> = 3

3 0
3 years ago
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