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DiKsa [7]
2 years ago
13

Given f ( x ) = 1 x + 10 , find the average rate of change of f ( x ) on the interval [ 8 , 8 + h ] . Your answer will be an exp

ression involving h .
Mathematics
1 answer:
son4ous [18]2 years ago
6 0

Answer: 1

Step-by-step explanation:

f(8+h)=8+h+10=18+h\\\\f(8)=8+10=18

So, the average rate of change is:

\frac{(18+h)-18}{(8+h)-8}=\frac{h}{h}=\boxed{1}

The question is wrong, the expression will not be in terms of h since it is a linear function.

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krek1111 [17]

Answer: 3/4 or 75%

Step-by-step explanation:

5 0
2 years ago
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Is a positive integer raised to a negative power always positive
Evgesh-ka [11]

Answer:

Yes

Step-by-step explanation:

Many students gets confused with these exponential problems, they often get misguided but understand that there is nothing to be confused of.

When you have a negative number, you just take the reciprocal of the whole exponent number.

What is a reciprocal?

Let's take an integer for example, let's take the number 3.

The reciprocal for 3 is (1/3)

Let's take an other number, let's take 2/3

The reciprocal for 2/3 is (3/2)

In conclusion, we just reverse the denominator and the numerator or just switch it.

We take 3 as (3/1) and that is the reason, the reciprocal would be (1/3)

Now, coming to the negative integers. Taking an example:

(2)⁻¹

This would be become (1/2¹) = (1/2)

Hence, the result of a negative integer is positive but would be a fraction.

Hope I helped!


3 0
3 years ago
10 POINTS !! :( ANSWER BOTH PLS !! DUE BEFORE 11:59 PM
fenix001 [56]

Answer:

Easy,

1) To find the area you need to multiply. Therefore 10 x 4 = 40 square centimeters.

2) 20, why? A calculator online for area of a triangle, it really helps! :) just look up triangle area in the future.

Step-by-step explanation:

8 0
3 years ago
Is this function linear or nonlinear?​
Ivahew [28]
Nonlinear. hope this helps :D
4 0
2 years ago
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3ab - 2bc = 12 solve for b
Nikitich [7]
(3ab/b)-(2bc/b)=12/b
3a-2c=12b^-1
(3a-2c)/12=(12b^-1)/12
(3a-2c)/12=b^-1
12/(3a-2c)=b
4 0
3 years ago
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