All categories

3 years ago

Given the function h(x)=-x^2-5x+14 determine the average rate of change of the function over the interval -9

Mathematics

1 answer:

Juli2301 [7.4K]3 years ago

8 0

Answer:

The average rate of change of the function over the interval is of 6.

Step-by-step explanation:

Average rate of change:

The average rate of change of a function h(x) over an interval [a,b] is given by:

$A = \frac{h(b)-h(a)}{b-a}$

In this question:

Over the interval [-9,-2], so $a = -9, b = -2, b - a = -2 -(-9) = 7$

The function is:

$h(x) = -x^2 - 5x + 14$

$h(-9) = -9^2 -5(-9) + 14 = -81 + 45 + 14 = -22$

$h(-2) = -2^2 -5(-2) + 14 = -4 + 10 + 14 = 20$

Then

$A = \frac{h(-2)-h(-9)}{7} = \frac{20-(-22)}{7} = \frac{42}{7} = 6$

The average rate of change of the function over the interval is of 6.

You might be interested in

Brainliest if correct

harkovskaia [24]

Answer:

3/8 & 4/8

Step-by-step explanation:

1/2 - 1/8 = 4/8 - 1/8 = 3/8

x - 1/8 = 3/8

x = 4/8

6 0

2 years ago

Find the height of the triangle in which A=192 cm^2

romanna [79]

Step-by-step explanation:

one second 12345678910

6 0

2 years ago

PLEASE HELP IF YOU NEED ME TO TELL YOU THE INFO FOR A.) I WILL

JulijaS [17]

28.34 can be estimated to the nearest number, which is 28. 26.59 can be estimated to the nearest number, which is 27. 33.11 can be estimated to the nearest number, which is 33. 28+27+33=88

3 0

3 years ago

Read 2 more answers

The exponential function y= a(b)^x passes through the points (3,25) and (8,10). Which of the following is closest to the value o

Yuki888 [10]

If the exponential function $y=ab^x$ passes through the points (3,25) and (8,10), the value of b is 0.833

The given exponential equation is:

$y=ab^x$

The function passes through the points (3, 25) and (8, 10)

Substitute x = 3 and y = 25 into the function $y=ab^x$

$25=ab^3$ ............................(1)

Substitute x = 8 and y = 10 into the function $y=ab^x$

$10=ab^8$ .............................(2)

Divide equation (2) by equation (1)

$\frac{10}{25}=\frac{ab^8}{ab^3} \\\\0.4=b^5\\\\b=0.4^\frac{1}{5} \\\\b=0.4^{0.2}\\\\b=0.833$

Therefore, if the exponential function $y=ab^x$ passes through the points (3,25) and (8,10), the value of b is 0.833

Learn more on exponential functions here: brainly.com/question/12940982

6 0

2 years ago

ACTIVITY 2 (19) Mr Duma recently inherited a rectangular plot, part of the estate left by his late father. The plot with the fol

seraphim [82]

The sum of the lengths of three sides of the rectangle, gives the length

of the fencing, while one third of the rectangle area is for the pavement.

Responses:

Project A: The formula for the length of the fencing is, L = 4·x - 1

Project B: Length of the fancy wall = 2·x + 1

$Project \ C:Area \ of \ the \ paving = \underline{ \dfrac{2\cdot x^2 }{3} - \dfrac{x }{3} - \dfrac{1}{3}}$

<h3>Which method can be used to find the length and area of paving from the given equations?</h3>

Project A: Let QP and SR represent the longest sides of the rectangle, we have;

PQ = SR = 2·x + 1

Given parameters are;

Length of the rectangular plot = 2·x + 1

Width of the rectangular plot = x - 1

Vertices of the rectangular plot are; QPSR

Project A: Let QP and SR represent the longest sides of the rectangle, we have;

PQ = SR = 2·x + 1

Which gives;

SP = QR = x - 1

The length of the fencing, L = SP + PQ + QR = x - 1 + 2·x + 1 + x - 1 = 4·x - 1

The formula for the length of the fencing is, L =<u> 4·x - 1</u>

Project B: The front side is SR

Therefore;

Length of the fancy wall = <u>2·x + 1</u>

Project C:

Area, A = Length × Width

Area of the plot, A = (2·x + 1) × (x - 1) = 2·x² - x - 1

$Area \ of \ the \ paving = \dfrac{1}{3} \times \left(2 \cdot x^2 - x - 1\right) = \underline{ \dfrac{2\cdot x^2 }{3} - \dfrac{x }{3} - \dfrac{1}{3}}$

Learn more about writing equations here:

brainly.com/question/24760633

6 0

2 years ago