In order to calculate the rate of change, take two points (in this example I'll use points (8, 2) and (10, 4)) and input them into this expression used to calculate rate of change: $\frac{y2-y1}{x2-x1}$

$\frac{4-2}{10-8}$

Solve:

4 - 2 = 2

10 - 8 = 2

2/2 = 1

The rate of change is 1.

Hope this helps :)

7 0

3 years ago

What is the average rate of change of f<br> over the interval (-5,0]?<br> Give an exact number.

Paraphin [41]

Answer:

Average rate of change = 0.6

Step-by-step explanation:

Average rate of change of a function between x = a and x = b is given by,

Average rate of change = $\frac{f(b)-f(a)}{b-a}$

From the graph attached,

f(-5) = 0

f(0) = 3

Average rate of change of the graph between x = -5 and x = 0,

Average rate of change = $\frac{3-0}{0-(-5)}$

= $\frac{3}{5}$

= 0.6

Therefore, average rate of change of the given function between x = -5 and x = 0 is 0.6.

8 0

4 years ago

Susie has 3 1/5 pounds Of jellybeans. Julie wants to combined his 2 1/3 Pound bag of licorice with Susie's jellybeans. What woul

Naily [24]

Answer:

The Total Weight of candy is $5\frac{8}{15} \ pounds$ .

Step-by-step explanation:

Given:

Amount of Jelly beans Susie have = $3\frac{1}{5}$ pounds

$3\frac{1}{5}$ can be rewritten as $\frac{16}{5}$

Also Given;

Julie has amount of licorice = $2\frac{1}{3}$ pounds

$2\frac{1}{3}$ can be Rewritten as $\frac{7}{3}$

Now Julie wants to combined her bag of licorice with Susie's jellybeans.

Hence the Total weight of candy will be the sum of Amount of Jelly beans and amount of licorice.

Total weight of candy = $\frac{16}{5} + \frac{7}{3}$

Now making the Denominator common we get;

Total weight of candy = $\frac{16\times3}{5\times 3} + \frac{7\times5}{3\times5}= \frac{48}{15} + \frac{35}{15}=\frac{48+35}{15}= \frac{83}{15} \ pounds$

$\frac{83}{15} \ pounds$ can be Rewritten as $5\frac{8}{15} \ pounds$

Hence The Total Weight of candy is $5\frac{8}{15} \ pounds$ .

6 0

3 years ago