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Goshia [24]
3 years ago
7

6. A number j increased by 8 *

Mathematics
1 answer:
e-lub [12.9K]3 years ago
8 0

Answer:

6. b. j+ 8

7. b. n - 2

8. a. Addition

9. d. y ÷ -5

10. c. 18 - 19

11. a. 10 + 6

12. b. Divided by

13. a. three added to 5

14. b. 9

15. b. 75

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Volunteers helped clean up 8.2 kg of trash in one neighborhood and 11 1/2 kg in another. They sent 1 1/4 kg to be recycled and t
Ivan

Answer: 18.45\ kg or 18\frac{9}{20}\ kg

Step-by-step explanation:

We can convert from mixed numbers to decimal numbers:

11\frac{1}{2}\ kg=(11+0.5)\ kg=11.5\ kg\\\\1\frac{1}{4}\ kg=(1+0.25)\ kg=1.25\ kg

Since they cleaned up 8.2 kilograms of trash in one neighborhood and 11.5 kilograms in another neighborhood, the total trash they cleaned up was:

Total=8.2\ kg+11.5\ kg\\\\Total=19.7\ kg

We know that they sent 1.25 kilograms to be recycled. Then, in order to calculate how many kilograms of trash they threw away, we must subtract the total kilograms of trash they cleaned up and the kilograms they sent to be recycled.

Then:

19.7\ kg-1.25\ kg=18.45\ kg or 18\frac{9}{20}\ kg

8 0
4 years ago
Read 2 more answers
Use the graph to answer the question what is the average rate from change from x=3 to x=11
expeople1 [14]

Answer:

-1/8

Step-by-step explanation:

7 0
4 years ago
Which graph shows a rate of change of 1/2 which is a fraction between –4 and 0 on the x-axis?
irakobra [83]

By definition, we have that the average rate of change is given by:

AVR = \frac{y2-y1}{x2-x1}

Substituting values we have:

AVR = \frac{3-1}{0-(-4)}

Rewriting we have:

AVR = \frac{2}{0+4}

AVR = \frac{2}{4}

AVR = \frac{1}{2}

Therefore, the graph that has an AVR of 1/2 is the graph of the linear function.

Answer:

graph shows a rate of change of 1/2 is the linear function

6 0
3 years ago
Read 2 more answers
Find an equation for a line that is normal to the graph of y=xe^x and goes through the origin
Marina CMI [18]
Y = xe^x
dy/dx(e^x x)=>use the product rule, d/dx(u v) = v*(du)/(dx)+u*(dv)/(dx), where u = e^x and v = x:
= e^x (d/dx(x))+x (d/dx(e^x))
y' = e^x x+ e^x
y'(0) = 1 => slope of the tangent
slope of the normal = -1
y - 0 = -1(x - 0)
y = -x => normal at origin
3 0
3 years ago
Students in a zoology class took a final exam. They took equivalent forms of the exam at monthly intervals thereafter. After t​
juin [17]

Answer:

a. 73; b. 48.9; c. 2; d. 33.8; e. 73

Step-by-step explanation:

Assume the function was

S(t)= 73 - 15 ln(t + 1), t  ≥ 0

a. Average score at t = 0

S(0) = 73 - 15 ln(0 + 1) = 73 - 15 ln(1) = 73 - 15(0) =73 - 0 = 73

b. Average score at t = 4

S(4) = 73 - 15 ln(4 + 1) = 73 - 15 ln(5) = 73 - 15(1.61) =73 - 24.14 = 48.9

c. Average score at t =24

S(24) = 73 - 15 ln(24 + 1) = 73 - 15 ln(25) = 73 - 15(3.22) =73 - 48.28 = 24.7

d. Percent of answers retained

At t = 0. the students retained 73 % of the answers.

At t = 24, they retained 24.7 % of the answers.

\text{Percent retention} = \dfrac{\text{24.7}}{\text{73}} \times 100 \, \% = \text{33.8 \%}\\\\\text{The students retained $\large \boxed{\mathbf{33.8 \, \%}}$ of their original knowledge after two years.}

e. Maximum of the function

The maximum of the function is at t= 0.

Max = 73 %

The graph below shows your knowledge decay curve. Knowledge decays rapidly at first but slows as time goes on.

 

6 0
3 years ago
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