Mathematics, 22.11.2020 02:50 robin6150

SOMEONE HELP? I’ll give brainliest

Answers: 1

Show answers

Answers

Answer from: alexamorantess

$(\pi 4^2 (9))-(\pi 2^2 (9))$

and

339.292 cm^3

Step-by-step explanation:

This question is simple once you break it down.

Let's concentrate on the outer cylinder.

First, find the volume of the outer cylinder.

The volume of a cylinder is expressed as: $V = \pi r^2 h$

Gathering information from the diagram..

d = 8 cm

h = 9 cm

You can get the radius from the diameter by just dividing the diameter by 2.

$r = \frac{8}{2}=4$

r = 4

Plug everything into the volume equation.

$V = \pi 4^2 (9)$

Don't solve just yet!

Now, find the volume of the inner cylinder.

Do the exact same thing as you did for the outer cylinder.

d = 4 cm

h = 9 cm

$r = \frac{4}{2}=2$

r = 2

Plug everything into the volume equation..

$V = \pi 2^2 (9)$

The question asks for an expression to represent the volume of the large cylinder after the cylindrical hole is removed. This means you need to subtract the volume of the cylindrical hole from the volume for the large cylinder.

Again, the volume for the large cylinder is $V = \pi 4^2 (9)$ and the volume for the cylindrical hole is $V = \pi 2^2 (9)$

This means..

$(\pi 4^2 (9))-(\pi 2^2 (9))$

That is your answer to the first part of the question.

For the second part, just solve.

$(\pi 4^2 (9))-(\pi 2^2 (9))$

$(\pi*16*9)-(\pi*4*9)$

$(\pi*144)-(\pi*36)$

$144\pi-36\pi$

$108\pi$

339.292

Don't forget the units.

339.292 cm^3

Hope this helped you!

Answer from: Quest

part a

answer: a reasonable domain is $0 \le n \le 10$ where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

=====================================

part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

=====================================

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).

Answer from: Quest

well what did you do it matters what you did : )

step-by-step explanation: