Ask question

Ask question

All categories

3 years ago

12

What is the surface area of a right circular cylindrical oil can, if the radius of its base is 4 inches and its height is 11 inc

hes? for 25 points

2 answers:

SpyIntel [72]3 years ago

7 0

Answer:

= 2πr ( r + h)

= 2 * 22 / 7 * 4 ( 4 + 11)

= 2 * 22/7 * 4 * 15

= 377.14 inches ²

Step-by-step explanation:

That's your answer your welcome :) hope this helps :)

Alla [95]3 years ago

4 0

Step-by-step explanation:

Given

Radius (r) = 4 inches

height (h) = 11 inches

Total surface area of cylindrical oil can is

= 2πr ( r + h)

= 2 * 22 / 7 * 4 ( 4 + 11)

= 2 * 22/7 * 4 * 15

= 377.14 inches ²

You might be interested in

When the same whole number is added to both the numerator and denominator of $\frac{2}{5}$, the value of the new fraction is $\f

AVprozaik [17]

Answer:

Correct answer: x = 4

Step-by-step explanation:

If I saw your fractions right it's like this:

(2 + x) / (5 + x) = 2 / 3 => 3 (2 + x) = 2 (5 + x) => 6 + 3 x = 10 + 2 x

3 x - 2 x = 10 - 6 => x = 4

We will check

2 + 4 / 5+ 4 = 6 / 9

When we divide both numerator and denominator

with 3 we get 2/3

God is with you!!!

3 0

3 years ago

A cyclist travels at distance of 400 meters in 120 seconds towards school, calculate his speed. (Show your work)

DochEvi [55]

we know that

The scalar magnitude of the velocity vector is the speed. The speed is equal to

$Speed=\frac{distance}{time}$

in this problem we have

$distance=400\ m \\time=120\ sec$

substitute in the formula

$Speed=\frac{400}{120}$

$Speed=3.5\frac{m}{sec}$

therefore

<u>the answer Part a) is</u>

the speed is equal to $3.5\frac{m}{sec}$

<u>Part b) </u>Find the velocity

we know that

<u>Velocity </u>is a vector quantity; both magnitude and direction are needed to define it

in this problem we have

the magnitude is equal to the speed

$magnitude=3.5\frac{m}{sec}$

$direction=North\ East\ (NE)$

therefore

<u>the answer Part b) is</u>

the velocity is $3.5\frac{m}{sec}\ North\ East\ (NE)$

Part c)

we know that

the acceleration is equal to the formula

$a=\frac{V2-V1}{t2-t1}$

in this problem we have

$V2=0 \\V1=3.5\frac{m}{sec}$

$t2=15\ sec\\t1=0$

substitute in the formula

$a=\frac{0-3.5}{15-0}$

$a=-\frac{3.5}{15}\frac{m}{sec^{2}}$

$a=-\frac{7}{30}\frac{m}{sec^{2}}$

therefore

<u>the answer Part c) is</u>

the acceleration is $-\frac{7}{30}\frac{m}{sec^{2}}$

This is an example of negative acceleration

7 0

3 years ago

Read 2 more answers

Please help!!!!!!!!!!!!

podryga [215]

Given : Fix amount charges = $200.

Each month additional charge = $50.

Let us assume number of months are taken by x.

So, we can setup a function as

Total charge = Fix charge + each month charge * number of months.

f(x) = 200 + 50*x

Or f(x) = 200 +5x.

Where function f is the total charge of x number of months.

The values of x's represents domain and function value represents range.

According to problem, it is said " first three months".

So, we need to take number of months as x=0,1,2,3.

And we need to find values of function for those x values.

On plugging x=0,1,2 and 3 we would get

f(0) = 200 +5(0) = 200 +0 = 200.

f(1) = 200 +5(1) = 200 +50 = 250.

f(2) = 200 +5(2) = 200 +100 = 300.

f(3) = 200 +5(3) = 200 +150 = 350.

We got function values 200,250,300 and 350.

Therefore, with respect to the given situation, the domain of the funtion is

{0,1,2,3} and Range { 200,250,300, 350}.

6 0

3 years ago

Tammy has begun an exercise program. She worked out five days the first week for 28 minutes, 32 minutes, 21 minutes, 38 minutes

Luba_88 [7]

28 rounds up to 30.
32 rounds down to 30.
21 rounds down to 20.
38 rounds up to 40.
42 rounds down to 40.
30 + 30 + 20 + 40 + 40 = 160

Answer: C. 160 min

5 0

3 years ago

Read 2 more answers

I have a 82.92 in anatomy class but I recieved a 60 percent on an exam.. what will this bring my grade to?

Semmy [17]

Answer:

-D

Step-by-step explanation:

5 0

3 years ago