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jok3333 [9.3K]
3 years ago
14

If the height of the cylinder is 3 inches, what is the volume of the cylinder? Write your answer in terms of p.

Mathematics
2 answers:
azamat3 years ago
7 0
Formula for Volume of Cylinder you need the radius
Harrizon [31]3 years ago
4 0

Answer:

if the radius is 5 inches we simply find the area first

pi x r^2 (squared)

3.4 x 5 x 5

a = 34cm^2

v = a x h

= 34 x 3inch

=102inch^3

Replace p values as pi

as pi is 3.4 102/3.4 =30

To look like this

30px= 34x * 3x - x

102x -x = 30px

we use and rid of parenthesis

102x=30p

30/102x = 3.14p

x=3.14p

Step-by-step explanation:

You will see the cylinder is a fraction length of 0.5 so we just divide by 10 if using 0.5 inches.

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Which statement below is an example of the
Hitman42 [59]

Answer:

B

Step-by-step explanation:

Let say, a=9,b=11,c=6

Then associative properties says that every element is related to each other, here "b" associates a &c.

Therefore, a is related to b+c

c is related to a+b

3 0
2 years ago
What transformations occur on the absolute value function for -4f(x+2)-5?
Temka [501]
-4fx + -8f - 5


I’m pretty sure that’s correct, might be wrong double check with someone else
7 0
3 years ago
Abby took an 8-question multiple-choice quiz. Suppose
Alenkasestr [34]
It is 0.0623 hope it helps
4 0
3 years ago
In ΔLMN, n = 510 cm, l = 820 cm and ∠M=20°. Find ∠L, to the nearest degree
Dafna11 [192]

The angular measure L is : 47°

<h3>What are angles?</h3>

Angles are formed when two lines meet.

Analysis:

Firstly, we calculate for m using cosine rule.

m^{2} = 510^{2} + 820^{2} -2(510)(820) cos 20

m^{2} = 260100 + 672400 -83600(0.9396)

m^{2} = 146618.56

m = \sqrt{146618.56} = 382.9cm

using sine rule to find ∠L

382.9/sin20 = 820/sinL

sinL = 820sin20/382.9

sinL = 820(0.342)/382.9

sinL = 0.732

L = sin inverse of 0.732 = 47°

In conclusion, ∠L is 47°

Learn more about sine and cosine rule: brainly.com/question/4372174

#SPJ1

8 0
2 years ago
In the figure above, the vertices of square
Olin [163]

(C) 6 + 3√3

<u>Explanation:</u>

Area of the square = 3

a X a = 3

a² = 3

a = √3

Therefore, QR, RS, SP, PQ = √3

ΔBAC ≅ ΔBQR

Therefore,

\frac{BQ}{BA} = \frac{QR}{AC}

\frac{BQ}{BA} = \frac{\sqrt{3} }{BA}

In ΔBAC, BA = AC = BC because the triangle is equilateral

So,

BQ = √3

So, BQ, QR, BR = √3 (equilateral triangle)

Let AP and SC be a

So, AQ and RC will be 2a

In ΔAPQ,

(AP)² + (QP)² = (AQ)²

(a)² + (√3)² = (2a)²

a² + 3 = 4a²

3 = 3a²

a = 1

Similarly, in ΔRSC

(SC)² + (RS)² = (RC)²

(a)² + (√3)² = (2a)²

a² + 3 = 4a²

3 = 3a²

a = 1

So, AP and SC = 1

and AQ and RC = 2 X 1 = 2

Therefore, perimeter of the triangle = BQ + QA + AP + PS + SC + RC + BR

Perimeter = √3 + 2 + 1 + √3 + 1 + 2 + √3

Perimeter = 6 + 3√3

Therefore, the perimeter of the triangle is 6 + 3√3

8 0
3 years ago
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