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-Dominant- [34]
3 years ago
9

The volume, V, of a cylinder is V= \pi^2h, where r is the radius of the cylinder and h is the height. Using rounding to the near

est whole number,which of the following is an estimate of the volume of a cylinder with radius of 2.75 inches and height of 5.21 inches?
A. 45 in.^3
B. 90 in.^3
C. 135 in.^3
D. 216 in.^3
Mathematics
1 answer:
mylen [45]3 years ago
7 0

To solve this problem you must apply the proccedure shown below:

You have that the radius of the cylinder is 2.75 inches and the height is 5.21 inches. Therefore, you only need to substitute these values into the formula for calculate the volume of a cylinder:

V=\pi r^{2}h\\V=\pi(2.75in)^{2}(5.21in)\\V=123.78in^{3}=124in^{3}

Therefore, as you can see, the answer is: The volume of the cylinder is 124 in³.

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Elenna [48]

Answer:

Circumference= 69.08

Area = 380.13

Step-by-step explanation:

4 0
3 years ago
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Find the value of YVU
tigry1 [53]
XV = 180
180 - 85 = 95
3 0
3 years ago
Five times the sum of a number and 27 is greater than or equal to six times the sum of that
AleksAgata [21]

The complete version of question:

<em>Five times the sum of a number and 27 is greater than or equal to six times the sum of that number and 26. What is the solution of this problem.</em>

Answer:

5\left(x+27\right)\ge \:6\left(x+26\right)\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x\le \:-21\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-21]\end{bmatrix}

Step-by-step explanation:

As the description of the statement is:

'<em>Five times the sum of a number and 27 is greater than or equal to six times the sum of that number and 26'.</em>

<em />

As

  • <em>Five times the sum of a number and 27  </em>is written as: 5(x + 27)
  • <em>greater than or equal </em>is written as:  \geq
  • <em>six times the sum of that number and 26'  </em>is written as: 6(x + 26)

so lets combine the whole statement:

<em />5\left(x\:+\:27\right)\ge \:6\left(x\:+\:26\right)<em />

solving

<em />5\left(x\:+\:27\right)\ge \:6\left(x\:+\:26\right)<em />

<em />5x+135\ge \:6x+156<em />

<em />5x+135-135\ge \:6x+156-135<em />

<em />5x\ge \:6x+21<em />

<em />5x-6x\ge \:6x+21-6x<em />

<em />-x\ge \:21<em />

<em />\mathrm{Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)}<em />

<em />\left(-x\right)\left(-1\right)\le \:21\left(-1\right)<em />

<em />x\le \:-21<em />

Therefore,

5\left(x+27\right)\ge \:6\left(x+26\right)\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x\le \:-21\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-21]\end{bmatrix}

6 0
3 years ago
For which inequality is x = 5 a solution?<br>A x&lt;2<br>B. x&gt;9 <br>C x&lt;18 <br>D. x&gt; 42​
marishachu [46]

Answer:

c

Step-by-step explanation:

because 5 is smaller then 18

7 0
3 years ago
A bus driver pays all tolls, using only nickels and dimes, by throwing one coin at a time into the mechanical toll collector.
lutik1710 [3]

Answer: a) An = An-1 + An-2

b) 55ways

Step-by-step explanation:

a) a nickel is 5 cents and a dime is 10cent so a multiple of 5 cents is the possible way to pay the tolls in both choices.

Let An represents the number of possible ways the driver can pay a toll of 5n cents, so that

An = 5n cents

Case 1: Using a nickel for payment which is 5 cents, the number of ways given as;

An-1 = 5( n-1)

Case 2: using a dime which is two 5 cents, the number of ways is given as;

An-2 = 5(n-2)

Summing up the number of ways, we have

An = An-1 + An-2

From the relation,

If n= 0, Ao= 1

n =1, A1= 1

b) 45 cents paid in multiples of 5cents will give us 9 ways(A9)

From the relation, we have that

Ao = 1

A1 = 1

An =An-1 + An-2

Ao = 1

A1 = 1

A2 = A1+Ao = 1+1= 2

A3 = A2 + A1 = 3

A4 = A3+A2=5

A5=A4+A3=8

A6=A5+A4=13

A7 =A6+A5 = 21

A8= A7+A6= 34

A9= A8+A7= 55

So there are 55ways to pay 45cents.

4 0
3 years ago
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