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gulaghasi [49]
3 years ago
14

Given a cylinder with

Mathematics
2 answers:
lara31 [8.8K]3 years ago
3 0
The formula for the volume of a cylinder is

V=πr²*h    where r=radius and h=height
plug in the data we know
V=π8²*2
V=π*64*2
V=128π
V=402.1cm³

Answer=402.1cm³
ale4655 [162]3 years ago
3 0

<span>402.1cm³ is correct!
Thanks for asking 
Have a nice day!
__PS____________________________________________
</span>The formula for the volume of a cylinder is

V=πr²*h  
V=π8²*2
V=π*64*2
V=128π
V=402.1cm³


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The domain and range is?
Molodets [167]

Answer:

In inequality notation:

Domain: -1 ≤ x ≤ 3

Range: -4 ≤ x ≤ 0

In set-builder notation:

Domain: {x | -1 ≤ x ≤ 3 }

Range: {y | -4 ≤ x ≤ 0 }

In interval notation:

Domain: [-1, 3]

Range: [-4, 0]

Step-by-step explanation:

The domain is all the x-values of a relation.

The range is all the y-values of a relation.

In this example, we have an equation of a circle.

To find the domain of a relation, think about all the x-values the relation can be. In this example, the x-values of the relation start at the -1 line and end at the 3 line. The same can be said for the range, for the y-values of the relation start at the -4 line and end at the 0 line.

But what should our notation be? There are three ways to notate domain and range.

Inequality notation is the first notation you learn when dealing with problems like these. You would use an inequality to describe the values of x and y.

In inequality notation:

Domain: -1 ≤ x ≤ 3

Range: -4 ≤ x ≤ 0

Set-builder notation is VERY similar to inequality notation except for the fact that it has brackets and the variable in question.

In set-builder notation:

Domain: {x | -1 ≤ x ≤ 3 }

Range: {y | -4 ≤ x ≤ 0 }

Interval notation is another way of identifying domain and range. It is the idea of using the number lines of the inequalities of the domain and range, just in algebriac form. Note that [ and ] represent ≤ and ≥, while ( and ) represent < and >.

In interval notation:

Domain: [-1, 3]

Range: [-4, 0]

3 0
3 years ago
Please Help!!
goldenfox [79]
Question 5 is B
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5 0
3 years ago
Read 2 more answers
Solve the equation ​
UNO [17]

Answer:

x = 10

Step-by-step explanation:

2x/3 + 1 = 7x/15 + 3

<em><u>(times everything in the equation by 3 to get rid of the first fraction)</u></em>

2x + 3 = 21x/15 + 9

<em><u>(times everything in the equation by 15 to get rid of the second fraction)</u></em>

30x+ 45 = 21x + 135

<em><u>(subtract 21x from 30x; subtract 45 from 135)</u></em>

9x = 90

<em><u>(divide 90 by 9)</u></em>

x = 10

<h2>Another solution:</h2>

2x/3 + 1 = 7x/15 + 3

<u><em>(find the LCM of 3 and 15 = 15)</em></u>

<u><em>(multiply everything in the equation by 15, then simplify)</em></u>

10x + 15 = 7x + 45

<u><em>(subtract 7x from 10x; subtract 15 from 45)</em></u>

3x = 30

<em><u>(divide 30 by 3)</u></em>

x = 10

4 0
3 years ago
What would she have had to pay at the regular ?
vlabodo [156]
Well it depends. what the problem
4 0
3 years ago
Read 2 more answers
G(b) = 5b- 9 h(b) = (b - 1)^2 Evaluate. (hog)(-6) =
laiz [17]

Given:

The two functions are:

g(b)=5b-9

h(b)=(b-1)^2

To find:

The value of (h\circ g)(-6).

Solution:

We have,

g(b)=5b-9

h(b)=(b-1)^2

We know that,

(h\circ g)(b)=h(g(b))

(h\circ g)(b)=h(5b-9)

(h\circ g)(b)=[(5b-9)-1]^2

(h\circ g)(b)=[5b-10]^2

Putting b=-6, we get

(h\circ g)(-6)=[5(-6)-10]^2

(h\circ g)(-6)=[-39-10]^2

(h\circ g)(-6)=[-49]^2

(h\circ g)(-6)=2401

Therefore, the value of (h\circ g)(-6) is 2401.

7 0
3 years ago
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