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jeyben [28]
3 years ago
9

What is the lateral area of a right cylinder if the radius of the base is 12 mm and the altitude of the cylinder is 5 times the

base radius? Round your answer to the nearest whole number.
A. 2715 mm2
B. 905 mm2
C. 3167 mm2
D. 4524 mm2
Mathematics
1 answer:
lakkis [162]3 years ago
5 0
The lateral area of a cylinder is given by:
Area=πrl+2πr^2
radius,r=12 mm
length,l=5*12=60mm
therefore the lateral area will be:
Area=π*12*60
Area=2,261.95 mm^2
The area of the bases will be:
A=2*π*12^2=904.78 mm^s
The lateral area will be:
2,261.95+904.78
=3,166.73
=3167 mm^2
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Which graph below represents a functional relationship? A. 10 10 8 8 6 6 4 2 2 X 10864 6 8 10 10 8 6 4 4 6 8 10 -21 4 -61 -8 -8
Anestetic [448]

Answer:

Graph (C)

Step-by-step explanation:

To find whether the graph\table represents a relationship or a function we have to analyze the input-output values given.

Graph A.

In this graph for every input value (x-value) there are two output values (y-values).

For x = -2, y = -2, 2

So the graph doesn't represent a function.

Graph B.

For every x value there are two y-values

For x = -5, y = -3, 3

So the graph doesn't represent a function.

Graph C.

For every input value there is a different y-value.

Therefore, graph represents a function.

Graph D.

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4 0
3 years ago
Please help me <br> Show your work <br> 10 points
Svet_ta [14]
<h2>Answer</h2>

After the dilation \frac{5}{3} around the center of dilation (2, -2), our triangle will have coordinates:

R'=(2,3)

S'=(2,-2)

T'=(-3,-2)

<h2>Explanation</h2>

First, we are going to translate the center of dilation to the origin. Since the center of dilation is (2, -2) we need to move two units to the left (-2) and two units up (2) to get to the origin. Therefore, our first partial rule will be:

(x,y)→(x-2, y+2)

Next, we are going to perform our dilation, so we are going to multiply our resulting point by the dilation factor \frac{5}{3}. Therefore our second partial rule will be:

(x,y)→\frac{5}{3} (x-2,y+2)

(x,y)→(\frac{5}{3} x-\frac{10}{3} ,\frac{5}{3} y+\frac{10}{3} )

Now, the only thing left to create our actual rule is going back from the origin to the original center of dilation, so we need to move two units to the right (2) and two units down (-2)

(x,y)→(\frac{5}{3} x-\frac{10}{3}+2,\frac{5}{3} y+\frac{10}{3}-2)

(x,y)→(\frac{5}{3} x-\frac{4}{3} ,\frac{5}{3}y+ \frac{4}{3})

Now that we have our rule, we just need to apply it to each point of our triangle to perform the required dilation:

R=(2,1)

R'=(\frac{5}{3} x-\frac{4}{3} ,\frac{5}{3}y+ \frac{4}{3})

R'=(\frac{5}{3} (2)-\frac{4}{3} ,\frac{5}{3}(1)+ \frac{4}{3})

R'=(\frac{10}{3} -\frac{4}{3} ,\frac{5}{3}+ \frac{4}{3})

R'=(2,3)

S=(2,-2)

S'=(\frac{5}{3} (2)-\frac{4}{3} ,\frac{5}{3}(-2)+ \frac{4}{3})

S'=(\frac{10}{3} -\frac{4}{3} ,-\frac{10}{3}+ \frac{4}{3})

S'=(2,-2)

T=(-1,-2)

T'=(\frac{5}{3} (-1)-\frac{4}{3} ,\frac{5}{3}(-2)+ \frac{4}{3})

T'=(-\frac{5}{3} -\frac{4}{3} ,-\frac{10}{3}+ \frac{4}{3})

T'=(-3,-2)

Now we can finally draw our triangle:

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