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vazorg [7]
3 years ago
5

The circumference of the base of a cylinder is 24π mm. A similar cylinder has a base with circumference of 60π mm. The lateral a

rea of the larger cylinder is 210π mm2.
What is the lateral area of the smaller cylinder?
Mathematics
2 answers:
labwork [276]3 years ago
8 0

Answer:

33.6π mm2

Step-by-step explanation:

answer is right on edge

earnstyle [38]3 years ago
6 0
Given that the figures are similar polygons, the ratio of their ratios should be equal to the square of the ratio of their circumference. If we let x be the lateral area of the smaller cylinder then,
                        (x/210π) = (24π/60π)²
The value of x from the equation is, 
                                     x = 33.6π
Thus, the area of the smaller cylinder is equal to 33.6π mm². 
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Nonamiya [84]

Answer:

The Answer would be option C. 3/4 books per hour

Step-by-step explanation:

You would flip 1/3 to 3/1 which makes it a whole number. then multiply

1/4 and 3.

Therefore your answer is 3/4

I hope this helped you,

8 0
3 years ago
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ValentinkaMS [17]

Step-by-step explanation:

As seen from the graph, f(x) and g(x) intersect each other when x = 3.

Hence, f(3) = g(3).

3 0
3 years ago
Simplify the expression. 10z - 10z​
nika2105 [10]

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7 0
2 years ago
What are the coordinates of the endpoints of the midsegment for DEF that is parallel to DE
Nutka1998 [239]

Answer:

\left(\dfrac{x_D+x_F}{2},\dfrac{y_D+y_F}{2}\right),  \left(\dfrac{x_E+x_F}{2},\dfrac{y_E+y_F}{2}\right).

Step-by-step explanation:

Let points D, E and F have coordinates (x_D,y_D),\ (x_E,y_E) and (x_F,y_F).

1. Midpoint M of segment DF has coordinates

\left(\dfrac{x_D+x_F}{2},\dfrac{y_D+y_F}{2}\right).

2. Midpoint N of segment EF has coordinates

\left(\dfrac{x_E+x_F}{2},\dfrac{y_E+y_F}{2}\right).

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6 0
3 years ago
Find equations of the line that is parallel to the z-axis and passes through the midpoint between the two points (0, −4, 9) and
Lorico [155]

Answer: x = -4, y = 0.5, z = 5 +t

Hi!

The line L whose direction is parallel to vector V a passes through point A

is parametrized

P_L = Vt + A

Where t, is a real number, and P_L is a any point on line L.

In this case the direction is that of the z-axis , so V = (0, 0, 1)

A is the midpoint between points B = (0, -4, 9) and C=(-8, 5, 1)

The midpoint is A = (B + C)/2 = (-4, 0.5, 5)

Then the line is:

P_L = (0,0,1) t + (-4, 0.5, 5)

The equations for each coordinate are:

x = -4\\y = 0.5\\z = 5 + t

7 0
3 years ago
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