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o-na [289]
3 years ago
11

A and B are similar solid cylinders base area of A : base area of B = 9 : 25

Mathematics
1 answer:
fomenos3 years ago
7 0

Answer:

Two figures are similar if the figures have the same shape but different sizes.

Then if we have two figures X and X'

Such that one dimension of X is D, the correspondent dimension on X' will be:

D' = k*D

Such that k is the scale factor that relates the figures.

1:k

Because all the dimensions will be rescaled by the same scale factor k, we can conclude that any surface on X will be related to the same surface in X' by:

S' = k^2*S

then the ratio of the surfaces is:

1:k^2

While the relation between the volumes will be:

V' = k^3*V

Here the ratio is:

1:k^3

Ok, in this case we have two cylinders

We know that the ratio between the base area ( a surface) is:

9:25

a) We want to find the ratio: curved surface area of A : curved surface area of B

Because again we have a surface area, the ratio should be exactly the same as before, 9:25

b) height of A : height of B

In this case, we have a single dimension.

Because in the rescaling of a surface we need to use k^2, then we can conclude that the ratios:

9:25

is related to k^2

Then the ratio, in this case, is given by applying the square root to both sides of the previous ratio, so we get:

√9:√25

3:5

This is the ratio of the heights.

Also from this we could get the value of k, that is the right value when we leave the left value equal to 1, we can get that if we divide both sides by 3.

(3/3):(5/3)

1:(5/3)

Then:

k = 5/3

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Naoya read a book cover to cover in a single session, at a rate of 555555 pages per hour. After 444 hours, he had 350350350 page
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Answer:

y=-55x+570

Step-by-step explanation:

The correct question is

Naoya read a book cover to cover in a single session, at a rate of 55 pages per hour. After 4 hours, he had 350 pages left to read.

Let y represent the number of pages left to read after x hours.

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Let

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point\ (4,350)

substitute

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8 0
4 years ago
A force of 12 lb is required to hold a spring stretched 8 in. beyond its natural length. How much work W is done in stretching i
sattari [20]

Answer:

Work done in stretching the spring = 7.56 lb-ft.

Step-by-step explanation:

Normal length of the spring = 8 in or \frac{8}{12} ft

= \frac{2}{3} ft

If the spring has been stretched to 11 inch then the stretched length of the spring is = 11 in

= \frac{11}{12} ft

Force applied to stretch the spring = 12 lb

By Hook's law,

F = kx [where k is the spring constant and x = length by which the spring is stretched]

12 = k(\frac{2}{3})

k = 18

Work done (W) to stretch the spring by \frac{11}{12} ft will be

W = \int\limits^\frac{11}{12} _0 {kx} \, dx

    = \int\limits^\frac{11}{12} _0 {(18x)} \, dx

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    = 9(\frac{11}{12})²

    = 7.56 lb-ft

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