Rectangle a has length 12 and width 8. rectangle b has length 15 and width 10. rectangle c has length 30 and width 15. is rectangle a a scaled copy of rectangle b? if so, what is the scale factor? is rectangle b a scaled copy of rectangle a? if so, what is the scale factor? explain how you know that rectangle c is not a scaled copy of rectangle b. is rectangle a a scaled copy of rectangle c? if so, what is the scale fact

For two rectangles, one of length L and width W, and other of length L' and width W', the second is a rescale of the first one only if exists a real number k such that:

L' = k*L

W' = k*W

Here we know:

Let's see if rectangle A is a scaled copy of rectangle B.

To see this, we just must see if the quotients between the lengths and between the widths are equal:

15/12 = 1.25

10/8 =  1.25

Then yes, rectangle A is a rescaled copy of rectangle B, and the scale factor is k = 1.25

Is rectangle B a rescaled copy of rectangle A?

Obviously yes. The scale factor will be the inverse of the previous one, we will get:

k = 1/1.25 = 0.8

How we do know that rectangle C is not a scaled copy of rectangle B?

Because the length of C is twice the length of B, but the width of C is not twice the width of B.

Is rectangle A a scaled copy of rectangle C?

No, as we already see that rectangle C is not a rescaled copy of rectangle B, and we know that rectangle A is a rescaled copy of rectangle B.

If you want to learn more you can read:

link

Rectangles are similar figures, thus if scaled copies of each other then the ratios of corresponding sides must be equal

compare ratios of lengths and widths

rectangles A and B

k = = ← ratio of lengths

k = = ← ratio of widths

scale factors are equivalent, hence rectangle A is a scaled copy of B

rectangles C and B

k = = ← ratio of lengths

k = = ← ratio of width

scale factors (k ) are not equal, hence C is not a scaled copy of B

rectangles A and C

k = = ← ratio of lengths

k = ← ratio of widths

the scale factors are not equal hence A is not a scaled copy of C

an advantage of the standard deviation is that it increases as the dispersion of the data increases. e. the interquartile range is preferred when the data are not skewed or no have outliers. an advantage of the standard deviation is that it uses all the observations in its computation.

answer: the price of one pound cabbage = $ 0.8

the point (10,8) represents that the cost of 10 pounds cabbage is $ 8

step-by-step explanation:

since, here is a proportional relation between the number of pounds of cabbage (x) and the price in dollars (y).

therefore, we can write,

⇒ y = k x

where k is the constant of proportionality.

now, point (5,4) is in the line.

thus, it will satisfies the above equation of line.

⇒ 4 = 5 k

⇒ k = 0.8

thus, the equation of the line is,

y = 0.8 x

a) for x = 1,

y = 0.8

therefore, the cost of one pound of cabbage is 0.8 dollars.

b) in the below graph (10,8) represents that the cost of 10 pounds of cabbage is 8 dollars.