Two clocks are circular. clock a has twice the area clock b. how many times greater is the radius of clock a than the clock b radius?

The radius of the circle A is times greater than the radius of circle B

Step-by-step explanation:

step 1

Find the scale factor

we know that

If two figures are similar, then the ratio of its areas is equal to the scale factor squared

Let

z-----> the scale factor

x----> the area of the clock A

y----> the area of the clock B

-----> equation A

we have

-----> equation B

substitute equation B in equation A

square root both sides

step 2

Find how many times greater is the radius of Clock A than the Clock B radius

we know that

If two figures are similar, then the ratio of its corresponding sides is equal to the scale factor

Let

z-----> the scale factor

x----> the radius of the clock A

y----> the radius of the clock B

we have

substitute

therefore

The radius of the circle A is times greater than the radius of circle B

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.