3 years ago

what are the coordinates of the image of point (-1,-2) under a dilation of 3/2 with respect to the origin?

Advanced Placement (AP)

1 answer:

serious [3.7K]3 years ago

7 0

Answer:

Point P '(-1,-2) is the image of point P(-2,3) under translation T

Explanation:

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1.32 x 10^8 X 2.34 10^4

zaharov [31]

Answer:

30888*10^8

Explanation:

concept used

law of exponent

$a^x * a ^y =a^{x+y}$

____________________________

1.32 x 10^8 X 2.34 10^4

multiplying 1.32 and 2.34 we have

1.32*2.34 = 3.0888

10^8 X 10^4 = 10^(8+4) = 10^12

thus,

1.32 x 10^8 X 2.34 10^4 = 3.0888*10^12

3.0888 can be written as 30888/10000

1.32 x 10^8 X 2.34 10^4 = 30888/10000*10^12

10000 = 10^4

1.32 x 10^8 X 2.34 10^4 = (30888/10^4)*10^12

concept used now

$a^x / a ^y =a^{x-y}$

1.32 x 10^8 X 2.34 10^4 = 30888*10^(12-4)

1.32 x 10^8 X 2.34 10^4 = 30888*10^8

thus, answer is 30888*10^8

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