Answer:
If line EF was dilated by a scale of 1/2 from the origin to get line E'F', then if line E'F' is dilated by a scale of 2 from the origin, EF is obtained. Also, the length of EF is double than the length of E'F'.
Dilation by a scale of 2 from the origin transforms point (x,y) into (2*x, 2*y)
E'(1,0) -> E(2,0)
F'(1, 3) ->F(2, 6)
length of E'F': 3
length of EF: 6
The dividend is the number being divided so the dividend is 27,438
The divisor is the one dividing the number so the divisor is 9
The quotient is your answer so you quotient is 3048
Your answer is C.
Hope this helps!
3/4 and 1/3
4x3 = 12 that will be the common denominator.
For 3/4 we first divide 12/4 = 3 and then multiply 3 x3 = 9 that's the numerator
For 1/3 we first divide 12/3 = 4 and then multiply 1x4 = 4 that's the numerator
The fractions will be rewrited as: 9/12 and 4/12
Answer:
1. No.

Since f(1)=f(3) and
then f isn't one-to-one.
2. No

Since f(0)=f(2) and
then f isn't one-to-one.
3. No

Since f(0)=f(2) and
then f isn't one-to-one.
4. Yes

Since
, then f is one-to-one
5. Since f(1)=f(3) and
then, f isn't one-to-one
Answer:
the answer for this is B 9