Can someone help me please this is due tomorrow
1 answer:
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Answer:
g(x) = 1/2*(4)^(–x) and
g(x) =1/2*(1/4)^(x)
Please, see attached picture.
Step-by-step explanation:
Your full question is attached in the picture below
To easily solve this problem, we can graph each equation and see, which one represents a reflection of the function over the y axis.
See, second image.
The answers are
g(x) = 1/2*(4)^(–x) and
g(x) =1/2*(1/4)^(x)
Answer:
FD≈25.94.. rounded = 26
Step-by-step explanation:
FD²=12²+(4x+11)²
FD²=144+16x²+88x+121
FD²=265+16x²+88x
also
FD²=12²+(13x-16)²
FD²=144+169x²-416x+256
FD²=400+169x²-416x
thus
265+16x²+88x = 400+169x²-416x
16x²-169x²+88x+416x+265-400 = 0
-153x²+504x-135 = 0
we will solve this quadratic equation by suing the quadratic formula to find x
x=(-504±sqrt(504²-4(-153)(-135)))/2(-153)
x=(-504±)/2(-153)
x=(-504±)/-306
x=(-504±)/-306
x=(-504±414)/-306
x=(-504+414)/-306 and x=(-504-414)/-306
x=-90/-306 and x=-918/-306
x= 5/17 , 3
substituting x by the roots we found
check for 5/17:
4x+11 = 4×(5/17)+11 = (20/17)+11 = (20+187)/17 = 207/17 ≈ 12.17..
13x-16 = 13×(5/17)-16 = (65/17)-16 = (65-272)/17 = -207/17 ≈ -12.17..
check for 3:
4x+11 = 4×3+11 = 12+11 = 23
13x-16 = 13×3-16 = 23
thus 3 is the right root
therfore
ED=23 and CD=23
FD²=FE²+ED² or FD²=FC²+CD²
FD²=12²+23²
FD²=144+529
FD²=673
FD=√673
FD≈25.94.. rounded = 26
The co ordinates of P' is (-7,-2) and Q' is (-16, -8)
<u><em>Explanation</em></u>
PQ is rotated 180 degrees clockwise about P. It means <u>P and P' are the same points</u>.
According to the graph, the coordinates of P is (-7, -2) and Q is (2, 4)
When PQ is rotated 180 degrees clockwise about P, then <u>P or P' will be the mid-point of Q and Q' </u>
Suppose, the co ordinate of Q' is (x, y)
Now according to the mid-point formula, the coordinate of P or P' will be: , which is actually at (-7, -2)
Thus.....
So, the co ordinates of P' is (-7,-2) and Q' is (-16, -8)