<u>ANSWER TO PART A</u>
The given triangle has vertices 
The mapping for rotation through 
counterclockwise has the mapping
Therefore
We plot all this point and connect them with straight lines.
ANSWER TO PART B
For a reflection across the y-axis we negate the x coordinates.
The mapping is
Therefore
We plot all this point and connect them with straight lines.
See graph in attachment
5/6(x-1)=4
(x-1)=(4*6)/5
x-1=24/5
x=24/5 + 1
least common multiple=5
x=(24+1*5)/5
x=29/5
Answer: x=29/5
To check:
5/6(x-1)=5/6(29/5 -1)=5/6[(29-5)/5]=5/6(24/5)=(5*24)/(6*5)=120/30=4
Answer:
1170450 yd^2
Step-by-step explanation:
The first thing is to calculate the necessary perimeter, which would be like this:
2 * a + b = 3060
if we solve for b, we are left with:
b = 3060-2 * a
Now for the area it would be:
A = a * b = a * (3060-2 * a
)
A = 3060 * a -2 * a ^ 2
To maximize the area, we calculate the derivative with respect to "a":
dA / da = d [3060 * a -2 * a ^ 2
]/gives
dA / day = 3060 - 4 * a
If we equal 0:
0 = 3060 - 4 * a
4 * a = 3060
a = 3060/4
a = 765 and d
Therefore b:
b = 3060 - 2 * a = 3060 - 1530 = 1530
A = a * b
A = 765 * 1530
A = 1170450 and d ^ 2
F(2)= 4(16) - 6(4) + 8(2) - 15
F(2)= 64 -24 + 16 - 15
F(2)= 41
