If you are proving a congruence of triangles then you need to match the corresponding parts to the triangle you are proving congruent.
BAC would then be congruent to MJR because of the corresponding angles
B is the same angle as M
A is the same angle as J
C is the same angle as R
So g(x) doesn't come in
ok
remember pemdas
inside first
evaluate f(4)
f(4)=4^2+1=16+1=17
now we have
[17]^2=289
answer s 289
X= 1.80/10 . x is the number of dimes. 1.80 is your total and 10 represents the worth of the dime
The dimensions of the rectangular pen should be 15 by 20 feet and the maximum area is 1200 square feet.
Let the area be y .
Area = (base) × (height)
Base = 2x
Height = h
Let the area of the rectangular pens be y .
∴ y = 2xh
Perimeter of all the fencing = 4x+3h
∴ 4x+3h = 120
now we solve for h
3h = 120-4x
h = 40 - 4/3 x
Now we will substitute this value in the above first equation:
y = 2xh
or, y = 2x (40 - 4/3 x)
or, y = 80x - 8/3 x²
Now for the maximum area we have to find the first order differentiation of y
now,
dy /dx = 80 - 16/3 x
At dy/dx = 0 we get the value of x for which y is maximum.
80 - 16/3 x = 0
or, - 16/3 x = -80
or, x = 15 feet
Hence height = 40 - 4/3 x = 40 - 20 = 20feet
Maximum area = 2xh = 2×15×40 = 1200 square feet
The dimensions of the rectangular pen should be 15 by 20 feet and the maximum area is 1200 square feet.
Disclaimer : The missing figure for the question is attached below.
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