It's a linear program; the extrema will be at the corners.
Let's enumerate them. 4 choose 2 gives 6 possible meets of 4 lines. One pair are parallel, down to five to try.
4x-y=1 intersects x=0 at y=-1, outside the domain y≥0.
4x-y=1 intersects y=0 at x=1/4, (1/4, 0)
4x-y=1 intersects x=5 at y=19, (5,19)
(0,0) is outside the domain 4(0)-0=0 which isn't ≥1.
(5,0) is a valid corner.
It's a triangular domain. Three points to try,
C(1/4,0) = 6(1/4) + 2(0) = 3/2
C(5,19) = 6(5) + 2(19) = 68
C(5,0) = 6(5) + 2(0) = 30
Answer: Maximum C=68 at (x,y)=(5,19)
Answer:
the sine of the angle of one corner equals the opposite side divided by the hypotenuse.
the cosine of the angle of one corner equals the adjacent side divided by the hypotenuse.
the tangent of the angle of one corner equals the opposite side divided by the adjacent side.
So first we divide 140 by 7 because our ratio is a total of 7 dollars to get 20
So we multiply both sides of the ratio by 20 to get a ratio of 40:100
We can check our answer and it is right because 100+40=140
In the problems A-F, You multiply each of the numbers by itself. Like 9²=9×9=81
3³=3×3×3=27 6³=6×6×6=216 and so on for the other problems
G-I is 36=6² 100=10² and so on the cube is the same except you multiply the number three times by itself. I hope this helped a bit. Just ask if you still don't understand.
Hack
for y=ax^2+bx+c
the xvalue of the vertex is -b/2a
and the y value is found by subsituting that for x
so
-b/2a
h(t)=2x^2+4x+7
a=2
b=4
-b/2a=-4/(2*2)=-4/4=-1
the y value
h(-1)=2(-1)^2+4(-1)+7
h(-1)=2(1)-4+7
h(1)=2+3
h(1)=5
(-1,5) is da vertex
