A bridge across a river is supported by a semielliptical arch with diameter (length of the major axis) of 1250 feet. The surface of the river lies on the major axis, and the center of the arch rises 175 feet above the river's surface. A horizontal roadway is 3 feet above the center of the arch. Find the vertical distance from the roadway to a point on the arch that is 600 feet (horizontally) from the center.

the answer is f(x)=2x-1

step-by-step explanation:

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

b.my answer is b.

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