The slope is 2. y2-y1/x2-x1
We either need to see a picture of this and/or get more information about the measurements of the triangle. In general, the area outside of the triangle will be the area of the semi-circle minus the area of the triangle itself, or: 1/2*49*3.14 - 1/2 b*h of the triangle. That first part, which is the area of the semi-circle, works out to 76.93 So based on the info we have, it becomes 76.93 - 1/2*b*h of the triangle = area outside of triangle.
Step-by-step explanation:
It seems here that they are asking us to solve for x
to do this we first need to factor
Since we can't factor this using the normal method we can instead do this
x^2 -4x-17=0
add 4 to both sides as it is a perfect square
x^2 - 4x + 4 = 17 +4
(x-2)^2 = sqrt 21
x-2 = ± 4.58
x -2 = 4.58 x-2 = -4.58
x= 6.58 x=-2.58
Or just say x=2+√21 or x=2−√21
