Both terms are divisible by 16 or -16, depends on how you want the final factored product to turn out
factor out a 16
79 because you just have to carry a few number over
F(x)=-5(x-4)^2+80
f(x)=-5(x²-8x+16)+80
f(x)=-5x²+40x-80+80
f(x)=-5x²+40x
Replace f(x) to y, now switch the location of the variables x and y. Solve for new y.
y=-5x²+40x
x=-5y²+40y
0=-5y²+40y-x
Use quadratic formula to find the value of y.
y=-40+-√(40^2 -4(-5)(-x)/2(-5)
f^-1(x)= (-40+-√1600-20x)/-10 <--- Not too sure about this one :/
Hope I helped :)
(3.75, 5.5)
(Work is attached )
When using substitution , you substitute with a variable that is already by itself. In this case, the variable “y” is already by itself and equal to x+ 1.75 . you substitute that for the the y in the other equation and solve for the other variable which is “x”. After you solve for x , to find your y value, you have to substitute the x value in for either equation.
The triangle inequality applies.
In order for ACD to be a triangle, the length of AC must lie between CD-DA=0 and CD+DA=8.
In order for ABD to be a triangle, the length of AC must lie between BC-AB=3 and BC+AB=9.
The values common to both these restrictions are numbers between 3 and 8. Assuming we don't want the diagonal to be coincident with any sides, its integer length will be one of ...
{4, 5, 6, 7}