A parabola is a quadratic function, and a quadratic can be expressed in vertex form, which is:
y=a(x-h)^2+k, where (h,k) is the vertex (absolute maximum or minimum point of the quadratic)
In this case we are given that (h,k) is (-5,80) so we have so far:
y=a(x--5)^2+80
y=a(x+5)^2+80, we are also told that it passes through the point (0,-45) so:
-45=a(0+5)^2+80
-45=25a+80 subtract 80 from both sides
-125=25a divide both sides by 25
-5=a, so now we know the complete vertex form is:
y=-5(x+5)^2+80
The x-intercepts occur when y=0 so:
0=-5(x+5)^2+80 add 5(x+5)^2 to both sides
5(x+5)^2=80 divide both sides by 5
(x+5)^2=16 take the square root of both sides
x+5=±√16 which is
x+5=±4 subtract 5 from both sides
x=-5±4 so the x-intercepts are:
x=-1 and -9
Answer:
61
Step-by-step explanation:
we kno that a triangle is 180 degrees so we just add the first two angles
37 + 82 = 119
then subtract that from 180
180 - 119 = 61
so x = 61 degrees
lmk if this wrong and ill fix it
hope this helps <3
One pair is 4.345+4.345 (which is the pair that does not involve grouping)
so hmmm then we know the slope of that line is -2/3, so we're really looking for the point-slope form of a line with a slope of -2/3 and that passes through (-3 , 8)
