Answer:
x^5
Step-by-step explanation:
Because the area of a square can be found by squaring one of its side lengths, we can set up this equation:
(x)^2 = x^10
Using properties of exponents, I know that I can turn the exponent of a similar base from 2 to 10 by multiplying it by 5. Therefore, I know that the side length of this square is x^5.
Mooooooon 83 zd.............
Answer:
A) (-8, -16)
B) (0, 48)
C) (-4, 0), (-12, 0)
Step-by-step explanation:
A) the vertex is the minimum y value.
extremes of a function we get by using the first derivation and solving it for y' = 0.
y = x² + 16x + 48
y' = 2x + 16 = 0
2x = -16
x = -8
so, the vertex is at x=-8.
the y value is (-8)² + 16(-8) + 48 = 64 - 128 + 48 = -16
B) is totally simple. it is f(0) or x=0. so, y is 48.
C) is the solution of the equation for y = 0.
the solution for such a quadratic equation is
x = (-b ± sqrt(b² - 4ac)) / (2a)
in our case here
a=1
b=16
c=48
x = (-16 ± sqrt(16² - 4×48)) / 2 = (-16 ± sqrt(256-192)) / 2 =
= (-16 ± sqrt(64)) / 2 = (-16 ± 8) / 2 = (-8 ± 4)
x1 = -8 + 4 = -4
x2 = -8 - 4 = -12
so the x- intercepts are (-4, 0), (-12, 0)
Subsitute 2 for x
f(2)=2(2)²+5√(2-2)
f(2)=2(4)+5√0
f(2)=8+5(0)
f(2)=8
