The zeroes of <span>f(x) = x^5-12x^2+32x can be found by factoring,
</span><span>f(x) = x^5-12x^2+32x=(x-8)(x-4)
By the zero product theorem, (x-8)=0 or (x-4)=0 which means
x=8 or x=4.
So the zeroes of f(x) are S={4,8}</span>
Ok so any number tat makes the denomenator 0 or makes the inside of a square root negative is restricted
we only have a denomenaor so
100v=0
v=0
therefor 0 is the excluded value since 0/0 doesn't make sense
Answer:
c
Step-by-step explanation:
1 + tan²theta = sec²theta
tan²theta = 3² - 1
tan²theta = 8
tan theta = sqrt(8)
Positive because Quadrant 1
sqrt(8) = sqrt(4×2) = sqrt(4)×sqrt(2)
= 2×sqrt(2)
Answer:
G
Step-by-step explanation:
3.6÷-0.5 is -7.2, so it's G
