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>
Answer:
C.
Step-by-step explanation:
think big it is pretty easy
Remember x is x and f(x) is same thing as y.
f(5) = 4
It is same thing as saying, when x = 5, y aka f(x) should be 4. Aka the point (5,4)
Let’s look at D, if x =5 then f(x)=5? No, it should equal 4. So this is wrong. It is also not linear. So NOT D
Lets look at C, if x =5 then
1/5(5)+3 =
5/5 + 3 =
1+3 =
4
So, f(x) aka y = 4. Is this correct? Yes when x =5, y indeed should equal 4, but it is not linear. So NOT C
Lets look at B, if x=5 then,
2^5 -28 =
32 - 28 =
4
So, f(x) aka y = 4. Is this correct? Yes when x =5, y indeed should equal 4, AND it is linear SO THE ANSWER IS B
Check A too, it gets y=14 so thats wrong as it is not 4.
Answer:
25y + 15x - 0.2y - 6 +(-2) =31.8
Answer:
What does it say in english
Step-by-step explanation:
ill answer it in comments if you tell me what it says in english
