The possible zeros of f(x) = 3x^6 + 4x^3 -2x^2 + 4 are 
<h3>How to determine the possible zeros?</h3>
The function is given as:
f(x) = 3x^6 + 4x^3 -2x^2 + 4
The leading coefficient of the function is:
p = 3
The constant term is
q = 4
Take the factors of the above terms
p = 1 and 3
q = 1, 2 and 4
The possible zeros are then calculated as:
So, we have:
Expand
Solve
Hence, the possible zeros of f(x) = 3x^6 + 4x^3 -2x^2 + 4 are
Read more about rational root theorem at:
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Answer:
I just learned this topic and have the same question, so I'm not sure if this right. However, I see a fellow student struggling and have to at least put in what I know. My equation was f(n) = 64(1/2)^n-1, then the question was write an expression to represent f(20).
Step-by-step explanation:
So, my answer was f(20) = 64(1/2)^19
f(20) is the number you plug into n. And because the exponent is n-1, I knew to subtract 1 from 20 to get 19.
Then I just wrote the equation out like the original expect with the numbers pluged in: f(20) = 64(1/2)^19
I hope this helps!
Answer:
TRIANGLES
Step-by-step explanation:
Answer:
Step-by-step explanation:
There are 6 ways we can roll doubles out of a possible 36 rolls (6 x 6), for a probability of 6/36, or 1/6, on any roll of two fair dice. So you have a 16.7% probability of rolling doubles with 2 fair six-sided dice.
