Answer:
33.195%
Step-by-step explanation:
AMDM: 221 students
Physics: 241 students
Taking both: 80 students
80/241 = ~33.195%
Answer: No, x+3 is not a factor of 2x^2-2x-12
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Explanation:
Let p(x) = 2x^2 - 2x - 12
If we divide p(x) over (x-k), then the remainder is p(k). I'm using the remainder theorem. A special case of the remainder theorem is that if p(k) = 0, then x-k is a factor of p(x).
Compare x+3 = x-(-3) to x-k to find that k = -3.
Plug x = -3 into the function
p(x) = 2x^2 - 2x - 12
p(-3) = 2(-3)^2 - 2(-3) - 12
p(-3) = 12
We don't get 0 as a result so x+3 is not a factor of p(x) = 2x^2 - 2x - 12
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Let's see what happens when we factor p(x)
2x^2 - 2x - 12
2(x^2 - x - 6)
2(x - 3)(x + 2)
The factors here are 2, x-3 and x+2
There are 16 ounces in a pound. If there are 4 pounds, then you take 16 and multiply it by 4 and then add 6.
The answer is: 70.
Answer:
The radical notation is ![3x\sqrt[3]{y^2z}](https://tex.z-dn.net/?f=3x%5Csqrt%5B3%5D%7By%5E2z%7D)
Step-by-step explanation:
Given
![\sqrt[3]{27 x^{3} y^{2} z}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B27%20x%5E%7B3%7D%20y%5E%7B2%7D%20z%7D)
Step 1 of 1
Write the expression using rational exponents.
![\sqrt[n]{a^{m}}=\left(a^{m}\right)^{\frac{1}{n}}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5E%7Bm%7D%7D%3D%5Cleft%28a%5E%7Bm%7D%5Cright%29%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D)
Simplify 
![$=3 x \sqrt[3]{y^{2} z}$](https://tex.z-dn.net/?f=%24%3D3%20x%20%5Csqrt%5B3%5D%7By%5E%7B2%7D%20z%7D%24)
Learn more about radical notation, refer :
brainly.com/question/15678734
