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
Answer:
top left x=70⁰
top right is 69⁰
bottom left is 125⁰
bottom right is 130⁰
Step-by-step explanation:
comment for explanation
1 kilogram (kg) is 1000 grams. So 3,560 grams is
3.56 kilograms
Let us assume that the center of the circle is (0,0)
(x-h)² + (y-k)² = r²
x² + y² = 5,625
r² = 5,625
r = √5,625
r = 75
Area of a circle = π * r²
A = 3.14 * (75mi)²
A = 3.14 * 5,625 mi²
A = 17,662.50 mi²