<span>1)What is f(3) if f(x) = -5x3 + 6x2 - x - 4?
a. -74
b. -88
c. 74
d. 182
f(3) = -5(3)^3 + 6(3)^2 - 3 - 4
f(3) = -5(27) + 6(9) - 7
f(3) = -135 + 54 - 7 = -88
(b.)
2)What is f(x + 1) if f(x) = 6x3 - 3x2 + 4x - 9?
a. 6x3 + 12x2 + 4x + 2
b. 6x3 + 3x2 + 8x + 6
c. 6x3 + 21x2 + 20x + 4
d. 6x3 + 15x2 + 16x - 2
f(x + 1) = 6(x + 1)^3 - 3(x + 1)^2 + 4(x + 1) - 9
f(x + 1) = 6(x^3 + 3x^2 + 3x + 1) - 3(x^2 + 2x + 1) + 4x + 4 - 9
f(x + 1) = 6x^3 + 18x^2 + 18x + 6 - 3x^2 - 6x - 3 + 4x + 4 - 9
f(x + 1) = 6x^3 + 15x^2 + 16x - 2
(d.)
3)What is 3[f(x + 2)] if f(x) = x3 + 2x2 - 4?
a. x3 + 8x2 + 20x + 12
b. 3x3 + 12x2 + 18x + 6
c. 3x3 + 24x2 + 60x + 36
d. 3x3 + 18x2 + 24x + 60
f(x + 2) = (x + 2)^3 + 2(x + 2)^2 - 4
f(x + 2) = x^3 + 6x^2 + 12x + 8 + 2x^2 + 8x + 8 - 4
f(x + 2) = x^3 + 8x^2 + 20x + 12
3[f(x + 2)] = 3x^3 + 24x^2 + 60x + 36
(c.)
4)Use synthetic division to determine which of the following is a factor of x3 - 3x2 - 10x + 24.
a. x - 2
b. x - 3
c. x + 4
d. x + 8
2|....1....-3....-10....24
.......1.....-1.....-12....0
(x - 2) works .... (a.)
5)Use synthetic division to determine which of the following is a factor of 2x3 - 13x2 + 17x + 12.
a. x - 2
b. x - 3
c. x + 4
d. x + 6
3|....2....-13....17....12
.......2.....-7.....-4....0
(x - 3) is a factor .... (b.)
6)What is the remainder when (6x3 + 9x2 - 6x + 2) ÷ (x + 2)?
a. -4
b. 0
c. 2
d. 74
-2|....6....9....-6....2
..........6.....-3.....0....2
(c.)
7)What is the remainder when (x3 - x2 - 5x - 3) ÷ (x + 1)?
a. -8
b. 0
c. 2
d. 4
-1|....1....-1....-5....-3
.........1.....-2.....-3....0
(b.)
8)What are the factors of x3 + 2x2 - x - 2?
a. (x - 1)(x + 1)(x - 2) = (x^2 - 1)(x - 2) = x^3 - 2x^2 - x + 2
b. (x - 2)(x + 2)(x - 1)
c. (x - 2)(x + 2)(x + 1)
d. (x - 1)(x + 1)(x + 2) = (x^2 - 1)(x + 2) = x^3 + 2x^2 - x - 2
(d.)
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Answer:
We need the picture.
Step-by-step explanation:
Unless the data is skewed (which means there is outliers) then you use mean as the most accurate. If you have a skewed graph or outliers then mean would mislead you to believe prices are higher (if you have high outliers) or low (if you have low outliers). Median is typically very accurate. Mode may be inaccurate if you have a lot of prices in a really high or low value.
Um please continue the question, thanks!
Answer:
Step-by-step explanation:
As you can see from the graph I attached you, the possible solutions in the interval from 0 to 2π are approximately:
So, it's useful to solve the equation too, in order to verify the result:
Taking the inverse sine of both sides:
Using this result we can conclude the solutions in the interval from 0 to 2π are approximately:
GED is 43
CEB and AED is 90
GEB and FEA is 47
hope this helps!
