What is 3 to the tenth power
1 answer:
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Answer:
Step-by-step explanation:
Given: trapezoid RSTU with vertices R(-1, 5), S(1, 8), T(7, -2), and U(2, 0).
Plot these points on the coordinate plane. As you can see, lines RU and ST are parallel lines and segments ST and RU are bases of the trapezoid.
The median of the trapezoid (the middle line) passes through the midpoints of the sides RS and TU. Find these midpoints:
The equation of the line MN is
Answer:
- a = 1
- b = -28
- P(x) = (x +2)(x -3)(x^2 +3x +10)
Step-by-step explanation:
a) Since x+2 is a factor, we know P(-2) = 0.
P(x) = (((x +2)x +a)x +b)x -60
Then the value of P(-2) is ...
P(-2) = 0 = (((0)(-2) +a)(-2) +b)(-2) -60 = (-2a +b)(-2) -60 = 4a -2b -60
We know the remainder from division by (x+3) is 60, so
P(-3) = 60 = (((-3+2)(-3) +a)(-3) +b)(-3) -60 = ((3+a)(-3) +b)(-3) -60
= (-9 -3a +b)(-3) -60 = 27 +9a -3b -60
93 = 9a -3b
These two equations can be put into standard form:
2a -b = 30
3a -b = 31
Then we have the solution ...
a = 1 . . . . . (by subtracting the first equation from the second)
-28 = b . . . by substituting into the first equation
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b) To show that (x-3) is a factor we need to evaluate P(3).
P(3) = (((3 +2)(3) +1)(3) -28)(3) -60 = (48 -28)(3) -60 = 0
The function value is 0, so (x -3) is a factor.
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c) We want to find Q(x) = x^2 +cx +d such that ...
(x +2)(x -3)Q(x) = P(x)
(x^2 -x -6)(x^2 +cx +d) = x^4 +2x^3 +x^2 -28x -60
x^4 +(c-1)x^3 +(-6-c+d)x^2 +(-6c-d)x -6d = x^4 +2x^3 +x^2 -28x -60
This gives rise to the equations ...
c -1 = 2 ⇒ c = 3
-6d = -60 ⇒ d = 10
Then P(x) can be factored as ...
P(x) = (x +2)(x -3)(x^2 +3x +10)
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<em>Comment on the attached graph</em>
I like to use a graphing calculator to find real roots of higher-degree polynomials. This graph shows the real zeros to be -2 and +3, so we know that (x +2) and (x -3) are factors. The green curve is P(x) with those factors divided out, so is a graph of Q(x). The vertex of that graph tells us that ...
Q(x) = (x +1.5)^2 +7.75 = x^2 +3x +10
