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kumpel [21]
3 years ago
9

100 POINT QUESTION!!! Please help!

Mathematics
2 answers:
jarptica [38.1K]3 years ago
6 0

Answer:

1) 2.5x³-3.25x²+3.5x-1.5

2) -2x⁶+7x⁴+3x³-3x²+11x+20

Step-by-step explanation:

1) (½x-¼)(5x²-2x+6)

2.5x³-x²+3x-2.25x²+0.5x-1.5

2.5x³-3.25x²+3.5x-1.5

2) (-2x³+x-5)(x³-3x-4)

-2x⁶+6x⁴+8x³+x⁴-3x²-4x-5x³+15x+20

-2x⁶+7x⁴+3x³-3x²+11x+20

Aleksandr-060686 [28]3 years ago
5 0

Answer: Quest 1 final result. (I looked all this up)

(2x -1) * (5x squared - 2x + 6)

_____________________.

4

I could not find the work for part two but I got the answer: x=3,7

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4 0
1 year ago
Which polynomial function has a leading coefficient of 1 and roots (7 i) and (5 – i) with multiplicity 1? f(x) = (x 7)(x – i)(x
Nuetrik [128]

It looks like you might have intended to say the roots are 7 + i and 5 - i, judging by the extra space between 7 and i.

The simplest polynomial with these characteristics would be

f(x) = (x - (7 + i)) (x - (5 - i))

but seeing as each of the options appears to be a quartic polynomial, I suspect f(x) is also supposed to have only real coefficients. In that case, we need to pair up any complex root with its conjugate to "complete" f(x). We end up with

f(x) = (x - (7 + i)) (x - (7 - i)) (x - (5 - i)) (x - (5 + i))

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5 0
2 years ago
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Answer:

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Step-by-step explanation:

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3 years ago
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Uhh, answer choices?

In any case, since \triangle KJL is inscribed in a semicircle, it is a right triangle (due to the inscribed angle theorem). Thus, by the Pythagorean theorem, KL=\sqrt{6^2+11^2}=\sqrt{157}. Now, the circumradius of a right triangle (the radius of the circle passing through all three of its vertices) is simply half its hypotenuse, so OJ=OK=OL=\frac{\sqrt{157}}{2}\approx \boxed{6.265}.

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3 years ago
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