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4 years ago

8

Kyla took her bike to a path that is 2 and 3/4 miles in length. Kyla completed 3 and 1/2 laps before she had to take a break. Ho

w many miles did she ride before she took a break?

1 answer:

worty [1.4K]4 years ago

4 0

She Took 5 and 3/8 Miles

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-3x⁵+11x⁴+33x³-26x²-36x-6 / x³+6x²-3x-5

olganol [36]

Answer:

$-3x^2+29x-150+\dfrac{946x^2-341x-756}{x^3+6x^2-3x-5}$

Step-by-step explanation:

You have to divide the polynomial $-3x^5+11x^4+33x^3-26x^2-36x-6$ by the polynomial $x^3+6x^2-3x-5:$

First, multiply the polynomial $x^3+6x^2-3x-5$ by $-3x^2$ and subtract the result from the polynomial $-3x^5+11x^4+33x^3-26x^2-36x-6:$

$-3x^5+11x^4+33x^3-26x^2-36x-6-(-3x^2)(x^3+6x^2-3x-5)\\ \\=-3x^5+11x^4+33x^3-26x^2-36x-6+3x^5+18x^4-9x^3-15x^2\\ \\=29x^4+24x^3-41x^2-36x-6$

Now, multiply the polynomial $x^3+6x^2-3x-5$ by $29x$ and subtract the result from the polynomial $29x^4+24x^3-41x^2-36x-6:$

$29x^4+24x^3-41x^2-36x-6-29x(x^3+6x^2-3x-5)\\ \\=29x^4+24x^3-41x^2-36x-6-29x^4-174x^3+87x^2+145x\\ \\=-150x^3+46x^2+109x-6$

At last, multiply the polynomial $x^3+6x^2-3x-5$ by $-150$ and subtract the result from the polynomial $-150x^3+46x^2-181x-6:$

$-150x^3+46x^2+109x-6-(-150)(x^3+6x^2-3x-5)\\ \\=-150x^3+46x^2+109x-6+150x^3+900x^2-450x-750\\ \\=946x^2-341x-756$

So,

$\dfrac{-3x^5+11x^4+33x^3-26x^2-36x-6}{x^3+6x^2-3x-5}=\\ \\=\dfrac{(-3x^2+29x-150)(x^3+6x^2-3x-5)+946x^2-341x-756}{x^3+6x^2-3x-5}=\\ \\=-3x^2+29x-150+\dfrac{946x^2-341x-756}{x^3+6x^2-3x-5}$

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3 years ago

Help with the second question. I'm stuck.

IRISSAK [1]

(1/12) / 10 = 1/x.....1/12 distance to 10 min = whole distance(12/12) or 1 to x min
cross multiply
(1/12)(x) = (10)(1)
1/12x = 10
x = 10 / (1/12)
x = 10 * 12
x = 120 minutes or 2 hrs

I am not sure if u do it this way...but this will get u the answer

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Step-by-step explanation:

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