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Oksanka [162]
3 years ago
7

Please help I will mark brainlyest​

Mathematics
1 answer:
IgorLugansk [536]3 years ago
3 0

Answer:

C.  x^5/2 + 6x^3/2 + 1.

Step-by-step explanation:

(x^3 + 6x^2 + x^1/2) / x^1/2

= x^(3 - 1/2) + 6 x(2-1/2) + 1

= x^5/2 + 6x^3/2 + 1.

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which of the following gives an equation of a line that passes through the point (6 over 5, -19 over 5) and is parallel to the l
victus00 [196]
One equation for this would be

y = \frac{41}{16} x-\frac{55}{8}

We start by finding the slope between the two points:

m=\frac{y_2-y_1}{x_2-x_1}=\frac{-12-\frac{-19}{5}}{-2-\frac{6}{5}}
\\
\\=(-12+\frac{19}{5}) \div (-2-\frac{6}{5})
\\
\\=(\frac{-60}{5}+\frac{19}{5}) \div (\frac{-10}{5}-\frac{6}{5})
\\
\\=\frac{-41}{5} \div \frac{-16}{5}=\frac{-41}{5} \times \frac{-5}{16}=\frac{41}{16}

A line parallel to this one will have the same slope.  We will use point-slope form to write our equation:

y-y_1=m(x-x_1)
\\
\\y-\frac{-19}{5}=\frac{41}{16}(x-\frac{6}{5})
\\
\\y+\frac{19}{5}=\frac{41}{16}x- \frac{41}{16} \times \frac{6}{5}
\\
\\y+\frac{19}{5}=\frac{41}{16}x-\frac{246}{80}
\\
\\y+\frac{304}{80}=\frac{41}{16}x-\frac{246}{80}
\\
\\y=\frac{41}{16}x-\frac{246}{80}-\frac{304}{80}
\\
\\y=\frac{41}{16}x-\frac{550}{80}
\\
\\y=\frac{41}{16}x-\frac{55}{8}
6 0
3 years ago
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