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Lyrx [107]
3 years ago
13

8ktar=r-2y solve for r

Mathematics
1 answer:
NARA [144]3 years ago
3 0

8ktar=r-2y\ \ \ |-r\\\\8ktar-r=-2y\\\\r(8kta-1)=-2y\ \ \ |:(8kta-1)\neq0\\\\r=\dfrac{-2y}{8kta-1}\to r=\dfrac{2y}{1-8kta}

Example from comment:

8k+ar=r-2y\ \ \ |-8k\\\\ar=r-2y-8k\ \ \ |-r\\\\ar-r=-2y-8k\\\\r(a-1)=-2y-8k\ \ \ |:(a-1)\neq0\\\\r=\dfrac{-2y-8k}{a-1}\to r=\dfrac{-(2y+8k)}{-(1-a)}\to r=\dfrac{2y+8k}{1-a}

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GEOMETRY 10TH GRADE NEED HELP ASAP
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\boxed{(6, -3)}

Step-by-step explanation:

Given the ratio 5:3

and Points A and B where A is located at (-6, 3), and B is located at (26, -13).

Let point A be (x_{1}, y_{1}), and let point B be (x_{2}, y_{2}).

(-6, 3) → (x_{1}, y_{1}).

(26, -13) → (x_{2}, y_{2}).

Let 5 be n, and 3 be m.

5:3 → n:m

(\frac{nx_{1} + mx_{2}}{n+m}, \frac{ny_{1} + my_{2}}{n + m}).

To solve, just substitute these variables into the expressions of these coordinates to get the answer.

(\frac{nx_{1} + mx_{2}}{n+m}, \frac{ny_{1} + my_{2}}{n + m}) →

(\frac{(5)x_{1} + (3)x_{2}}{(5)+(3)}, \frac{(5)y_{1} + (3)y_{2}}{(5) + (3)}) →

(\frac{(5)(-6)+ (3)(26)}{(5)+(3)}, \frac{(5)(3)+ (3)(-13)}{(5) + (3)}) →

(\frac{-30 + 78}{8}, \frac{15+ -39}{8}) →

(\frac{78 – 30}{8}, \frac{15 – 39}{8}) →

(\frac{48}{8}, \frac{-24}{8})

→

(\frac{6}{1}, \frac{-3}{1})

→

(6, -3).

Thus the coordinates of B are:

\boxed{(6, -3)}

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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