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Alisiya [41]
3 years ago
13

1 Solve: -8–2k=-2(k+4)

Mathematics
1 answer:
Olegator [25]3 years ago
4 0

Answer:

zero=0

Explanation:

-8-2k=-2(k+4)

-8-2k=-2k-8

-2k+2k=-8+8

zero=0

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kiruha [24]

Answer: 975.

A 28% discount makes 702 72% of the original price, so 702/.72 gives 975

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8 0
2 years ago
Please need help on this not too sure on this
kifflom [539]



option 3

they are corresponding angles and are congruent

8 0
3 years ago
4/5 divided by -3/5 any answers because right now I’m really stuck
MatroZZZ [7]

Answer:

-1 1/3 as a mixed number (if they ask for it in simplest form, choose this one)

-4/3 as an improper fraction

Step-by-step explanation:

1. Keep, change, flip

4/5 x -5/3

2. Cross cancel the fives

4/1 x -1/3

3. Simplify

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7 0
3 years ago
Please help I’ll give you brainlist or something
harkovskaia [24]

Answer:

1 slope of f is 5

slop of g is 2

2. connor

initial value is 200

rate of change is -10

Pilar

initial value is 242

rate of change is -8

3.

y intercept of f is -3

y intercept of g is -1

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
The points A(1, 4), B(5,1) lie on a circle. The line segment AB is a chord. Find the equation of a diameter of the circle.
tangare [24]

Check the picture below.

well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.

bearing in mind that <u>perpendicular lines have negative reciprocal</u> slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.

\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}

so, it passes through the midpoint of AB,

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{5+1}{2}~~,~~\cfrac{1+4}{2} \right)\implies \left(3~~,~~\cfrac{5}{2} \right)

so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)

\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{\frac{5}{2}}) \stackrel{slope}{m}\implies \cfrac{4}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{\cfrac{5}{2}}=\stackrel{m}{\cfrac{4}{3}}(x-\stackrel{x_1}{3})\implies y-\cfrac{5}{2}=\cfrac{4}{3}x-4 \\\\\\ y=\cfrac{4}{3}x-4+\cfrac{5}{2}\implies y=\cfrac{4}{3}x-\cfrac{3}{2}

4 0
3 years ago
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