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kipiarov [429]
3 years ago
12

3х - у = 3 what's the value of y?

Mathematics
2 answers:
mel-nik [20]3 years ago
4 0
Y=-3+3x. You get this answer for y because you move all terms to the right that don’t contain y, and then you solve.
Licemer1 [7]3 years ago
3 0
3x - y = 3

add y to both sides
3x = 3 + y

subtract 3 from both sides
3x - 3 = y

rewrite
y = 3x - 3
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Which expressions are equivalent to log4 (1/4x^2),​
Vlada [557]

Step-by-step explanation:

2 log 4 (1/4i)

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2 years ago
Two friends bring hamburger meat to your cookout. One brings 2.7 pounds, and the other brings 3.54 pounds. How much hamburger me
inessss [21]
Answer:
They bought 6.24 pounds of hamburger meat

Explanation:
We are given that:
The first friend brought 2.7 pounds of hamburger meat
The second friend brought 3.54 pounds of hamburger meat

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Hope this helps :)
4 0
3 years ago
Please help I will reward brainly
valentina_108 [34]
I believe the answer is 9.8
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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}

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3 years ago
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