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worty [1.4K]
3 years ago
6

(-5/6) + c = (-11/12)

Mathematics
2 answers:
melomori [17]3 years ago
7 0
Remember you can do anything to an equation as long as you doit t both sides

-5/6+c=-11/12
add 5/6 to both sides
5/6-5/6+c=-11/12+5/6
0+c=-11/12+5/6
c=-11/12+5/6
how do we add?
we make same denomenator

11/12 and 5/6
6 times 2=12 so
5/6 times 2/2=10/12

c=-11/12+5/6
c=-11/12+10/12
c=-1/12
rosijanka [135]3 years ago
4 0
<span>(-5/6) + c = (-11/12)
c=-11\12+5\6
c=19\12 or 1 7\12</span>
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Help me to solve this please​
scZoUnD [109]

Answer:

-4/9

Step-by-step explanation:

We first calculate the first bracket.

[2/3-(-4/9)]

= (2/3+4/9) (we cancel out the negatives)

= (6/9+4/9) (common denominator)

= 10/9

Then we convert -2 1/2 into an improper fraction.

-2 1/2

= -5/2

Finally we calculate the division by swapping the numerator and denominator of -5/2.

10/9 / -5/2

= 10/9 * -2/5

= -20/45

= -4/9 (simplify the terms)

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2 years ago
What is -2n - 24 = -8(2n-4) , and how do I do it? Thanks in advance.
myrzilka [38]
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7 0
3 years ago
Read 2 more answers
A student was given the line y=2/3x-1 and the point (-7, 1/2) and asked to find an equation that went through the given point an
Colt1911 [192]

keeping in mind that perpendicular lines have <u>negative reciprocal</u> slopes,  hmmm what's the slope of y=2/3x-1 anyway?


\bf \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}~\hspace{7em}y=\stackrel{\stackrel{m}{\downarrow }}{\cfrac{2}{3}}x-1 \\\\[-0.35em] ~\dotfill

\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\cfrac{2}{3}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{3}{2}}\qquad \stackrel{negative~reciprocal}{-\cfrac{3}{2}}}


so, notice, "one of his mistakes" is that he used 3/2 as the slope, not -3/2.

so, we're really looking for a line whose slope is -3/2 and runs through (-7, 1/2).


\bf (\stackrel{x_1}{-7}~,~\stackrel{y_1}{\frac{1}{2}})~\hspace{10em} slope = m\implies -\cfrac{3}{2} \\\\\\ \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-\cfrac{1}{2}=-\cfrac{3}{2}[x-(-7)]\implies y-\cfrac{1}{2}=-\cfrac{3}{2}(x+7) \\\\\\ y-\cfrac{1}{2}=-\cfrac{3}{2}x-\cfrac{21}{2}\implies y=-\cfrac{3}{2}x-\cfrac{21}{2}+\cfrac{1}{2}\implies y=-\cfrac{3}{2}x-10

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ElenaW [278]

Answer:

Step-by-step explanation:

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Answer:

Step-by-step explanation:

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