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polet [3.4K]
2 years ago
11

Rewrite the equation so that it does not have fractions

Mathematics
1 answer:
olya-2409 [2.1K]2 years ago
7 0

Answer:

\frac{2 - 3}{4x}  =   \frac{5}{12}  \\ 4x(5) = 12(2 - 3) \\ 20x = 24 - 36 \\ 20x = 12

<h2><em><u>hope</u></em><em><u> it</u></em><em><u> helps</u></em><em><u> you</u></em><em><u><</u></em><em><u>3</u></em></h2>

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What is the value of n n+11=13​
gtnhenbr [62]

Answer:

n=2

Step-by-step explanation:

8 0
4 years ago
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Granny has taken up deep-sea fishing! Last week, she caught a fish so big that she had to cut it into 3 pieces (head, body and t
Marianna [84]

Answer:

54kg

Step-by-step explanation:

Body weight = B

Tail weight = T

Head weight = H

Total weight = F

Use equation = B + T + H = F

We are told that T = 9kg, H = T + (1/3) x B, B = H + T

We have 3 equations and 3 unknowns, solve the system of equations to find W.  

B = H + 9

H = 9 + 1/3 x (H+9)=9+H/3+3 --> 2/3 x H = 12  --> H = 18

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4 0
3 years ago
Which triangle would be most helpful in finding the distance between the points (-4, 3) and (1,-2) on the coordinate
earnstyle [38]

Answer:

On a coordinate plane, a triangle has points (negative 4, 3), (negative 4, negative 2), (1, negative 2).

Step-by-step explanation:

The points (-4,3), (1,2) and (-4, -2) would form a right triangle when graphed and connected by lines.

(-4,3), (1,2) and (1,3) would also work as well

5 0
2 years ago
Please help with these questions
AveGali [126]
5 ft.................
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3 years ago
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How do you do this problem? I need to know how you found the answer.
Alexxandr [17]

to get the equation of a line, we simply need two points, say for the Red one ... notice in the graph the lines passes through (0,2) and (-1,6), so let's use those


\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{-1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-2}{-1-0}\implies \cfrac{4}{-1}\implies -4 \\\\\\ \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-2=-4(x-0) \\\\\\ y-2=-4x\implies \blacktriangleright y=-4x+2 \blacktriangleleft


now, for the Blue one, say let's use hmmm it passes through (0,2) and (1.6)


\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-2}{1-0}\implies \cfrac{4}{1}\implies 4 \\\\\\ \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-2=4(x-0) \\\\\\ y-2=4x\implies \blacktriangleright y=4x+2 \blacktriangleleft

3 0
3 years ago
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