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goldfiish [28.3K]
3 years ago
12

7a-3b-a+8-9b Pls help!

Mathematics
2 answers:
iVinArrow [24]3 years ago
7 0

Answer:

6a-12b+8

Step-by-step explanation:

Im pretty sure this is the answer but if it's not im so sorry

hram777 [196]3 years ago
4 0

Answer:

6  −  1 2  +  8

Step-by-step explanation:

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Find the area of the following rectangle.<br> length = 12 ft.<br> width = 8 ft.
iragen [17]

Answer:

96 ft

Step-by-step explanation:

A= bh

A = 12 x 8

A = 96

4 0
3 years ago
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1.)<br> What is the solution to the system of equations below?<br> y = 2x + 8<br> 3(-2x + y) = 19
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Answer:

y = 2x + 8

6x + 3y = 19

6x + 3(2x + 8) = 19

6x + 6x + 24 = 19

12x = -5

x = -5/12

y = 2(-5/12) + 8

y = -5/6 + 48/6

y = 43/6

3 0
3 years ago
Solve for b.<br> 7<br> b<br> 12<br> b= ✓ [?]<br> Pythagorean Theorem: a2 + b2 = c2<br> Enter
fomenos

Answer: b=\sqrt{95}

Step-by-step explanation:

To solve for b, we want to use the Pythagorean Theorem as given.

b and 7 are the legs, and 12 is the hypotenuse.

7^2+b^2=12^2      [exponent]

49+b^2=144     [subtract both sides by 49]

b^2=95               [square root both sides]

b=\sqrt{95}

Now we know b=\sqrt{95}.

6 0
3 years ago
Zoom in if beaded and pls help pronto
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8 0
3 years ago
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What equation is graphed in this figure?
Ksivusya [100]

to get the equation of any straight line, we simply need two points off of it, let's use those two points in the picture below.

(\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{5}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{5}-\stackrel{y1}{2}}}{\underset{run} {\underset{x_2}{1}-\underset{x_1}{0}}}\implies \cfrac{3}{1}\implies 3 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_2=m(x-x_2) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_2}{5}=\stackrel{m}{3}(x-\stackrel{x_2}{1})

keeping in mind that for the point-slope form, either point will do, in this case we used the second one, but the first one would have worked just the same.

7 0
2 years ago
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