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Thepotemich [5.8K]
3 years ago
5

A student created a table of values to help solve the following equation: y =4x +24 and y = 2x+6...what x-value is a solution to

the equation?
Mathematics
1 answer:
Yuki888 [10]3 years ago
6 0

Answer:

x = negative 9 or x=-9

Step-by-step explanation:

4x+24=2x+6. We can subtract 6 from both sides to get    

4x+18=2x. After that, we can switch the sides to get    

2x-18=4x, so negative 18 (-18) is equivalent to 2x. We get

2x=-18, divide both sides by 2 to get x=(-9).

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Trough: (2,2) parallel to y=x+4
Vlada [557]

Answer:

  y = x

Step-by-step explanation:

The slopes of parallel lines are the same, so we know the equation will be ...

  y = x + constant

We can find the constant by using the given point's values for x and y:

  2 = 2 + constant

Obviously, the constant is zero.

The equation of the parallel line through (2, 2) is y = x.

7 0
3 years ago
4x + 5y = 12 and 3x + 4y = 9.25 . Solve for systems of equations
masha68 [24]

Answer:

y=1

x=7/4

Step-by-step explanation:

4x + 5y = 12

3x + 4y = 9.25

4x + 5y = 12

12x + 16y =37

-12x -15 y =-36

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y=1

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8 0
3 years ago
Read 2 more answers
Which of the following possibilities will form a triangle?
kvasek [131]

B i think im not sure sorry

6 0
3 years ago
Nikolai is taking a taxi to school. He has to
adoni [48]
7 because the first mile will be 3 dollars then if you add the rest of the miles it will be seven because you are adding four more miles
5 0
3 years ago
Find the area of the shaded region ​
o-na [289]

so hmmm let's get the area of the whole hexagon, and then get the area of the circle inside it, then <u>subtract the area of the circle from that of the hexagon's</u>, what's leftover is what we didn't subtract, namely the shaded part.

\textit{area of a regular polygon}\\\\ A=\cfrac{1}{4}ns^2\cot\stackrel{\stackrel{degrees}{\downarrow }}{\left( \frac{180}{n} \right)}~ \begin{cases} n=\textit{number of sides}\\ s=\textit{length of a side}\\[-0.5em] \hrulefill\\ n=\stackrel{hexagon}{6}\\ s=\frac{9}{2} \end{cases}\implies A=\cfrac{1}{4}(6)\left( \cfrac{9}{2} \right)^2 \cot\left( \cfrac{180}{6} \right)

A=\cfrac{1}{4}(6)\cfrac{9^2}{2^2} \cot(30^o)\implies A=\cfrac{243}{8}\cot(30^o)\implies A=\cfrac{243\sqrt{3}}{8} \\\\[-0.35em] ~\dotfill\\\\ \textit{area of circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=\frac{4}{5} \end{cases}\implies A=\pi \left( \cfrac{4}{5} \right)^2\implies A=\cfrac{16\pi }{25} \\\\[-0.35em] ~\dotfill

\stackrel{\textit{area of the hexagon}}{\cfrac{243\sqrt{3}}{8}}~~ - ~~\stackrel{\textit{area of the circle}}{\cfrac{16\pi }{25}}\implies \cfrac{6075\sqrt{3}-128\pi }{200}

5 0
2 years ago
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