MATHEMATICS HIGH SCHOOL

Which of these is a simplified form of the equation 8y + 6 = 7 + 2y + 3y?3y = 1
8y = 6
13y = 13
5y = 13

Answers

Answer 1
Answer: 8y+6=7+2y+3y

8y+6=7+5y

3y+6=7

3y=1

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Good evening guys

I nee your help , can you ?

Answers

5h^3: 5 * h * h * h
20h^2: 2 * 2 * 5 * h * h
GCF(5h^3, 20h^2) = 5h^2

4h: 2 * 2 * h
16: 2 * 2 * 2 * 2
GCF(4h, 16) = 4

The answer is C.

Explain how to solve 3x+4=19

Answers

Answer:

X = 5

Step-by-step explanation:

3x + 4 = 19

    -4.      -4

3x = 15

divide both sides by 3 to solve for x

x = 5

A sports ball has a diameter of 16 cm. Find the volume of the ball.

Answers

Answer:

4/3 pi radius cubed

Step-by-step explanation:

4/3 pi x (8x8x8)

512 x 4/3 pi

2144.660585

the point-slope form of the equation of the line that passes through (–4, –3) and (12, 1) is y – 1 = (x – 12). What is the standard form of the equation for this line?

Answers

                   y - 1 = 0.25(x - 12)
                   y - 1 = 0.25(x) - 0.25(12)
                   y - 1 = 0.25x - 3
                     + 1             + 1
                        y = 0.25x - 2
          -0.25x + y = 0.25x - 0.25x + 2
          -0.25x + y = 2
    -1(-0.25x + y) = -1(2)
-1(-0.25x) - 1(y) = -2
           0.25x - y = -2

Solve for e:(-7)=e/3+14? show your work

Answers

-7 = (e)/(3) + 14
take 14 to the other side
-7 -(14) = (e)/(3)
-21 = (e)/(3)
Multiply by 3 on either sides to isolate e and to find a value

-21 × 3 = (e)/(3) × 3
3 and 3 cancels out

-63 = e

A design engineer is mapping out a new neighborhood with parallel streets. If one street passes through (6, 4) and (5, 2), what is the equation for a parallel street that passes through (−2, 6)?A. y = 1 half x + 5
B. y - 1 half x + 1
C. y = 2x + 10
D. y = 2x − 14

Answers

keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the first street.

(\stackrel{x_1}{6}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{2}) ~\hfill~ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{2}-\stackrel{y1}{4}}}{\underset{\textit{\large run}} {\underset{x_2}{5}-\underset{x_1}{6}}} \implies \cfrac{ -2 }{ -1 } \implies 2

so we are really looking for the equation of a line whose slope is 2 and it passes through (-2 , 6)

(\stackrel{x_1}{-2}~,~\stackrel{y_1}{6})\hspace{10em} \stackrel{slope}{m} ~=~ 2 \n\n\n \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\n \cline{1-1} \n y-y_1=m(x-x_1) \n\n \cline{1-1} \end{array}\implies y-\stackrel{y_1}{6}=\stackrel{m}{2}(x-\stackrel{x_1}{(-2)}) \implies y -6 = 2 ( x +2) \n\n\n y-6=2x+4\implies {\Large \begin{array}{llll} y=2x+10 \end{array}}
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