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Nastasia [14]
3 years ago
15

Solve -3x+y=-7 by slope intercept form

Mathematics
1 answer:
marin [14]3 years ago
8 0

can u add picture?

if u dont add picture i cant solve sorry...

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DaniilM [7]
Y=-3/2 -3 is the slope intercept
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What is the value of x using the Pythagorean theorem?
Vesnalui [34]
A^2+b^2= c^2

Plug in the points. the longest side is always c.

95.2^2+ b^2= 168^2.

Square the numbers.

9063.04+ b^2= 28224

Subtract 9063.04 on both sides.

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The equation y=mx+b is in form
hichkok12 [17]

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slope intercept form

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Select the equations whose graphs are parallel to the graph of the equation y=23x+5.
iren [92.7K]

Answer:

2x-3y=9    and    4x-6y=30

Step-by-step explanation:

hello :  

note y=(2/3)x+5  no y=23x+5

2/3 is the slope....... all lines parallel to this line has same slope.

1)   2x-3y=9   means : 3y = 2x-9   divide by 3 : y=(2/3)x-3..same slope.

2)  4x-6y=30  means : 6y= 4x-30 divide by 6 : y=(4/6)x -30/6

but 4/6 = 2/3 so : y=(2/3)x-5..same slope.

6 0
3 years ago
Find the solution to this system of equations <br> 3x+2y+3z=3 4x-5y+7z=1 2x+3y-2z=6 <br> x=? Y=? Z=?
Ede4ka [16]

Answer:

The solution is x=2,\ y=0,\ z=-1.

Step-by-step explanation:

You are given the system of three equations:

\left\{\begin{array}{l}3x+2y+3z=3\\4x-5y+7z=1\\2x+3y-2z=6\end{array}\right.

Multiply the first equation by 4, the second equation by 3 and subtract them. Then multiply the third equation by 2 and subtract it from the second equation:

\left\{\begin{array}{l}3x+2y+3z=3\\4(3x+2y+3z)-3(4x-5y+7z)=4\cdot 3-3\cdot 1\\4x-5y+7z-2(2x+3y-2z)=1-2\cdot 6\end{array}\right.\Rightarrow \\\\\left\{\begin{array}{l}3x+2y+3z=3\\12x+8y+12z-12x+15y-21z=12-3\\4x-5y+7z-4x-6y+4z=1-12\end{array}\right.

So,

\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\-11y+11z=-11\end{array}\right.\Rightarrow \left\{\begin{array}{l}3x+2y+3z=3\\23y-9z=9\\y-z=1\end{array}\right.

Multiply the third equation by 23 and subtract it from the second equation:

\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\23y-9z-23(y-z)=9-23\cdot 1\end{array}\right.\Rightarrow \left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\23y-9z-23y+23z=9-23 \end{array}\right.

Hence,

\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\14z=-14 \end{array}\right.\Rightarrow z=-1

Substitute it into the second equation:

23y-9\cdot (-1)=9\Rightarrow 23y+9=9\\ \\23y=0\\ \\y=0

Substitute them into the first equation:

3x+2\cdot 0+3\cdot (-1)=3\Rightarrow 3x-3=3\\ \\3x=6\\ \\x=2

The solution is x=2,\ y=0,\ z=-1.

3 0
4 years ago
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