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Vladimir [108]
2 years ago
15

Can you find the slope -intercept equation of each line code ?​

Mathematics
1 answer:
ikadub [295]2 years ago
5 0

Answer:

You can find and write the equations in Slope-Intercept form due to the slope and y-intercept.

Step-by-step explanation:

The Slope-Intercept Form Formula: y=mx+b

The m represents the slope while the b represents the y-intercept

You can also use y2 - y1 / x2 - x1 to find the slope which is the m in your formula.

Lastly, to find a y-intercept, you can plug in your coordinate for the x and y to do two-step linear equations.

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-2-(3(x+6)-9-(7-x)-3x
olchik [2.2K]

Answer:

- 24 - 5x

Step-by-step explanation:

1) Remove Parentheses

-2 - 3x - 6 - 9 - 7 + x - 3x

2) Collect like terms

( -2 - 6 - 9 - 7) + (-3x+x-3x)

3) Simplify

7 0
3 years ago
A/75=15/100<br> help me in this :(
Svetach [21]

a / 75 = 15 / 100

100a = 75 * 15       [cross multiplication]

a = 11.25

i hope this helps!!! :D

4 0
3 years ago
Read 2 more answers
If g(x)=2x-1 and h(x)= swuare root x, find (g o h) (9)
AnnZ [28]

Answer:

5

Step-by-step explanation:

To evaluate (g ○ h)(9), evaluate h(9) , then use this value to evaluate g(x)

h(9) = \sqrt{9} = 3, then

g(3) = 2(3) - 1 = 6 - 1 = 5

8 0
3 years ago
Use the Quadratic Formula to solve x2 + 20x + 98 = 0
Lubov Fominskaja [6]
ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\if\ \Delta \ \textless \  0\ then\ no\ solution\\\\if\ \Delta =0\ then\ one\ solution\ x_0=\dfrac{-b}{2a}\\\\if\ \Delta \ \textgreater \  0\ then\ two\ solutions\ x_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\-----------------------------

x^2+20x+98=0\\a=1;\ b=20;\ c=98\\\\\Delta=20^2-4\cdot1\cdot98=400-392=8 \ \textgreater \  0\\\sqrt\Delta=\sqrt8=\sqrt{4\cdot2}=\sqrt4\cdot\sqrt2=2\sqrt2\\\\x_1=\dfrac{-20-2\sqrt2}{2\cdot1}=\dfrac{-20-2\sqrt2}{2}=-10-\sqrt2\\\\x_2=\dfrac{-20+2\sqrt2}{2\cdot1}=\dfrac{-20+2\sqrt2}{2}=-10+\sqrt2
3 0
3 years ago
A set of equations is given below:
Zanzabum
The answer is:  [A]:  "No solution" .
_______________________________________________
Note:  Given:
_______________________________________________
  y = 4x + 8 ;
  y = 4x + 2 ;
_______________________________________________

Since:  "y = y" ;

4x + 8 = 4x + 2 ; which is not true;
______________________________
 4x  +  8  ≠  4x  +  2 ;  since,  "8  ≠ 2" ;

Note:  Suppose:  "4x + 8 = 4x + 2" ; 

→ Subtract "4x" from each side of the equation:

4x + 8 − 4x  = 4x + 2 − 4x ;

to get:
_________________
 " 8 = 2 " ;  not true ; since  "8 ≠ 2" ;
______________________________
Also, suppose:
_______________________________
 4x + 8 = 4x + 2 ;

→ Subtract "2" from BOTH SIDES of the equation;

4x + 8 − 2 = 4x + 2 − 2 ;

to get:

4x + 6 = 4x

(Note:  "4x + 0 = 4x" ;  but "4x + 6 ≠ 4x") l

But even if we have:

4x + 6 = 4x ;

→ Subtract "4x" from EACH side of the equation:

4x + 6 − 4x  = 4x − 4x ;

to get:

   6 = 0 ; which is not true;  "6 ≠ 0" . 
____________________________________
The correct answer is:  [A]:  "No solution" .
______________________________________

4 0
3 years ago
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