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What is the slope of the line that is perpendicular to the line that passes through (-2, 7) and (4, 9)? Type a numerical answer

kodGreya [7K]

Answer:

Therefore $m_{2}$ = -3.

Step-by-step explanation:

i) let us say the slope of the line passing through (-2,7) and (4,9) is $m_{1}$ .

ii) to find the slope of the line passing through two points ( $x_{1}$ , $y_{1}$ ) and ( $x_{2}$ , $y_{2}$ ) we use the formula m = $\dfrac{y_{2} - y_{1} }{x_{2} - x_{1} }$ . Therefore we can say that $m_{1}$ = $\dfrac{9 - 7}{4 - (-2)}$ = $\dfrac{1}{3}$ .

iii) let us say the slope of the perpendicular line to the line passing through (-2,7) and (4,9) is $m_{2}$

iv) from the formula for the slopes of a line and the line perpendicular to it which is given by $m_{1}$ × $m_{2}$ = -1, therefore $m_{2}$ = -1 ÷ $m_{1}$ = -3. Therefore $m_{2}$ = -3

6 0

4 years ago

Mario has save $200 towards the cost of buying a $1000 computer in the weeks to come he passes seven additional $30 per week cre

EastWind [94]

Answer:

30w+200=1000

Step-by-step explanation:

3 0

2 years ago

Please help me!!!!!!!! Factor z2 – 3z – 18

Vladimir [108]

First, the factor pairs for 18 are:
1, 18
2, 9
3, 6

Since the middle term is -3, the two factors must be 3 apart. The only pair that is 3 values apart is 3, 6.
Since the middle value is negative, the larger number is negative. This means that the 6 is negative.
z = -6
z = 3

Set each equal to 0;
z = -6
z + 6 = 0

z = 3
z - 3 =0
So your factors are (z+6) and (z-3), or answer B.

4 0

3 years ago

What is the equation of the line

Arada [10]

Answer:

y = -7

Step-by-step explanation:

The line represents the y-coordinate, because it is a straight line over the x axis, indicating it cant be an x coordinate

Because it is 7 units below the x-axis, it is -7

y = -7

If my answer is incorrect, pls correct me!

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-Chetan K

3 0

3 years ago

Use properties to rewrite the given equation. Which equations have the same solution as 3/5x +2/3 + x = 1/2– 1/5x? Check all tha

vodomira [7]

we have

$\frac{3}{5}x+ \frac{2}{3}+x=\frac{1}{2}-\frac{1}{5}x$

Combine like terms in both sides

$(\frac{3}{5}x+ x)+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$

we know that

$(\frac{3}{5}x+ x)=(\frac{3}{5}x+ \frac{5}{5}x)=\frac{8}{5}x$

substitute in the expression above

$\frac{8}{5}x+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$ -----> equation A

Multiply equation A by $5*3*2=30$ both sides

$30*(\frac{8}{5}x+\frac{2}{3})=30*(\frac{1}{2}-\frac{1}{5}x)$

$48x+20=15-6x$ ---------> equation B

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$48x+6x=15-20$

$54x=-5$ ---------> equation C

Solve for x

$x=-\frac{5}{54} =-0.09$

We are going to proceed to verify each case to determine the solution.

Case a) $\frac{8}{5}x+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$

the case a) is equal to the equation A

so

the case a) have the same solution that the given equation

Case b) $18x+20+30x=15-6x$

Combine like terms in left side

$(18x+30x)+20=15-6x$

$(48x)+20=15-6x$

the case b) is equal to the equation B

so

the case b) have the same solution that the given equation

Case c) $18x+20+x=15-6x$

Combine like terms in left side

$(18x+x)+20=15-6x$

$(19x)+20=15-6x$

$19x+6x=15-20\\25x=-5\\x=-0.20$

$-0.20\neq -0.09$

therefore

the case c) not have the same solution that the given equation

Case d) $24x+30x=-5$

Combine like terms in left side

$54x=-5$

the case d) is equal to the equation C

so

the case d) have the same solution that the given equation

Case e) $12x+30x=-5$

Combine like terms in left side

$42x=-5$

$x=-5/42=-0.12$

$-0.12\neq -0.09$

therefore

the case e) not have the same solution that the given equation

therefore

the answer is

case a) $\frac{8}{5}x+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$

case b) $18x+20+30x=15-6x$

case d) $24x+30x=-5$

7 0

4 years ago

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