All categories

2 years ago

Please help asap!!! DX DX DX

Mathematics

2 answers:

Gnoma [55]2 years ago

8 0

I believe the answer is C becuz i had that same answer and question so I HOPED I HELPED AND SMASH THAT THANKS BUTTON AND ADD ME AS A FRIEND

fiasKO [112]2 years ago

8 0

The answer would be c

You might be interested in

Explain to me in words how you would find the slope of this line

Alchen [17]

Answer:

Step-by-step explanation: From one point, go over to the top/bottom of the other point then go down/up to the point (go from one point to the other) then the rise you did (1) over the run across (3) then the formula is rise over run. then divide.

3 0

2 years ago

HELP ME ASAP!!!!!!!!!!

MAXImum [283]

Answer: x(2) -9

I’m pretty sure that it

4 0

2 years ago

Read 2 more answers

60 POINTS FOR THE THISS!!!

Goryan [66]

Yes because they are constant throughout the table.

4 0

2 years ago

Read 2 more answers

write an equation for the perpendicular bisector of the line joining the two points. PLEASE do 4,5 and 6

myrzilka [38]

Answer:

4. The equation of the perpendicular bisector is y = $\frac{3}{4}$ x - $\frac{1}{8}$

5. The equation of the perpendicular bisector is y = - 2x + 16

6. The equation of the perpendicular bisector is y = $-\frac{3}{2}$ x + $\frac{7}{2}$

Step-by-step explanation:

Lets revise some important rules

The product of the slopes of the perpendicular lines is -1, that means if the slope of one of them is m, then the slope of the other is $-\frac{1}{m}$ (reciprocal m and change its sign)
The perpendicular bisector of a line means another line perpendicular to it and intersect it in its mid-point
The formula of the slope of a line is $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
The mid point of a segment whose end points are $(x_{1},y_{1})$ and $(x_{2},y_{2})$ is $(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$
The slope-intercept form of the linear equation is y = m x + b, where m is the slope and b is the y-intercept

∵ The line passes through (7 , 2) and (4 , 6)

- Use the formula of the slope to find its slope

∵ $x_{1}$ = 7 and $x_{2}$ = 4

∵ $y_{1}$ = 2 and $y_{2}$ = 6

∴ $m=\frac{6-2}{4-7}=\frac{4}{-3}$

- Reciprocal it and change its sign to find the slope of the ⊥ line

∴ The slope of the perpendicular line = $\frac{3}{4}$

- Use the rule of the mid-point to find the mid-point of the line

∴ The mid-point = $(\frac{7+4}{2},\frac{2+6}{2})$

∴ The mid-point = $(\frac{11}{2},\frac{8}{2})=(\frac{11}{2},4)$

- Substitute the value of the slope in the form of the equation

∵ y = $\frac{3}{4}$ x + b

- To find b substitute x and y in the equation by the coordinates

of the mid-point

∵ 4 = $\frac{3}{4}$ × $\frac{11}{2}$ + b

∴ 4 = $\frac{33}{8}$ + b

- Subtract $\frac{33}{8}$ from both sides

∴ $-\frac{1}{8}$ = b

∴ y = $\frac{3}{4}$ x - $\frac{1}{8}$

∴ The equation of the perpendicular bisector is y = $\frac{3}{4}$ x - $\frac{1}{8}$

∵ The line passes through (8 , 5) and (4 , 3)

- Use the formula of the slope to find its slope

∵ $x_{1}$ = 8 and $x_{2}$ = 4

∵ $y_{1}$ = 5 and $y_{2}$ = 3

∴ $m=\frac{3-5}{4-8}=\frac{-2}{-4}=\frac{1}{2}$

- Reciprocal it and change its sign to find the slope of the ⊥ line

∴ The slope of the perpendicular line = -2

- Use the rule of the mid-point to find the mid-point of the line

∴ The mid-point = $(\frac{8+4}{2},\frac{5+3}{2})$

∴ The mid-point = $(\frac{12}{2},\frac{8}{2})$

∴ The mid-point = (6 , 4)

- Substitute the value of the slope in the form of the equation

∵ y = - 2x + b

- To find b substitute x and y in the equation by the coordinates

of the mid-point

∵ 4 = -2 × 6 + b

∴ 4 = -12 + b

- Add 12 to both sides

∴ 16 = b

∴ y = - 2x + 16

∴ The equation of the perpendicular bisector is y = - 2x + 16

∵ The line passes through (6 , 1) and (0 , -3)

- Use the formula of the slope to find its slope

∵ $x_{1}$ = 6 and $x_{2}$ = 0

∵ $y_{1}$ = 1 and $y_{2}$ = -3

∴ $m=\frac{-3-1}{0-6}=\frac{-4}{-6}=\frac{2}{3}$

- Reciprocal it and change its sign to find the slope of the ⊥ line

∴ The slope of the perpendicular line = $-\frac{3}{2}$

- Use the rule of the mid-point to find the mid-point of the line

∴ The mid-point = $(\frac{6+0}{2},\frac{1+-3}{2})$

∴ The mid-point = $(\frac{6}{2},\frac{-2}{2})$

∴ The mid-point = (3 , -1)

- Substitute the value of the slope in the form of the equation

∵ y = $-\frac{3}{2}$ x + b

- To find b substitute x and y in the equation by the coordinates

of the mid-point

∵ -1 = $-\frac{3}{2}$ × 3 + b

∴ -1 = $-\frac{9}{2}$ + b

- Add $\frac{9}{2}$ to both sides

∴ $\frac{7}{2}$ = b

∴ y = $-\frac{3}{2}$ x + $\frac{7}{2}$

∴ The equation of the perpendicular bisector is y = $-\frac{3}{2}$ x + $\frac{7}{2}$

8 0

2 years ago

What are the constants in the expression 7r - 3 + 2s + 5

slega [8]

B. -3 and 5

constants are the numbers without variables (the letters) and coefficients are the numbers with variables

6 0

2 years ago