Urgent: What is an equation of the line perpendicular to y=-x-2 and through (-2,4)?

Mathematics

2 answers:

Aleksandr [31]3 years ago

8 0

Point slope form- y-4=(x+2)
Slope intercept- y=x+6

sasho [114]3 years ago

6 0

Perpendicular line means the gradient will be negative reciprocal. The gradient of the line that is given is -1.

Therefore the gradient of the perpendicular line is -1 / -1 = 1

Now we know the equation is y=x+c

We want to find out what c is.

To do this, we substitute in the coordinates given:

4 = 1(-2) + c

4 = -2 + c

C= 6

Therefore the perpendicular line equation is:

y = x + 6

I hope this has helped you today,

Any questions please ask me in the comments below.

Have a great day!

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Which is the slope of the line y=−3x+2?y=-3x+2?

Morgarella [4.7K]

The slope is -3.
Finding slope can be easy if you just remember that the formula is set up as y=mx+b.
The number in the 'm' place is always the slope.
Example: y=4x+1, the slope would be 4 because it's in the 'm' place.
If you have trouble remembering which place the slope is, trying thinking about skiing or sledding; you can only sled or ski down a Mountain slope. Mountain starts with M.
M=slope.
Hope this helps :)
If you have any questions, just let me know!

5 0

3 years ago

I need to show work and Idk how for 2 or 3. The -5,-2 and 8,3 is the answers just need work shown thanks

prisoha [69]

#2
2x - 4y = -2
2x + 3y = -16
------------------subtract
-7y = 14
y = 14/-7
y = -2

2x - 4y = -2
2x - 4(-2) = -2
2x + 8 = -2
2x = -2 -8
2x = -10
x = -10/2
x = -5
answer x = -2 and y = -5

check:

2(-5)- 4(-2) = -2
-10 +8 = -2
-2 = -2.....true

2x + 3y = -16
2(-5)+3(-2) = -16
-10 + (-6) = -16
-16 = -16...true

3 0

3 years ago

PLEASE HELP

lisabon 2012 [21]

Step-by-step explanation:

The figure below shows a portion of the graph of the function $j\left(x\right) \ = \ 4^{x-2}$ , hence the average rate of change (slope of the blue line) between the $x$ and $x+h$ is

$\text{Average rate of change} \ = \ \displaystyle\frac{\Delta y}{\Delta x} \\ \\ \rule{3.7cm}{0cm} = \dsiplaystyle\frac{f\left(x+h\right) \ - \ f\left(x\right)}{\left(x \ + \ h \right) \ - \ x} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{f\left(x + h\right) \ - \ f\left(x\right)}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{4^{x+h-2} \ - \ 4^{x-2}}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{4^{x-2+h} \ - \ 4^{x-2}}{h}$

$\\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{\left(4^{x-2}\right)\left(4^{h}\right) \ - \ 4^{x-2}}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{\left(4^{x-2}\right)\left(4^{h} \ - \ 1 \right)}{h}$