ok hi im here to help not rlly just want my points up your useless
Problem:
find line perpendicular to 4x+7y+3=0 passing through (-2,1).
Solution:
We will first find the general form of the perpendicular line.
The perpendicular line has the form
7x-4y+k=0 .........................(1)
by switching the coefficients of x and y, and switching the sign of one of the two coefficients. This way, the slope of (1) multiplied by that of the original equation is -1, a condition that the two lines are perpendicular. The value of k is to be determined from the given point (-2,1).
To find k, we substitute x=-2, y=1 into equation (1) and solve for k.
7(-2)-4(1)+k=0
=>
k=14+4=18
Therefore the required line is
7x-4y+18=0
Answer:
Divide by 7 on both sides
Step-by-step explanation:
7x = 28
To get the variable (x), alone we must divide by 7 on both sides
Value of x: x = 4
Hope this helps :)
Answer:
1. D,F
2. E,F,G
3. C
4. B,C
5. A,B,C
Step-by-step explanation:
