For lines to be perpendicular their slopes must be negative reciprocals of one another, mathematically:
m1*m2=-1
So we first need to find the slope of the reference line.
m=(y2-y1)/(x2-x1)=(7-2)/(-1--5)=5/4
So the perpendicular line will have a slope of:
5m/4=-1
m=-4/5
So our perpendicular line so far is:
y=-4x/5+b, now we can use point (3,-1) to solve for the y-intercept, "b"
-1=-4(3)/5+b
-1=-12/5+b
-5/5+12/5=b
7/5=b
So the line is:
y=-4x/5+7/5 or more neatly
y=(-4x+7)/5
y=-0.8x+1.4
Answer:
x=4
Step-by-step explanation:
subtract 2 from both sides---
5x + 2 - 2 = 22 - 2
simplify ---
5x=20
Divide both sides by 5 -----
5x/5 = 20/5
simplify -----
x-4
hope this helps :)
Answer:
False
Step-by-step explanation:
If W and Z are vector spaces, then no, that is not necessarily true. That's not generally true for any circumstance unless the vector space W is a subspace of Z, or vice versa.
Answer:
Wdym?
Step-by-step explanation:
Im confused
Answer:
47
Step-by-step explanation:
Plug the numbers in and solve
5 ^ 2 + 2 ( 5 + 6 )
25 + 22
47
also plz mark me brainiest
