Answer:
y=-
x+
Step-by-step explanation:
First, calculate the slope of the line that is perpendicular to the equation of line we are asked to find
m=(y2-y1)/(x2-x1)
=(2-(-4))/(-2-1)
=6/-3
=-2
in this equation the slope is 2, and to find the first equation, use y=mx+b
use the point (1, -4) to find b
-4=(2)(1)+b
-4=2+b
b=-6
the first equation of the line is y=2x-6
to find the x intercept of that line substitute 0 for y
0=2x-6
2x=6
x=3
the slope of a line perpendicular to this would be the opposite reciprocal of the slope which would be equal to -1/2
for the second equation of the line to pass thorugh the x-intercept of the first line, it must pass through (3, 0), so substitute and solve for b
y=mx+b
0=(-1/2)(3)+b
b=3/2
thus the equation of the line that is perpendicular to the line through (1,-4) and (-2, 2) and passes through the x intercept of that line is y=-x+3/2
Answer:
Step-by-step explanation:
30 minutes UoU
Every 10 minutes he finishes 20% of the book case.
so In 10 minutes he will have 50% done, In 20 minutes 60%, in 40 minutes 100%
8z=4(2z+1)
First you would distribute the constant into the numbers in the parentheses. so
8z=8z+4
then you would combine like terms which in this case would result in a zero.
8z-8z=0 So the z would be zero or no solution.
Step-by-step explanation:
I am not sure what exactly you mean.
do you mean the complete square of an expression or
term ?
if so, then by multiplying this term by itself, and that means in general, every part is multiplied by every part and the part results are added considering the signs involved.
e.g.
squaring a+b
(a+b)(a+b) = a×a + a×b + b×a + b×b = a² + 2ab + b²
remember that multiplication and addition are commutative (you can flip the right and left sides with each other and still get the same result : a+b = b+a, a×b = b×a).
squaring a-b
(a-b)(a-b) = a×a + a×-b + -b×a + -b×-b = a² - 2ab + b²
remember that
+×- = -×+ = -
-×- = +
+×+ = +
a more complex example ?
squaring a-b+c
(a-b+c)(a-b+c) =
= a×a + a×-b + a×c + -b×a + -b×-b + -b×c + c×a + c×-b + c×c =
= a² - 2ab - 2bc + 2ac + b² + c²
