If x + y = 6, then solve for y to get: y = 6 - x.
Now replace y with 6 - x in both equations.
(5x)/3 + 6 - x = c
2(6 - x) = c - 4x
The upper equation is solved for c.
Now we solve the lower equation for c.
c = 2(6 - x) + 4x
c = 12 - 2x + 4x
c = 2x + 12
Since we have two equations solved for c, we substitute to get
(5x)/3 + 6 - x = 2x + 12
This is an equation in only x, so we can solve for x.
(5x)/3 - 3x = 6
5x - 9x = 18
-4x = 18
x = -9/2
Now we solve for y.
x + y = 6
-9/2 + y = 6
y = 9/2 + 12/2
y = 21/2
Now we solve for c.
c = (5x)/3 + y
c = (5 * (-9/2))/3 + 21/2
c = -45/6 + 21/2
c = -15/2 + 21/2
c = 6/2
c = 3
Answer: c = 3
Take -5x + 3x = -2x
because there is a -3x and a 3x they cancel each other out.
and the -17 is left
-2x = -17
divide -2 by -17 = 8.5
Assuming linear.
if k is at (3,4) and j is at (-8,7)
set. (y1,x1) and(y2,x2)
[(y2-y1),(x2-x1)]
[(-8-3),(7-4)]
distance from k to j is (-11,3)
now take coordinates of j and distance from. k->j and add them to get coordinate for L
[(-11+(-8)),(7+3)]= (-19,10)=L
Answer:
A is not 3, 7
Step-by-step explanation:because it isint
Answer:
In the explanation
Step-by-step explanation:
Going to start with the sum identities
sin(x+y)=sin(x)cos(y)+sin(y)cos(x)
cos(x+y)=cos(x)cos(y)-sin(x)sin(y)
sin(x)cos(x+y)=sin(x)cos(x)cos(y)-sin(x)sin(x)sin(y)
cos(x)sin(x+y)=cos(x)sin(x)cos(y)+cos(x)sin(y)cos(x)
Now we are going to take the line there and subtract the line before it from it.
I do also notice that column 1 have cos(y)cos(x)sin(x) in common while column 2 has sin(y) in common.
cos(x)sin(x+y)-sin(x)cos(x+y)
=0+sin(y)[cos^2(x)+sin^2(x)]
=sin(y)(1)
=sin(y)
