Answer:
Step-by-step explanation:
Let's solve for 
in the first equation and then solve for 
in the second equation.
I will then use the following identity to get right of the parameter, 
:
(Pythagorean Identity).
Let's begin with .
Subtract 2 on both sides:
Divide both sides by -3:
Now time for the second equation, .
Subtract 1 on both sides:
Divide both sides by 4:
Now let's plug it into our Pythagorean Identity:
Answer:
point a is positive but point b is negative
Step-by-step explanation:
Your maths problem
<
1
5
?
.
8
0
.
7
0
.
1
0
0
.
1
1
0
Notice that the 2 expressions have 2 common terms.
(r-s) is just (s-r) times (-1)
similarly
(t-s) is just (s-t) times (-1)
this means that :
(r-s) (t-s) + (s-r) (s-t)=-(s-r)[-(s-t)]+(s-r) (s-t)
the 2 minuses in the first multiplication cancel each other so we have:
-(s-r)[-(s-t)]+(s-r) (s-t)=(s-r) (s-t)+(s-r) (s-t)=2(s-r) (s-t)
Answer:
d)<span>2(s-r) (t-s) </span>
Answer:
Step-by-step explanation:
1,8
5,7
2,3
