1 and 146
2 and 73
and that's about it
Answer:
<em><u>x=-6, y=-2</u></em>, (As a point) (-6, -2).....The point form is not necessary unless you want to solve the system (of equations) by graphing.
Step-by-step explanation:
By substitution:
x-y=-4 By adding y on both sides,
x=y-4
Now you can substitute x for the expression (y-4)
Plug the (y-4) as x in the other equation.
So -2x+3y=6 becomes
-2(y-4)+3y=6
Now solve:
-2(y-4)+3y=6 distributes out to be
-2y+8+3y=6 Now combining like terms
y+8=6 Subtract 8 on both sides to isolate the variable
<u><em>y=-2</em></u>
Now plug the value -2 in where the y is in any equation (preferably the easier/less complicated one) and solve for x.
So x-y=-4 becomes
x-(-2)=-4
=x+2=-4
=<u><em>x=-6</em></u>
Hey :) it’s (2,5) it’s C !!
Assume 0 < <em>x</em>/2 < <em>π</em>/2. Then
tan²(<em>x</em>/2) + 1 = sec²(<em>x</em>/2) ===> sec(<em>x</em>/2) = √(1 - tan²(<em>x</em>/2))
===> cos(<em>x</em>/2) = 1/√(1 - tan²(<em>x</em>/2))
===> cos(<em>x</em>/2) = 1/√(1 - <em>t</em> ²)
We also know that
sin²(<em>x</em>/2) + cos²(<em>x</em>/2) = 1 ===> sin(<em>x</em>/2) = √(1 - cos²(<em>x</em>/2))
Recall the double angle identities:
cos(<em>x</em>) = 2 cos²(<em>x</em>/2) - 1
sin(<em>x</em>) = 2 sin(<em>x</em>/2) cos(<em>x</em>/2)
Then
cos(<em>x</em>) = 2/(1 - <em>t</em> ²) - 1 = (1 + <em>t</em> ²)/(1 - <em>t</em> ²)
sin(<em>x</em>) = 2 √(1 - 1/(1 - <em>t</em> ²)) / √(1 - <em>t</em> ²) = 2<em>t</em>/(1 - <em>t</em> ²)
