Answer:
(-3,-1)
Step-by-step explanation:
-x+y=2
3x-5y=-4
in the top equation, we can get y by itself on one side of the equation:
-x+y=2
y=x+2
now, since we know that y=x+2, in the bottom equation we can substitute x+2 for y (substitution method) and solve:
3x-5y=-4
3x-5(x+2)=-4
3x-5x-10=-4
-2x-10=-4
-2x=6
x=-3
since we now know that x=-3, we can use it to find y by plugging in -3 for x in the top equation (you can do it in either equation, but doing it in the top equation is easier):
y=x+2
y=-3+2
y=-1
answer: (-3, -1)
Answer:
B
Step-by-step explanation:
So remember that a quarter of a circle is 90 degrees, a half of a circle is 180 degrees, and a whole circle is 360 degrees.
Looking at the image shown, it must be half a circle, whihc is 180 degrees. The image tells us that the 1st angle is 7 degrees. Now, we must find the 2nd angle.
Heres what we know however:
angle 1 + angle 2 = 180 degrees.
How do we know this?
Well, there are only 2 angles in this 180 degrees. We know that the first one is 7 degrees.
Lets input that into our nice lil equation to get an answer:
7 degrees+angle 2 = 180 degrees
Now lets solve.
Subtract 7 on both sides, and your left with:
angle 2 = 173 degrees.
So the answer must be:
<u>173 degrees</u>
Hope this helps! ;)
Answer:
y = 56
Step-by-step explanation:
the midsegment SV is half the length of the side TU , that is
y - 9 = 
(y + 38) ← multiply both sides by 2 to clear the fraction
2y - 18 = y + 38 ( subtract y from both sides )
y - 18 = 38 ( add 18 to both sides )
y = 56
Answer:
FALSE
Step-by-step explanation:
<E in ∆AED ≅ <E in ∆CEB.
Both are 90°.
Side ED ≅ Side EB
Side AD ≅ Side CB.
Thus, two sides (ED and AD) and a non-included angle (<E) of ∆AED are congruent to corresponding two sides (EB and CB) and a non-included angle (<E) of ∆CEB. Therefore, by A-S-S Congruence Theorem, both triangles are congruent to each other not by SSS.
I think that it is D or A
