Given quadrilateral ABCD is congruent to Quadrilateral A'B'C'D', What is the measure of B'C'?
1 answer:
You might be interested in
Answer:
+
Step-by-step explanation:
We sure can write 2^2/5+ 3^1/10 as
// use the to transform the expression
Convert to
=
(here a = 2, n = 5, m = 2)
Then convert to
=
(here a = 3, n = 10, m = 1)
Now we get =
+
. This is the simplified form, which is the answer.
We have to set up 2 different equations if we are to solve for 2 unknowns. The first equation is x = y + 4. One number (x) is (=) 4 more than another (y + 4). Since we have determined that x is larger (cuz it's 4 more than y), when we set up their difference, we are going to subtract y from x cuz x is bigger. The second equation then is 
. In our first equation we said that x = y + 4, so let's sub that value in for x in the second equation: 
. Expand that binomial to get 
. Of course the y squared terms cancel each other out leaving us with 8y + 16 = 64. Solving for y we get that y = 6. Subbing 6 in for y in our first equation, x = 6 + 4 tells us that x = 10. Yay!