M∠A′B′C′ = m∠ABC would be the answer for this. if you could give me a pic, I could answer. I don't see Triangle ABC
Answer:
y + 1 + y + 2 + y + 3 =3y+6
Step-by-step explanation:
study hard:)
Sounds as tho' you have an isosceles triangle (a triangle with 2 equal sides). If this triangle is also a right triangle (with one 90-degree angle), then the side lengths MUST satisfy the Pythagorean Theorem.
Let's see whether they do.
8^2 + 8^2 = 11^2 ???
64 + 64 = 121? NO. This is not a right triangle.
If you really do have 2 sides that are both of length 8, and you really do have a right triangle, then:
8^2 + 8^2 = d^2, where d=hypotenuse. Then 64+64 = d^2, and
d = sqrt(128) = sqrt(8*16) = 4sqrt(8) = 4*2*sqrt(2) = 8sqrt(2) = 11.3.
11 is close to 11.3, but still, this triangle cannot really have 2 sides of length 8 and one side of length 11.
It would most likely be 41 because when you would round 41 after you would × 41 times 10 equals 410 then you would round it to 400
Solving expression 
we get 
Step-by-step explanation:
We need to solve the expression:
Solving:
Multiplying 5 with terms inside the bracket:
So, Solving expression we get
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