Answer:
<u>y = w and ΔABC ~ ΔCDE</u>
Step-by-step explanation:
Given sin(y°) = cos(x°)
So, ∠y + ∠x = 90° ⇒(1)
And as shown at the graph:
ΔABC is aright triangle at B
So, ∠y + ∠z = 90° ⇒(2)
From (1) and (2)
<u>∴ ∠x = ∠z </u>
ΔCDE is aright triangle at D
So, ∠x + ∠w = 90° ⇒(3)
From (1) and (3)
<u>∴ ∠y = ∠w</u>
So, for the triangles ΔABC and ΔCDE
- ∠A = ∠C ⇒ proved by ∠y = ∠w
- ∠B = ∠D ⇒ Given ∠B and ∠D are right angles.
- ∠C = ∠E ⇒ proved by ∠x = ∠z
So, from the previous ΔABC ~ ΔCDE by AAA postulate.
So, the answer is <u>y = w and ΔABC ~ ΔCDE</u>
Answer:
Step-by-step explanation:
Here is a tip: When you have something that looks like this - ( - , turn it into a plus sign. It looks kind of like a plus sign anyway. So if you have 53- (-32), turn it into 53 + 32. Does that make sense?
So 53 + 32 = 85.
Another tip to do it in your head, is to do 50 + 30 = 80, then do 3 + 2 = 5, and add them. It's easier than trying to juggle the odd numbers. :)
we know that
Imaginary roots will come in pairs, and so the degree must be even.
therefore
the answer is
options


Its 11 cause I’m built different