Hello from MrBillDoesMath!
Answer:
See Discussion below
Discussion:
(sinq + cosq)^2 = => (a +b)^2 = a^2 + 2ab + b^2
(sinq)^2 + (cosq)^2 + 2 sinq* cosq => as (sinx)^2 + (cosx)^2 = 1
1 + 2 sinq*cosq (*)
Setting a = b = q in the trig identity:
sin(a+b) = sina*cosb + cosa*sinb
sin(2q) = (**)
sinq*cosq + cosq*sinq => as both terms are identical
2 sinq*cosq
Combining (*) and (**)
(sinq + cosq)^2 = 1 + 2sinq*cosq => (**) 2sinq*cosq = sqin(2q)
= 1 + sin(2q)
Hence
(sinq + cosq)^2 = 1 + sin(2q) => subtracting 1 from both sides
(sinq + cosq)^2 - 1 = sin(2q)
The last statement is what we are trying to prove.
Thank you,
MrB
<span>152.5 is 250% of 61
Hope this helps! (tried to do an explanation and my browser crashed...) </span>
Answer:
1. Area of a square = (x^2)^2 = x^4
(each side is x^2, area of a square = side^2
so (x^2)(x^2) = x^4 )
2. when x = 5
area = 5^4 = 625
3. when x = 10
area = 10^4 = 10000
Step-by-step explanation:
Hope its right!
I think it would be 449.88
Distribute 1/4 to all terms within the parenthesis
9 - 5 = 4
1/4(16x + 60) = (16x + 60)/4 = 4x + 15
4 = 4x + 15
Isolate the x. Note the equal sign. What you do to one side, you do to the other.
Subtract 15 from both sides
4 = 4x + 15
4 (-15) = 4x + 15 (-15)
4 - 15 = 4x
-11 = 4x
Isolate the x. Divide 4 from both sides
(-11)/4 = (4x)/4
x = -11/4
hope this helps
