Depends on how much space you have on your device
Answer:
y
Step-by-step explanation:
that easy just isolate the variables
We have that
[√(2x+1)]+3=0
for x=4
[√(2*4+1)]+3=0
[√(9)]+3=0
3+3=0----------6 is not zero
therefore
the solution is not correct for x=4
[√(2x+1)]+3=0--------> [√(2x+1)]=-3---------> <span>There is no real solution for that equation
</span>Because (2x+1) >= 0
the solution is with complex numbers
LHS ⇒ RHS:
Identities:
[1] cos(2A) = 2cos²(A) - 1 = 1 - 2sin²(A)
[2] sin(2A) = 2sin(A)cos(A)
[3] sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
[4] cos(A + B) = cos(A)cos(B) - sin(A)sin(B)
cos(x) - cos(x + 2Θ)
= cos(x) - (cos(x)cos(2Θ) - sin(x)sin(2Θ)) [4]
= cos(x) - cos(x)(1 - 2sin²(Θ)) + sin(x)(2sin(Θ)cos(Θ)) [1] [2]
= cos(x) - cos(x) + 2sin²(Θ)cos(x) + 2sin(Θ)sin(x)cos(Θ)
= 2sin²(Θ)cos(x) + 2sin(Θ)sin(x)cos(Θ)
= 2sin(Θ)(sin(Θ)cos(x) + sin(x)cos(Θ))
= 2sin(Θ)sin(x + Θ)
Boys : Girls = 2:7
let the no. of boys be x
so no of girls = 250+x
so, x:(250+x) = 2:7
x/(250+x) = 2/7
7x = 500+2x
7x-2x = 500
x = 500/5 =100
