Answer:
2 solutions
Step-by-step explanation:
Let
x--------> the first <span>odd integer
x+2-----> </span>the second odd integer
x+4-----> the third odd integer
we know that
6(x+2)-2x=20+(x+2)+(x+4)
6x+12-2x=26+2x
4x-2x=26-12
2x=14
x=7
the answer is
the numbers are
7, 9 and 11
Arcsin x + arcsin 2x = π/3
arcsin 2x = π/3 - arcsin x
sin[arcsin 2x] = sin[π/3 - arcsin x] (remember the left side is like sin(a-b)
2x = sinπ/3 cos(arcsin x)-cosπ/3 sin(arc sinx)
2x = √3/2 . cos(arcsin x) - (1/2)x)
but cos(arcsin x) = √(1-x²)===>2x = √3/2 .√(1-x²) - (1/2)x)
Reduce to same denominator:
(4x) = √3 .√(1-x²) - (x)===>5x = √3 .√(1-x²)
Square both sides==> 25x²=3(1-x²)
28 x² = 3 & x² = 3/28 & x =√(3/28)
We will be using the Point Slope formula which is 
The information we are given is that m = -4 and the point is (2,-5), so we can input it into the equation.
This simplifies into 
Then we can subtract 5 from each side to get the answer:
The sides of the rectangle are:
xy = 39
2x + 2y
Solve by simultaneous equation:
ysquared -17y + 30 = 0
Solution:
The sides are equal to 15 and 2
