I hope this helps you
9/9+18=x/x+10
9/27=x/x+10
1/3=x/x+10
3x=x+10
x=5
Answer:
Step-by-step explanation:
d= 2r
therefore the radius is 1
Answer:
(3, -8) and (2, -10)
Step-by-step explanation:
Given
Required
Select the true coordinate points in (3, -8) (2, 5) (-5, 1) (10, 3) (2, -10)
(3, -8)
x = 3 and y = -8
--- True
--- True
(2, 5)
x = 2 and y = 5
--- False (No need to check the other inequality)
(-5, 1)
x = -5 and y = 1
--- True
--- False
(10, 3)
x = 10 and y = 3
--- False (No need to check the other inequality)
(2, -10)
x = 2 and y = -10
--- True
--- True
Hence, the solution to the inequalities are:
(3, -8) and (2, -10)
0.10r + 0.20b = 24
3r = b + 20....b = 3r - 20
0.10r + 0.20(3r - 20) = 24
0.10r + 0.60r - 4 = 24
0.70r = 24 + 4
0.70r = 28
r = 28/0.70
r = 40 <=== there are 40 reds
0.10r + 0.20b = 24
0.10(40) + 0.20b = 24
4 + 0.20b = 24
0.20b = 24 - 4
0.20b = 20
b = 20/0.20
b = 100 <=== there are 100 blues
For cos(2x) * (2cos(x) + 1) = 0, use the double angle identity for cos(2x), which is cos^2 x - sin^2 x = cos^2 x - (1-cos^2) = 2cos^2 x - 1.
So we have (2cos^2 x - 1)(2cos x + 1) = 0. So 2cos^2 x -1 = 0 or x = 0 and 2pi.
For 2sec^2 x + tan^2 x - 3 = 0, use the identity sec^2 x = tan^2 x + 1, so we have
2(tan^2 x + 1) + tan^2 x - 3 = 0 or
<span>2tan^2 x + tan^2 x - 1 = 0 or
</span>3 tan^2 x = 1.
So x = pi/2, pi/2 + pi = 3pi/2.
