X = 20
First, you move the 40 to the other side.
-2x=-40
Then, you divided -40 by -2.
x= -40/-2
x= 20
x is 20 because the negatives cancel out making it positive.
Hey there!
Use The Pythagorean theorem. It’s a^2+b^2=c^2. Plug in 11 and 60. The equation would be 11^2+60^2=c^2. Simplify to get 121+3600=c^2. Simplify again to get 3721=c^2. Sqaure root each side to get c=61. The answer is the last one, c=61.
I hope this helps!
Answer:
Step-by-step explanation:
Given the equation
Sin(5x) = ½
5x = arcSin(½)
5x = 30°
Then,
The general formula for sin is
5θ = n180 + (-1)ⁿθ
Divide through by 5
θ = n•36 + (-1)ⁿ30/5
θ = 36n + (-1)ⁿ6
The range of the solution is
0<θ<2π I.e 0<θ<360
First solution
When n = 0
θ = 36n + (-1)ⁿθ
θ = 0×36 + (-1)^0×6
θ = 6°
When n = 1
θ = 36n + (-1)ⁿ6
θ = 36-6
θ = 30°
When n = 2
θ = 36n + (-1)ⁿ6
θ = 36×2 + 6
θ = 78°
When n =3
θ = 36n + (-1)ⁿ6
θ = 36×3 - 6
θ = 102°
When n=4
θ = 36n + (-1)ⁿ6
θ = 36×4 + 6
θ = 150
When n =5
θ = 36n + (-1)ⁿ6
θ = 36×5 - 6
θ = 174°
When n = 6
θ = 36n+ (-1)ⁿ6
θ = 36×6 + 6
θ = 222°
When n = 7
θ = 36n + (-1)ⁿ6
θ = 36×7 - 6
θ = 246°
When n =8
θ = 36n + (-1)ⁿ6
θ = 36×8 + 6
θ = 294°
When n =9
θ = 36n + (-1)ⁿ6
θ = 36×9 - 6
θ = 318°
When n =10
θ = 36n + (-1)ⁿ6
θ = 36×10 + 6
θ = 366°
When n = 10 is out of range of θ
Then, the solution is from n =0 to n=9
So the equation have 10 solutions in the range 0<θ<2π
A score of 85 would be 1 standard deviation from the mean, 74. Using the 68-95-99.7 rule, we know that 68% of normally distributed data falls within 1 standard deviation of the mean. This means that 100%-68% = 32% of the data is either higher or lower. 32/2 = 16% of the data will be higher than 1 standard deviation from the mean and 16% of the data will be lower than 1 standard deviation from the mean. This means that 16% of the graduating seniors should have a score above 85%.
