Sin³ x-sin x=cos ² x
we know that:
sin²x + cos²x=1 ⇒cos²x=1-sin²x
Therefore:
sin³x-sin x=1-sin²x
sin³x+sin²x-sin x-1=0
sin³x=z
z³+z²-z-1=0
we divide by Ruffini method:
1 1 -1 -1
1 1 2 1 z=1
-------------------------------------
1 2 1 0
-1 -1 -1 z=-1
--------------------------------------
1 1 0 z=-1
Therefore; the solutions are z=-1 and z=1
The solutions are:
if z=-1, then
sin x=-1 ⇒x= arcsin -1=π+2kπ (180º+360ºK) K∈Z
if z=1, then
sin x=1 ⇒ x=arcsin 1=π/2 + 2kπ (90º+360ºK) k∈Z
π/2 + 2kπ U π+2Kπ=π/2+kπ k∈Z ≈(90º+180ºK)
Answer: π/2 + Kπ or 90º+180ºK K∈Z
Z=...-3,-2,-1,0,1,2,3,4....
Answer: 11
Step-by-step explanation:
There are
total ways to draw any 6 numbers from the range 1 to 51, regardless of order.
Given 6 selected numbers that match those drawn by the lottery, there are
ways of rearranging them. So the probability of winning 1st prize is
Next, given 6 selected numbers of which 5 match those drawn by the lottery, there are
ways of rearranging those 5 matching numbers. There are 46 remaining numbers that didn't get drawn, so the probability of winning 2nd prize is
It’s -1/3 because you subtract 1-5 and -3-9, you get -4/-12 then simplify to get -1/3.
Let Natalie be x, and Fred, y.
So based on 1st condition, we get
x + y =43 - Equation 1
â´ x = 43-y - Equation 2
Now based on 2nd condition, we get
44x+y=88
Using x from Equation 2, we get
44(43-y) + y = 88
â´ 44x43 - 44y + y = 88
â´ 1892 -44y + y = 88
â´ -44y + y = 88 - 1892
â´ 43y =1804
â´ y = 1804/43
â´ y = 41.95
Using the value of y in Equation 2
x = 43 - 41.95
â´ x = 1.05
So Natalie is 1.05 yrs and Fred is 41.95 yrs old.
