<span>3-2(Cosx)^2 - 3Sinx = 0.
Recall (Sinx)^2 + (Cosx)^2 = 1.
Therefore (Cosx)^2 = 1 - (Sinx)^2
Substitute this into the question above.
</span><span>3-2(Cosx)^2 - 3Sinx = 0
3 - 2(1 - (Sinx)^2) - 3Sinx = 0 Expand
3 - 2 + 2(Sinx)^2 - </span><span><span>3Sinx = 0</span>
1 + 2(</span><span>Sinx)^2 - 3Sinx = 0 Rearrange
2(Sinx)^2 </span><span><span>- 3Sinx + </span>1 = 0
Let p = Sinx
2p^2 - 3p + 1 = 0 Factorise the quadratic expression
2p^2 - p - 2p +1 = 0
p(2p -1) - 1(2p -1) = 0
(2p-1)(p -1) = 0
Therefore 2p-1=0 or (p-1) = 0
2p=0+1 or (p-1) = 0
2p=1 or p = 0 +1.
p=1/2 or p = 1 Recall p = Sinx
Therefore Sinx = 1/2 or 1.
For 0<u><</u>x<u><</u>360
Sinx =1/2, x = Sin inverse (1/2) , x = 30,
(180-30)- 2nd Quadrant = 150 deg
Sinx = 1, x = Sin inverse (1) , x = 90
Therefore x = 30,90 & 150 degrees.
Cheers.</span>
The answer is C. because 3/16 is close to 3 because how it’s estimated of the germs and gasoline is how it’s going to hit africa and south america and dacota. hope this helps!!
So what this is is
many words
assuming year 0 is 2017
so compound first thing till 2020, take out 30000
the remaining is copmpounded til 2022, take out 50000
remaining is compounded for 1 more year and that is equal to 80000
so from 2017 to 2020, that is 5 years
from 2020 to 2022 is 2 years
from 2022 to 2023 is 1 year
work backwards
A=P(r+1)^t
last one
A=80000
P=?
r=0.08
t=1 year
80000=P(1.08)^1
divide both sides by 1.08
I would leave in fraction
20000000/27=P
now that is the remaining after paying 50000, after 2 years of compounding
so
50000+(2000000/27)=P(1.08)^2
solve using math
about
106374=P
now reverse back
5 years
paid 30000
30000+106374=P(1.08)^5
solve using math
92813.526=P
round
$92813.53
put $92813.53 in the fund
Answer:
Exponent laws:
1. Product law
In product law if bases are same then we add their respective powers.But if bases are different we can't add their powers.
x=base, a,b,c=exponent
If x=2 and a=3, b=5 , and c=10, then
2.Product raised to a power
1. ![[x^{a}]^{c}=x^{ac}](https://tex.z-dn.net/?f=%5Bx%5E%7Ba%7D%5D%5E%7Bc%7D%3Dx%5E%7Bac%7D)
2. ![[x^{a}\times x^{b}]^{c}=[x^{a+b}]^{c}=x^{ac+bc}](https://tex.z-dn.net/?f=%5Bx%5E%7Ba%7D%5Ctimes%20x%5E%7Bb%7D%5D%5E%7Bc%7D%3D%5Bx%5E%7Ba%2Bb%7D%5D%5E%7Bc%7D%3Dx%5E%7Bac%2Bbc%7D)
If product is raised to a certain power , keeping the base same , we just multiply the powers.for example
![[2^{3}]^{4}=2^{3\times4}=2^{12}](https://tex.z-dn.net/?f=%5B2%5E%7B3%7D%5D%5E%7B4%7D%3D2%5E%7B3%5Ctimes4%7D%3D2%5E%7B12%7D)
and
![[2^{3}\times3^{2}]^{2}=[2^{3}]^2 \times[3^{2}]^{2}=2^{6}\times3^{4}](https://tex.z-dn.net/?f=%5B2%5E%7B3%7D%5Ctimes3%5E%7B2%7D%5D%5E%7B2%7D%3D%5B2%5E%7B3%7D%5D%5E2%20%5Ctimes%5B3%5E%7B2%7D%5D%5E%7B2%7D%3D2%5E%7B6%7D%5Ctimes3%5E%7B4%7D)
![[2^{3}\times2^{2}]^{2}=[2^{3+2}]^{2}=[2^{5}]^{2}=2^{10}](https://tex.z-dn.net/?f=%5B2%5E%7B3%7D%5Ctimes2%5E%7B2%7D%5D%5E%7B2%7D%3D%5B2%5E%7B3%2B2%7D%5D%5E%7B2%7D%3D%5B2%5E%7B5%7D%5D%5E%7B2%7D%3D2%5E%7B10%7D)
I am not sure if the “+” was supposed to go there or if that was just a typo, so I do apologize if this is wrong... but if it doesn’t belong in the equation at all, 20 = 2X is a fairly simply multiplication problem. You are to divide both sides by the 2 and you should get X = 10, because 2 goes into 20, 10 times; and the 2 goes into 2 only once, therefore the 2 now gets cancelled out.
Answer: 20 divided by 2 = 2x divided by 2 is now 10 = X.
Really hope this helps you, have a good day :)
