Answer:
-2
Step-by-step explanation:
The population seems like a young age group.
Answer:
35.68°
Step-by-step explanation:
In the picture attached, the triangle is shown.
From sine definition:
sin(y) = opposite/hypotenuse
The leg measuring 7 units is opposite to angle y, and the hypotenuse measures 12 units, therefore:
sin(y) = 7/12
y = arcsin(7/12)
y = 35.68°
Answer:
n squared + 3n + 1
Step-by-step explanation:
5,11,19,29
Firstly look at the difference between each number. The first difference is 6 then 8 then 10 etc. After that you look at your created sequence - 6,8,10 etc. The difference is 2 each time. Then applying rules, you have to do the constant difference divided by 2 to get a coefficient of n squared. So in this case it's n squared because 2/2 = 1 so you don't have to place a 1 in front of the n squared. After you create a sequence from the n squared. That would be 1,4,9 etc. Then you need to see how to get from the sequence: 1,4,9 etc to your original sequence: 5,11,19 etc. So if you calculate it you will get 4,7,10 because firstly 5-1 = 4 then 11-4 = 7 etc. The sequence 4,7,10 is a linear sequence so the constant difference is 3 each time. So to get a nth term of a linear sequence you will start off as 3n then you will substitute 1 then 2 then 3 into the 3n. Therefore that would be 3,6 etc. So if you take the first substituted term, that would be 3 as said before then you will have to see how to get from the 3 to 4 so that is just adding 1. So the nth term of this linear sequence is 3n + 1. Check if it works at the end. So the overall nth term of the quadratic sequence is n squared as said before + 3n + 1.
Answer: 
Step-by-step explanation:
Given
Julien's parents want $64,000 at the end of 3 years
The rate of interest is 10% annually
Suppose they invested P amount initially
So, compound interest, compounded monthly is
![\Rightarrow A=P\left( 1+\frac{r}{n}\right)^{nt}\quad [\text{n=Numner of compounding periods}]\\\\\text{Insert the values}\\\Rightarrow 64,000=P\left( 1+\frac{0.10}{12}\right)^{12\times 3}\\\\\Rightarrow 64,000=P\left( 1+\frac{10}{12}\right)^{36}\\\\\Rightarrow P=\dfrac{64000}{1.3481}\\\\\Rightarrow P=\$\ 47,471.37\approx \$ 47,471](https://tex.z-dn.net/?f=%5CRightarrow%20A%3DP%5Cleft%28%201%2B%5Cfrac%7Br%7D%7Bn%7D%5Cright%29%5E%7Bnt%7D%5Cquad%20%5B%5Ctext%7Bn%3DNumner%20of%20compounding%20periods%7D%5D%5C%5C%5C%5C%5Ctext%7BInsert%20the%20values%7D%5C%5C%5CRightarrow%2064%2C000%3DP%5Cleft%28%201%2B%5Cfrac%7B0.10%7D%7B12%7D%5Cright%29%5E%7B12%5Ctimes%203%7D%5C%5C%5C%5C%5CRightarrow%2064%2C000%3DP%5Cleft%28%201%2B%5Cfrac%7B10%7D%7B12%7D%5Cright%29%5E%7B36%7D%5C%5C%5C%5C%5CRightarrow%20P%3D%5Cdfrac%7B64000%7D%7B1.3481%7D%5C%5C%5C%5C%5CRightarrow%20P%3D%5C%24%5C%2047%2C471.37%5Capprox%20%5C%24%2047%2C471)
