let's bear in mind that sin(θ) in this case is positive, that happens only in the I and II Quadrants, where the cosine/adjacent are positive and negative respectively.
![\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{6}}\qquad \impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-5^2}=a\implies \pm\sqrt{36-25}\implies \pm \sqrt{11}=a \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20sin%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B5%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B6%7D%7D%5Cqquad%20%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Badjacent%20side%7D%7D%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%5Cpm%5Csqrt%7Bc%5E2-b%5E2%7D%3Da%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20%5Cpm%5Csqrt%7B6%5E2-5%5E2%7D%3Da%5Cimplies%20%5Cpm%5Csqrt%7B36-25%7D%5Cimplies%20%5Cpm%20%5Csqrt%7B11%7D%3Da%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)

The answer to Q1 along with an explanation is shown in the picture;
Having a look at all the questions, they are all essentially the same type so can all be solved similarly;
So by following the method and explanation, you should be able to do the rest of the questions.
Here I'll Simplify it to show the work.
You need PEMDAS
9(2*2+4÷2)
9(4+4÷2)
9(8÷2)
9(4)
9(4) = 9 × 4 = 36
The Answer is 36.
Answer:
1/9
Step-by-step explanation:
(-3)^0 = 1
(-3)^2 = 9
so, 1/9
Answer:
1.98 × 10^10
Step-by-step explanation:
First solve for how many text messages were sent out per one month:
22,000,000 × 900
19,800,000,000
Convert into scientific notation:
1.98 × 10^10
In scientific notation, there can only be one number before the decimal. The rest of the numbers must be after the decimal; hence, why I put the numbers 9 and 8 after the decimal. If the rest of the numbers are zeros, you do not have to add them after the decimal. Instead, starting from right to left, count the number places until you reach your decimal. The number you have counted will be your exponent. I counted nine 0's, one 8, and one 9 which made this exponent 10 as there were 10 decimal placements.