Answer:
what ,boy
Step-by-step exp lanation:
His sum should be more than 48.
I know because if you throw away the decimal pieces
and just add up the whole numbers 45 and 3 ,
right there you would have 48 . So when you include
the decimal pieces, they'll make it even bigger.
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)
Hi there!
So let's see, we have a die and need to know the probability of rolling a number less than or equal to 4. Let's list the numbers that are less than or equal to 4: 1, 2, 3, 4. Now, since we know that there are 6 numbers on a die and 4 of them are less than or equal to 4, we can set up a fraction to find the percentage. The fraction would be 4/6 because 4 out of the 6 numbers on the die are less than or equal to 6. We can simplify 4/6 to 2/3 as well. To find the percentage, all we need to do is divide the numerator by the denominator. This leaves us with approximately 66.66%.
Hope this helps!! :)
If there's anything else that I can help you with, please let me know!
<span>S = 2Πrh + 2Πr2
Manipulating the equation for h.
Step 1. subtract </span><span>2Πr2 on both sides.
</span>S - 2Πr2 = 2Πrh + 2Πr2 - 2Πr2
S - 2Πr2 = <span>2Πrh
</span>
Step 2 . divide <span>2Πr on both sides
</span>
(S - 2Πr2)/2Πr = 2Πrh/<span>2Πr
</span><span>h = </span>(S - 2Πr2)/2Πr
