You roll a number cube twice. Find the probability of the event. Rolling a 3 then a 3
1 answer:
Answer:
2/12
Step-by-step explanation:
A number cube has 6 sides.
Only one side has a "3".
So the probability of rolling a 3 would be 1/6
But you have to find it for 2 roles of " 3"
So you multiply 1/6* 1/6 to get 2/12 which simplifies to 1/6
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Answer:
-16
Step-by-step explanation:
y = mx
-4 = 1/4x
Divide both sides by 1/4.
-16 = x
Answer:
Step-by-step explanation:
Yes because it has a constant rate of change
check the picture below.
![\bf (\stackrel{a}{-20},\stackrel{b}{-21})\impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=\stackrel{adjacent}{-20}\\ b=\stackrel{opposite}{-21}\\ \end{cases} \\\\\\ c=\sqrt{(-20)^2+(-21)^2}\implies c=\sqrt{841}\implies c=29 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill cos(\theta )=\cfrac{\stackrel{adjacent}{-20}}{\stackrel{hypotenuse}{29}}~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%28%5Cstackrel%7Ba%7D%7B-20%7D%2C%5Cstackrel%7Bb%7D%7B-21%7D%29%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Bhypotenuse%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%20c%3D%5Csqrt%7Ba%5E2%2Bb%5E2%7D%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3D%5Cstackrel%7Badjacent%7D%7B-20%7D%5C%5C%20b%3D%5Cstackrel%7Bopposite%7D%7B-21%7D%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20c%3D%5Csqrt%7B%28-20%29%5E2%2B%28-21%29%5E2%7D%5Cimplies%20c%3D%5Csqrt%7B841%7D%5Cimplies%20c%3D29%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20cos%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Badjacent%7D%7B-20%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B29%7D%7D~%5Chfill)
Answer:
The midsegment is one-half the length of the base
The answr is 2 okay that’s the answer
