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Artemon [7]
3 years ago
13

Show your work please

Mathematics
1 answer:
castortr0y [4]3 years ago
4 0
Hey buddy here is your answer
Mark it brainliest

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A number cube has sides that are labeled 1 to 6. Jamal rolls the number cube. What is the probability that he will roll a 22?
Tems11 [23]

-- The probability of rolling a 22 is zero.  That result is impossible, because the sides are labeled with single digits 1 through 6 .  Since 22 is not printed anywhere on the cube, it can never come up.

-- The probability of rolling a<em> 2</em> , however, is <em> 1/6</em> . <em>(B)</em>

<em></em>

The -probability of rolling something you want is always

<em>(the number of different possible results that you like) </em>

divided by

<em>(the total number of different possible results)</em>

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3 years ago
30 = y - 2<br> and work plzzz
9966 [12]

Answer:

Step-by-step explanation:

y - 2 = 30

y = 32

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3 years ago
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What is the domain of the function f (x)=x^2+1
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Alright, so since anything can be squared, the domain is (-inf, inf)
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3 years ago
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7 0
3 years ago
Suppose z equals f (x comma y ), where x (u comma v )space equals space 2 u plus space v squared, y (u comma v )space equals spa
barxatty [35]

z=f(x(u,v),y(u,v)),\begin{cases}x(u,v)=2u+v^2\\y(u,v)=3u-v\end{cases}

We're given that f_x(6,1)=3 and f_y(6,1)=-1, and want to find \frac{\partial z}{\partial v}(1,2).

By the chain rule, we have

\dfrac{\partial z}{\partial v}=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial v}

and

\dfrac{\partial x}{\partial v}=2v

\dfrac{\partial y}{\partial v}=-1

Then

\dfrac{\partial z}{\partial v}(1,2)=\dfrac{\partial z}{\partial x}(6,1)\dfrac{\partial x}{\partial v}(1,2)+\dfrac{\partial z}{\partial y}(6,1)\dfrac{\partial y}{\partial v}(1,2)

(because the point (x,y)=(6,1) corresponds to (u,v)=(1,2))

\implies\dfrac{\partial z}{\partial v}(1,2)=3\cdot2\cdot2+(-1)\cdot(-1)=\boxed{13}

4 0
3 years ago
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