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Brilliant_brown [7]
3 years ago
11

john and o'niel are playing a board game in which they roll two number cubes. John needs to get a sum of 8 on the number cubes t

o win. oniel needs a sum of 11. If they take turns rolling the number cube who is more likely to win
Mathematics
1 answer:
Musya8 [376]3 years ago
3 0

Answer:

John

Step-by-step explanation:

John either needs

2+6

3+5

4+4

5+3

6+2

Oniel eith needs

5+6

6+5

John is more likely

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Rewrite the equation in Ax+By=C form.<br>Use integers for A, B, and C.<br><br>y-1=-5(x-5)​
Varvara68 [4.7K]

Answer:

5x + y = 26

Step-by-step explanation:

Step 1: Distribute

y - 1 = -5x + 25

Step 2: Add 1 to both sides

y = -5x + 26

Step 3: Move 5 to the other side

5x + y = 26

7 0
4 years ago
Plz with steps .. it's very hard can anyone plz
liubo4ka [24]

Answer:

Step-by-step explanation:

\displaystyle\  \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\frac{0}{0} \\\\we\ can \ use\ Hospital's\ Rule\\\\\\f(x)=\sqrt{2x}-\sqrt{3x-a}  \qquad  f'(x)=\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}} \\\\g(x)=\sqrt{x} -\sqrt{a}  \qquad g'(x)=\dfrac{1}{2\sqrt{x}} \\\\\\\displaystyle\  \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\lim_{n \to a} \dfrac{\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}}  }{\dfrac{1}{2\sqrt{x}} }\\\\

\displaystyle \lim_{n \to a} \dfrac{2\sqrt{x} }{\sqrt{2x}} -\dfrac{3*\sqrt{x} }{\sqrt{3x-a}}  =\lim_{n \to a} \dfrac{2 }{\sqrt{2}} -\dfrac{3*\sqrt{x} }{\sqrt{3x-a}}\\\\\\=\dfrac{2}{\sqrt{2}} -\dfrac{3*\sqrt{a} }{\sqrt{2a}}\\\\\\=\dfrac{2}{\sqrt{2}} -\dfrac{3}{\sqrt{2}}\\\\\\=-\ \dfrac{1}{\sqrt{2}}\\\\

7 0
3 years ago
Fanga ran 28 miles per in 4 hours She ran at a rate of how many miles per hour
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5 0
3 years ago
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a team played 18 games during the season. the team lost 8 games and won the rest. what is the ratio of wins to losses
Westkost [7]

8 out of 18 games. That 8/18 simplify to 4/9 ?

7 0
3 years ago
Help me solve them plsss...both of the questions :)
zaharov [31]

\bold{\text{Answer:}\quad \dfrac{\sqrt2}{144}}

<u>Step-by-step explanation:</u>

When you plug in 3 directly to the equation, you get 0/0

Since it is indeterminate, you have to use L'Hopital's Rule.

That means that you find the limit of the DERIVATIVE of the numerator and the DERIVATIVE of the denominator.

\dfrac{d}{dx}(\sqrt{5+\sqrt{2x+3}}-2\sqrt2)\quad = \dfrac{1}{2\sqrt{5+\sqrt{2x+3}}(\sqrt{2x+3})}\\\\\\f'(3)\ \text{for the numerator}\ =\dfrac{1}{12\sqrt2}\\\\\\\\\dfrac{d}{dx}(x^2-9) = 2x\\\\\\f'(3) \text{for the denominator}\quad = 6\\\\\\\dfrac{f'(3)\ \text{numerator}}{f'(3)\ \text{denominator}}\quad = \dfrac{\dfrac{1}{12\sqrt2}}{6}\quad = \dfrac{1}{72\sqrt2}\quad \rightarrow \large\boxed{\dfrac{\sqrt2}{144}}

4 0
3 years ago
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