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olasank [31]
4 years ago
9

Richordo has 2 cases of video games with the same number of games in each case he gives for games to his brother richordo has 10

games left how many video games were in each case
Mathematics
1 answer:
stepladder [879]4 years ago
3 0
X=Amount of games in each case.

He has 2 cases of videos games with the same number of games in each case:

2x+2x

He gives 4 away to his brother:

2x+2x-4

Now he has 10 games left:

2x+2x-4=10
4x-4+4=10+4
4x/4=14/4
X=3.5. round up to 4 because you can’t have .5 of a game.

Richordo had 4 video games in each case.
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Which equation(s) represents a proportional relationship?
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Answer:

A and C are proportional relationships.

Step-by-step explanation:

Recall that a "proportional relationship" has no constant term.  Thus, we eliminate Answers B and D immediately.

Answer A represents a proportional relationship:  y varies directly with the square of x.

Answer C also represents a proportional relationship:  y varies directly with x.

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3 years ago
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matrenka [14]

Answer: K=2

Step-by-step explanation:

-8k+6=-10k+10

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3 years ago
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Toko follows a budget to buy supplies to paint his room. He spends 1/2 of his budget on a roller, a tray, and tape. He spends $1
Natasha_Volkova [10]
The answer is $36.

B - budget
So the calculation is as following:
B= \frac{1}{2}B + 12.27 +  \frac{3.82}{ \frac{2}{3} } = \frac{1}{2}B + 12.27 + \frac{3.82*3}{2}=   \frac{1}{2}B + 12.27 +5.73
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3 0
3 years ago
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3 years ago
How many terms are there in the sequence 1, 8, 28, 56, ..., 1 ?
BabaBlast [244]

Answer:

9 terms

Step-by-step explanation:

Given:  

1, 8, 28, 56, ..., 1

Required

Determine the number of sequence

To determine the number of sequence, we need to understand how the sequence are generated

The sequence are generated using

\left[\begin{array}{c}n&&r\end{array}\right] = \frac{n!}{(n-r)!r!}

Where n = 8 and r = 0,1....8

When r = 0

\left[\begin{array}{c}8&&0\end{array}\right] = \frac{8!}{(8-0)!0!} = \frac{8!}{8!0!} = 1

When r = 1

\left[\begin{array}{c}8&&1\end{array}\right] = \frac{8!}{(8-1)!1!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8

When r = 2

\left[\begin{array}{c}8&&2\end{array}\right] = \frac{8!}{(8-2)!2!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} =2 8

When r = 3

\left[\begin{array}{c}8&&3\end{array}\right] = \frac{8!}{(8-3)!3!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56

When r = 4

\left[\begin{array}{c}8&&4\end{array}\right] = \frac{8!}{(8-4)!4!} = \frac{8!}{4!3!} = \frac{8 * 7 * 6 * 5 * 4!}{4! *4*3* 2 *1} = \frac{8 * 7 * 6*5}{4*3 *2 *1} = 70

When r = 5

\left[\begin{array}{c}8&&5\end{array}\right] = \frac{8!}{(8-5)!5!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56

When r = 6

\left[\begin{array}{c}8&&6\end{array}\right] = \frac{8!}{(8-6)!6!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} = 28

When r = 7

\left[\begin{array}{c}8&&7\end{array}\right] = \frac{8!}{(8-7)!7!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8

When r = 8

\left[\begin{array}{c}8&&8\end{array}\right] = \frac{8!}{(8-8)!8!} = \frac{8!}{8!0!} = 1

The full sequence is: 1,8,28,56,70,56,28,8,1

And the number of terms is 9

3 0
3 years ago
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