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den301095 [7]
3 years ago
11

The quotient of a number and 3 decreased by 121 is -164. Select the solution to the sentence shown.

Mathematics
1 answer:
docker41 [41]3 years ago
6 0
Its algebra. The original equation is \frac{x}{3} -121 = -164

To solve for a variable, we reverse the order of operations, beginning with addition/subtraction, and then multiplication/division. To remove a number from one side, we must do the opposite to the other side. In this case, to get rid of the -121 we must add 121 to the -164. This gives us -43. Then, to get the x by itself, we must multiply the other side by 3. -43*3=129

When we are doing the opposite of an operation to the other side, we are really reversing the operation and, to keep both sides equal, we must do whatever we have done to one side to the other side. So when we have -121, we add 121 as it equals 0, therefore it is gone. Since a equation must be balanced, we have to do what we did to the other side (adding 121).



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How many terms are there in the sequence 1, 8, 28, 56, ..., 1 ?
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Step-by-step explanation:

Given:  

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

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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

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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

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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

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