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skelet666 [1.2K]
3 years ago
14

OMG PLEASE HELP ME PLESE

Mathematics
1 answer:
slava [35]3 years ago
7 0

Answer:

use demos

Step-by-step explanation:

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Which number line shows the solutions of 13x−8≥x+6?
Leviafan [203]

13x - x \geqslant  6 + 8 \\ 12x \geqslant 14 \\ x \geqslant  \frac{14}{12}  =  \frac{7}{6 }  = 1.167 \\ x \geqslant 1.167

i hope it helps u

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3 years ago
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X y = x² y = 2x<br> 0 0 1<br> 1 <br> 2
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2

Step-by-step explanation:

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Select ALL of the true statements about the line and it’s equation. ( don’t provide a link it does not work when looking for ans
mr Goodwill [35]

Answer: I need points to give out for answers-

Step-by-step explanation:

7 0
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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