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2 years ago

Can anyone answer please I don't know what to do.

Mathematics

1 answer:

ZanzabumX [31]2 years ago

5 0

Answer: 2 7/10

Step-by-step explanation:

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Make a function table to represent his total savings for 2,4,6,8 and 10 months.

OlgaM077 [116]

Well, let's say that bananas cost $1.60 per pound. That means that at step 0 the value is 0, at step 1 the value is$1.60, at step 2 the value is $$3.20 and so on. To make a table from this we'd need two columns and four rows. On the top row we right step followed by amount. In the step column, we list the numbers 0-2. In the amount column we list 0, $1.60, and $3.20. In the end, a function table looks like this:

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2 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

2 years ago

Level B

Darya [45]

The number has 2 digits - let's call them x and y. The positive difference between the two combinations (lowest-highest and highest-lowest) is always equal to the positive difference of x and y, multiplied by 9.

For example:

76 and 67: 76 - 67 = 9 and ((7-6)*9) = 1*9 = 9
73 and 37: 73 - 37 = 36 and ((7-3)*9) = 4*9 = 36

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2 years ago

Math test pls help me

Natalija [7]

Answer:

D. $\frac{45}{x^{10}}$

Step-by-step explanation:

$(3 {x}^{ - 4} )^{2} (5 {x}^{ - 2} ) \\ \\ = (9 {x}^{ - 8} )(5 {x}^{ - 2} ) \\ \\ = 9 \times 5 \times {x}^{ - 8 - 2} \\ \\ = 45 {x}^{ - 10} \\ \\ = \frac{45}{ {x}^{10} }$

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1 year ago

Liam has 3 games and 10 songs on his computer. How many outcomes are possible when randomly choosing a song and a game?

cupoosta [38]

3 outcomes are possible because you only have 3 games on it, and it is mandatory to only have 1 game and 1 song at a time.

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2 years ago