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2 years ago

Which of the following is the graph of the absolute-value parent function?

1 answer:

4 0

Should have the ordered pairs of
(-4,4) (-3,3) (-2,2) (-1,1) (0,0) (1,1) (2,2) (3,3).

Will look like a V when graphed. Reason being is if you take the absolute value of anything you are basically just dropping the negative sign as it is how far away it is from 0.

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a grocery store display has 2 boxes of cereal on the top row,3 boxes in the next row,4 boxes in the next row, and so on. How man

sveticcg [70]

There ya go! The answer is 8 rows. Added a picture to show you my work :)

8 0

2 years ago

Read 2 more answers

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

How to use a 12th grade calculator

kondaur [170]

Click random buttons until you figure it out, or even ask a parent or teacher

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2 months ago

Don randomly draws two cards from a standard deck of 52 cards. He does not replace the first card. What is the probability that

kramer

Easy
so remember
4 aces in a deck
probability =(desired outcomes)/(total possible)

total possible=52

4/52=1st card
then since he picked 1 out the total possible is and assuming he drwe an ace 3/51
so the answer is 4/52 times 3/51=1/221

7 0

2 years ago

Read 2 more answers

Which steps show how to use the distributive property to evaluate 7 . 82?

ryzh [129]

Answer is sus/11:,677(,//linkS.org

5 0

2 years ago