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2 years ago

Do u all work on algebra reasoning

Mathematics

2 answers:

Marysya12 [62]2 years ago

4 0

What is the question ?

Mice21 [21]2 years ago

3 0

Answer:

whats the question?

Step-by-step explanation:

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Help please this i will mark brainliest please

solniwko [45]

Step-by-step explanation:

please complete your question

6 0

3 years ago

At CD Shack, all used CDs are the same price. Jesse spent $54 and Mimi spent $66 on used CDs. What is the most each used CD coul

Alinara [238K]

Hey there
__________
The correct answer is
Whatever each CD costs, what each person paid is that cost times the number of CDs purchased (no sales tax for this problem).
So, the price of one CD is a factor of $66 (a number of $ that divides $66 evenly).
In theory, it could be $1, $2, $3, $6, $11, $22, $66.
It could even be $0.50, $0.25, $0.20, $0.10, $0.05,...
Also, the price of one CD must be a factor of $54. such as $54,$27,$18,$9,$6,$3,$2,$1,... .
You are looking for the most that price could be.

The grew greatest price that is in both lists is $6.

How can you make those lists?
You can start with the total price, then the price divided by 2, by 3, by whatever whole number you can divide it.

Otherwise, you could find the greatest common factor of 66 and 54
from the prime factorization of both numbers.
___________________
Hope this helps you

5 0

3 years ago

Which table represents a linear function?

GalinKa [24]

Answer:

The top right option.

Step-by-step explanation:

3 0

3 years ago

What is 122/11 equivalent to as a decimal??

kipiarov [429]

divide it two numbers to get 11.090909 which would most likely to be 11.10

3 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

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