Ask question

Ask question

All categories

tekilochka [14]

3 years ago

15

11.

1 answer:

Ad libitum [116K]3 years ago

6 0

Answer:

he got 8 books free and now has 24 books total

Step-by-step explanation:

16/2=8

that would be the answer for how many books he got for free because if there is a box of 16 books and for every2 books you get one free then you would divide 16 and 2 and get 8 then

16+8=24

add the number of books he got for free with the original 16 books he already had and get 24 total books.

You might be interested in

HELP! Screen-printing a batch of shirts requires 4 minutes per shirt in addition to 6 minutes of initial set-up time. If it take

Artist 52 [7]

46-6(initial set up time)=40. 40/4=10 shirts

7 0

3 years ago

Read 2 more answers

Select the choice that translates the following verbal phrase correctly to algebra: y less than 21

Romashka [77]

Y (y) less than (minus) 21 (21
y-21

3 0

3 years ago

Read 2 more answers

3(x+y)^2-(xy)2 x=3 y=-5

diamong [38]

The value of the expression when x = 3 and y = -5 is -213

<h3>What are algebraic expressions?</h3>

Algebraic expressions are expressions made up of variables and constants with mathematical operations such as addition, subtraction, division, multiplication, etc

From the expression given;

3(x+y)^2-(xy)2

Where:

x = 3
y = -5

Substitute the values into the expression

3 ( 3 + -5)^2 - ( 3 × -5 )^2

3 ( -2)^2 - ( -15)^2

3(4) - ( 225)

-12

Thus, the value of the expression when x = 3 and y = -5 is -213

Learn more about algebraic expressions here:

brainly.com/question/4344214

#SPJ1

4 0

2 years ago

The legs of a right triangle have lengths of 4 and 5 units. Find the length of the hypotenuse.

Law Incorporation [45]

The square root of 41
or
6.40312424

5 0

3 years ago

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar

erma4kov [3.2K]

Answer:

$\frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}$

Step-by-step explanation:

Given

5 tuples implies that:

$n = 5$

$(h,i,j,k,m)$ implies that:

$r = 5$

Required

How many 5-tuples of integers $(h, i, j, k,m)$ are there such that $n\ge h\ge i\ge j\ge k\ge m\ge 1$

From the question, the order of the integers h, i, j, k and m does not matter. This implies that, we make use of combination to solve this problem.

Also considering that repetition is allowed: This implies that, a number can be repeated in more than 1 location

So, there are n + 4 items to make selection from

The selection becomes:

$^{n}C_r => ^{n + 4}C_5$

$^{n + 4}C_5 = \frac{(n+4)!}{(n+4-5)!5!}$

$^{n + 4}C_5 = \frac{(n+4)!}{(n-1)!5!}$

Expand the numerator

$^{n + 4}C_5 = \frac{(n+4)!(n+3)*(n+2)*(n+1)*n*(n-1)!}{(n-1)!5!}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5!}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5*4*3*2*1}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}$

<u><em>Solved</em></u>

6 0

3 years ago