Ask question

Ask question

All categories

2 years ago

13

Please help!

1 answer:

Zarrin [17]2 years ago

5 0

<span>They use 9 blocks each, a total of 18 blocks.

Bill: 4b + 2
Ron: 8b - 5

To look for the value of b; I assume that both used the same number of blocks.

4b + 2 = 8b - 5
4b - 8b = -5 -2
-4b = -7
-4b/-4 = -7/-4
b = 1.75</span>

Bill: 4b + 2 where b = 1.75 ; then, 4(1.75) + 2 = 7 + 2 = 9
<span>Ron: 8b - 5 where b = 1.75; then, 8(1.75) -5 = 14 - 5 = 9</span>

You might be interested in

A pair of shoes cost $40 is marked up by 25%. What is the RETAIL PRICE?

Nitella [24]

The retail price is $50 because 25% of 40 is 10 and 10+40=50.

8 0

3 years ago

I need help if you answer this your littearly a pro hacker pls

Anit [1.1K]

Answer:

24 meters cubed

HOPE THIS HELPS

- Todo ❤️

Step-by-step explanation:

V=Bh

V=6*4

V=24

6 0

2 years ago

Read 2 more answers

Buying a 30 dollar shirt for $30 and getting a second shirt of equal value for 50%off

Goshia [24]

Answer:

you saved 15 bucks if you get the second one

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

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

2 years ago

HELPPP ASAP !!! it’s for a test

Marina86 [1]

Answer:

-4, 1, 2, 3

Step-by-step explanation:

I put them from least to greatest.

4 0

2 years ago

Read 2 more answers