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3 years ago

How do you solve the fraction 2 1/2 - 6

Mathematics

1 answer:

Slav-nsk [51]3 years ago

8 0

You could easily change the 2 1/2 to a 2.5 and do 2.5-6=-3.5

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You have $40 in your wallet, but you do not want to spend all of it. You want to have at least some money left. You find a shirt

lara31 [8.8K]

Answer:

Step-by-step explanation:

40-5=35

35 is left over so it is yout leftover change or money

6 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

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

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

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

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

Solved

6 0

2 years ago

Factorize: x² + 3x + 2

zvonat [6]

Step-by-step explanation:

x² + 3x + 2

x² + x + 2x + 2 because 1+2=3 and 2×1=2

By taking out common terms, we get

x(x+1)+2(x+1)

(x+1)(x+2)

7 0

3 years ago

4. Find the solution of y = 2x + 5 for x = 1. A (1, 7) B (3, 1) C (1, 12) D (1, -2)

Sati [7]

Answer:

Step-by-step explanation:

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3 years ago

What shape has eight edges?

wlad13 [49]

Answer:

A rectangular Pyramid

7 0

3 years ago

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