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2 years ago

How do I find the quotient 7/8÷2/9

2 answers:

6 0

7/8 / 2/9 = 7/8 x 9/2
You solve is by multiplying the first fraction by the flipped version of the second fraction :)

Mandy Woodland

Lena [83]2 years ago

5 0

$\dfrac{\frac{7}{8}}{\frac{2}{9}}=\frac{7}{8}\cdot\frac{9}{2}=\frac{63}{16}$

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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

Please help!! 15 points!!

vampirchik [111]

$- 1 = - 3 - 4 \cos(4θ + \frac{2\pi}{3} ) \\ - 1 + 4 \cos(4θ + \frac{2\pi}{3} ) = - 3 \\ 4 \cos(4θ + \frac{2\pi}{3} ) = - 3 + 1 \\ 4 \cos(4θ + \frac{2\pi}{3} ) = - 2 \\ \cos(4θ + \frac{2\pi}{3} ) = - \frac{1}{2} \\ \cos(4θ + \frac{2\pi}{3} ) = - \frac{1}{2} \\ \cos(2\pi - (4θ + \frac{2\pi}{3} )) = - \frac{1}{2} \\ 4θ + \frac{2\pi}{3} = arccos( - \frac{1}{2} ) \\ \cos( \frac{4\pi}{3} - 4θ) = - \frac{1}{2}$

3 0

2 years ago

Read 2 more answers

Find x 25 7 х I need help

Murljashka [212]

Answer:

24 = x

Step-by-step explanation:

You use Pythagorean theorem. Since the normal equation is $a^{2} + b^{2} =c^{2}$ , you already have the $c^{2}$ and another side which can be $a^{2}$ or $b^{2}$ .