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3 years ago

6

In the diagram below, OP is circumscribed about quadrilateral ABCD. What is

1 answer:

a_sh-v [17]3 years ago

8 0

The answer is B. Hope this helps

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Who can give me some extra points I need it please

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3 years ago

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Find the product of 4 2/7 x 3 1/2

Fiesta28 [93]

Answer:

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4 years ago

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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

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

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

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

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

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

6 0

3 years ago

A=1/4 and b=6.<br><br> −8ab=

podryga [215]

Answer:

When $a=\frac{1}{4}$ and $b=6$ :

$-8ab=-12$

Step-by-step explanation:

-8ab can be seen as -8×a×b. Insert the given values:

$-8*\frac{1}{4} *6$

Simplify multiplication from left to right:

$-8*\frac{1}{4} \\\\\frac{-8}{1} *\frac{1}{4} =\frac{-8}{4} =-2$

Insert and solve:

$-2*6=-12$

:Done

3 0

3 years ago

Solve the equation -4x+12=-4 for x .

mihalych1998 [28]

-4x +12 -12 = -4 -12
-4x = -16
x = 4

4 0

3 years ago

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