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grigory [225]
3 years ago
6

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

Mathematics
1 answer:
a_sh-v [17]3 years ago
8 0
The answer is B. Hope this helps
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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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The product of these two fractions would equal 15.

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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 thatn\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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