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Butoxors [25]
3 years ago
13

One interior angle of a triangle is 30⁰. The other angle measures are x⁰ and y⁰. Which of the following could be the values of x

and y? Select ALL that apply
Mathematics
1 answer:
avanturin [10]3 years ago
8 0

Step-by-step explanation:

One interior angle of a triangle = 30⁰ .

The measure of other two are x and y .

We know that the angle sum property of a triangle is 180⁰ .

=> x + y + 30⁰ = 180⁰

=> x + y = 180⁰ - 30⁰.

=> x + y = 150⁰.

Possible solution set of x and y are ,

(1,149) (2,148 ) ...... (148+2) & (149+1)

Basically we are here decreasing y and increasing x to obtain the values .

Hope it helped !

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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
sleet_krkn [62]

This question is incomplete, the complete question is;

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 are written in increasing order but are not necessarily distinct.

In other words, how many 5-tuples of integers  ( h, i , j , m ), are there with  n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1 ?

Answer:

the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Step-by-step explanation:

Given the data in the question;

Any quintuple ( h, i , j , m ), with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1

this can be represented as a string of ( n-1 ) vertical bars and 5 crosses.

So the positions of the crosses will indicate which 5 integers from 1 to n are indicated in the n-tuple'

Hence, the number of such quintuple is the same as the number of strings of ( n-1 ) vertical bars and 5 crosses such as;

\left[\begin{array}{ccccc}5&+&n&-&1\\&&5\\\end{array}\right] = \left[\begin{array}{ccc}n&+&4\\&5&\\\end{array}\right]

= [( n + 4 )! ] / [ 5!( n + 4 - 5 )! ]

= [( n + 4 )!] / [ 5!( n-1 )! ]

= [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Therefore, the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

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The quadratic equation 2x^2 + bx + 18 = 0 has a double root. Find all possible values of b.
Doss [256]
When a quadratic equation ax^2+bx+c has a double root, the discriminant,
D=b^2-4ac=0
Here 
a=2,
b=b,
c=18
and 
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solve for b
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=> b= ± sqrt(144)= ± 12

So in order that the given equation has double roots,  the possible values of b are ± 12.
4 0
4 years ago
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