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sdas [7]
3 years ago
12

Which is equal to 9 x 8?

Mathematics
2 answers:
Rufina [12.5K]3 years ago
5 0

Answer:

A

Step-by-step explanation:

Because

B= 54

C=45

D= 201

svet-max [94.6K]3 years ago
4 0
Are you have to do for the answer is just multiply those two numbers and then figure out if one of those multiple-choice questions have the same answers than that’s answer
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Find the x and y intercepts for the following equation. 4x - 2y = 16
irinina [24]

Answer:

( 4, 0 ) and ( 0,-8 )

Step-by-step explanation:

→ Rearrange into y = mx + c format

4x - 2y = 16

→ Minus 4x from both sides

-2y = -4x + 16

→ Divide everything by 2

-y = -2x + 8

→ Multiply everything by -1

y = 2x - 8

→ Substitute x = 0 to find the y-intercept and y = 0 to find the x intercept

x = 0 then y = -8 so ( 0,-8 )

y = 0 then x = 4 so ( 4, 0 )

4 0
3 years ago
Read 2 more answers
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

4 0
3 years ago
X^2-8x+10=-4 quadratic equation written in standard form
ASHA 777 [7]
Rewriting the formula in standard form would be x^2-8x+14=0
3 0
3 years ago
What is the value of the tangent of
kiruha [24]

Answer:

Step-by-step explanation:

With reference < H

perpendicular (p) = 12

base (b) = 5

so now

tangent of < H

= p / b

= 12 / 5

hope it helps :)

7 0
3 years ago
Read 2 more answers
(2x+5)/(3x)-:(2x-1)/(x+1)
andreyandreev [35.5K]

Answer:

2x(power of two)+7x+5

-------------over-------------------

6x(power of two) -3x

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
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