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faltersainse [42]
3 years ago
7

What is 10% of 1/4 please help and explain

Mathematics
2 answers:
azamat3 years ago
8 0
To get the answer you divide 1/4 by 10 and get:
fraction form- 1/40
decimal form- 0.025

makvit [3.9K]3 years ago
5 0
10/100=0.1
1/4=0.25
0.1*0.25=0.025
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1/8 divided by 1/3 please show your work
Ahat [919]

Step-by-step explanation:

\frac{1}{8}  \div  \frac{1}{3}

\frac{1}{8}  \times 3

\frac{3}{8}

Alternate form

0.375

6 0
3 years ago
Read 2 more answers
Please include work!
aleksandrvk [35]

Answer:

A) y = x^2 -1

B) -y = 2x^2 +1  We multiply B) by -1

B) y = -2x^2 -1

We can then say x^2 -1 = -2x^2 -1

3 x^2 = 0

x = 0

************* Double-Check: ******************

Equation A) y = -1 and

Equation B) y = -1

Step-by-step explanation:

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4 years ago
If f(x) is a third degree polynomial function, how many distinct complex roots are possible
inessss [21]
2.
You will have on real root, and two complex which derives from the 3rd degree polynomial function.
7 0
3 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
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
I mean this. when write 1.
yawa3891 [41]
If the question is 7 x 200+50+6 it equals 1792
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