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salantis [7]
4 years ago
6

Show that the cube of positive integer is 6q+r ,where q is an integer & r=0,1,2,3,4,5

Mathematics
1 answer:
den301095 [7]4 years ago
4 0

Answer:

6(6)² + 0 = 6³

6(0) + 1 = 1³

6(1) + 2 = 8 = 2³

6(4) + 3 = 27 = 3³

6(10) + 4 = 64 = 4³

6(20) + 5 = 125 = 5³

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Pls help due in 5 min
Ainat [17]

Answer:

Enlargement? or try Translation

5 0
3 years ago
Which of the following statements is true about the number 30/2?
olga nikolaevna [1]

Answer:

It's A :)

Step-by-step explanation:

30/2 is 15

15 is a whole number so it applies to everything except for irrational numbers

Hope this helps :)

5 0
2 years ago
What is 2. 3/6 +1/6 +1 2/3
Kaylis [27]
3/6+1/6+1 2/3
Make 1 2/3 into an improper fraction:
5/3
Give them all common denominators by multiplying both sides of 5/3 by 2:
10/6
Then the question becomes:
3/6+1/6+10/6
Now you add the numerators and the denominator stays the same:
14/6
Now you can simplify by dividing both sides by 2:
7/3
Now you can make it a mixed number:
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Hope this helps :)
8 0
3 years ago
Read 2 more answers
PLEASE HELP!!<br> Find the missing side lengths. Round decimals to the nearest tenth.
Brrunno [24]

Answer: 36.2

Step-by-step explanation:

7 0
2 years ago
Vicki started jogging the first time she ran she ran 3/16 mile the second time she ran 3/8 mile and the third time she ran 9/16
IrinaK [193]

Answer:

Jogging 6th time.

Step-by-step explanation:

We have been given that Vicki started jogging the first time she ran she ran 3/16 mile the second time she ran 3/8 mile and the third time she ran 9/16 mile.

We can see that the distance Vicki covers each time forms a arithmetic sequence, where 1st term is 3/16.

We know that an arithmetic sequence is in form a_n=a_1+(n-1)d, where,

a_n = nth term of sequence,

a_1 = 1st term of sequence,

n =  Number of terms in sequence,

d = Common difference.

Let us find common difference of our given sequence as:

\frac{3}{8}-\frac{3}{16}\Rightarrow \frac{6}{16}-\frac{3}{16}=\frac{3}{16}

Since Vicki needs to cover more than 1 mile, so we nth term of sequence should be greater than 1.

1

Let us solve for n.

1

1

1\cdot \frac{16}{3}

5.333

n>5.333

We can also write next terms of our sequence as:

\frac{3}{16},\frac{6}{16}, \frac{9}{16},\frac{12}{16},\frac{15}{16},\frac{18}{16}

Therefore, Vicki will run more than 1 mile when she is jogging for 6th time.

7 0
3 years ago
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