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3 years ago

Find six rational numbers between 9 and 10

Mathematics

1 answer:

olga55 [171]3 years ago

4 0

Answer:

To find 6 rational numbers between 9 and 10 are

9x7\1x7 10x7\1x7

63\7 70\7

so, 6 rational numbers are:

64\7,65\7,66\7,67\7,68\7and 69\7

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Can someone help with this!

Sveta_85 [38]

In the given image, $a_{23}$ is 5. The correct option is the first option 5

<h3>Matrix notation</h3>

From the question, we are to determine which entry in the given matrix represents $a_{23}$

NOTE: For any entry denoted as $a_{mn}$ , it represents the entry in the <em>m</em>-th row and <em>n</em>-th column.

Thus,

$a_{23}$ represents the entry in the 2nd row and 3rd column.

In the given image, the entry in the 2nd row and 3rd column is 5.

Hence, in the given image, $a_{23}$ is 5. The correct option is the first option 5

Learn more on Matrix notation here: brainly.com/question/2382978

#SPJ1

5 0

2 years ago

7. Find the complex conjugate of 3i+4.

Jet001 [13]

4-3i is the complex conjugate

5 0

3 years ago

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 that $n\ge h\ge i\ge j\ge k\ge m\ge 1$

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

Identify each function that has a remainder of -3 when divided x+6

Sergeu [11.5K]

Answer:

Step-by-step explanation:

According to remainder theorem, you can know the remainder of these polynomials if you plug in x = -6 into them.

<em>So we will plug in -6 into x of all the polynomials ( A through D) and see which one equals -3.</em>

<em />

$x^5 + 2x^2 - 30x + 30\\=(-6)^5 + 2(-6)^2 - 30(-6) + 30\\=-7494$

For B:

$x^4 + 4x^3 - 21x^2 - 53x + 12\\=(-6)^4 + 4(-6)^3 - 21(-6)^2 - 53(-6) + 12\\=6$

For C:

$x^3 - 10x^2 - 7\\=(-6)^3 - 10(-6)^2 - 7\\=-583$

For D:

$x^4 + 6x^3 - 10x - 63\\=(-6)^4 + 6(-6)^3 - 10(-6) - 63\\=-3$

The only function that has a remainder of -3 when divided by x + 6 is the fourth one, answer choice D.

5 0

3 years ago

Meenu bought two fans for ₹1200 each. She sold one at a loss of 5% and the other at a profit of 10%. Find the selling price of e

Vinvika [58]

<h3><u>Solution: </u></h3>

Overall<u> </u>CP of each fan = ₹1200 .

One is sold at a loss of 5% .

( This means if CP is ₹100, SP is ₹95 ) .

• Therefore,When CP is ₹1200 , Then SP is ₹ 1140.

$= > \frac{95}{100} \times 1200 = 1140$

Also,Second fan is sold at a profit of 10% .

It means , If CP is ₹100 , SP is ₹110.

Therefore , When CP is ₹1200 , Then SP is ₹1320.

<u>• We need to find the combined CP and SP to say whether there was an overall profit or Loss.</u><u>.</u>

Total CP = ₹ 1200 + ₹ 1200 = ₹ 2400.
Total SP = ₹ 1140 + ₹ 1320 = ₹ 2460.

Since total SP > total CP , A profit of ₹ ( 2460 - 2400 ) or ₹60 has been made ..

<h3>Hope this helps you :)</h3>

4 0

3 years ago

Find six rational numbers between 9 and 10​

Find six rational numbers between 9 and 10