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3 years ago

Find the number of three digit natural numberswhich are divisible by 11

Mathematics

2 answers:

mixer [17]3 years ago

4 0

Answer:

90.

Step-by-step explanation:

1000 / 11

= 90.9.

USPshnik [31]3 years ago

3 0

Answer:

121. three digits. 121/11=11

Step-by-step explanation:

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2 × (2,230 + 829) Which of the following is true about the expression above? A. The given expression is 2 more than (2,230 + 829

goblinko [34]

Answer:

B - the expression given is 2 timex as large at (2,230 + 829)

Step-by-step explanation:

Multiply 2 by each set of numbers then add.

2X 2230 = 4460

2x829 = 1658

Add them together

4460+1658 = 6118

3 0

3 years ago

There are 80 people waiting to tour the science museum. The tour guide takes 4 groups of 16 people each. The remaining people wi

Anna11 [10]

Answer:

16 people

Step-by-step explanation:

Given that :

Number of people to tour = 80

Number per group = 16

Number of groups = 4

Total number of people who make up the entire groups :

Number per group * number of groups

16 * 4 = 64

The number of people in the last tour group :

80 - 64

= 16 people

7 0

3 years ago

What is 2.99 as a fraction

Serga [27]

2.99

2 is a whole number

0.99 = 99/100

your fraction is:

mixed number = 2 99/100
improper fraction = 299/100 (multiply whole number to denominator & then add numerator)

hope this helps

8 0

3 years ago

Read 2 more answers

Help me what the answers

Marina86 [1]

If it is perimeter do this L+W×2
If it's area do this L×W

3 0

3 years ago

I need help with this problem from the calculus portion on my ACT prep guide

LenaWriter [7]

Given a series, the ratio test implies finding the following limit:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=r$

If r<1 then the series converges, if r>1 the series diverges and if r=1 the test is inconclusive and we can't assure if the series converges or diverges. So let's see the terms in this limit:

$\begin{gathered} a_n=\frac{2^n}{n5^{n+1}} \\ a_{n+1}=\frac{2^{n+1}}{(n+1)5^{n+2}} \end{gathered}$

Then the limit is:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=\lim _{n\to\infty}\lvert\frac{n5^{n+1}}{2^n}\cdot\frac{2^{n+1}}{\mleft(n+1\mright)5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert$

We can simplify the expressions inside the absolute value:

$\begin{gathered} \lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert \\ \lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert=\lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert \\ \lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert=\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert \end{gathered}$

Since none of the terms inside the absolute value can be negative we can write this with out it:

$\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert=\lim _{n\to\infty}\frac{2}{5}\cdot\frac{n}{n+1}$

Now let's re-writte n/(n+1):

$\frac{n}{n+1}=\frac{n}{n\cdot(1+\frac{1}{n})}=\frac{1}{1+\frac{1}{n}}$

Then the limit we have to find is:

$\lim _{n\to\infty}\frac{2}{5}\cdot\frac{n}{n+1}=\lim _{n\to\infty}\frac{2}{5}\cdot\frac{1}{1+\frac{1}{n}}$

Note that the limit of 1/n when n tends to infinite is 0 so we get:

$\lim _{n\to\infty}\frac{2}{5}\cdot\frac{1}{1+\frac{1}{n}}=\frac{2}{5}\cdot\frac{1}{1+0}=\frac{2}{5}=0.4$

So from the test ratio r=0.4 and the series converges. Then the answer is the second option.

8 0

1 year ago

Find the number of three digit natural numberswhich are divisible by 11​

Find the number of three digit natural numberswhich are divisible by 11