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3 years ago

We have 5 circles surrounding a circle, Is there enough room to fit one more circle

Mathematics

1 answer:

Alexus [3.1K]3 years ago

4 0

Answer:

Definitely

Step-by-step explanation:

In order to completely surround a circle, you need six circles to do that while here in the question it is mentioned that currently there are only five circles surrounding the circle, hence there is enough room to easily fit one more circle.

I just hope that you are satisfied with the answer, Best of Luck.

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Can somebody prove this mathmatical induction?

Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

$\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2$

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

3 step:

Check for n=k+1 whether the statement

$\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)$

is true.

Start with the left side:

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)$

According to the 2nd step,

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

Substitute it into the $\ast$

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)$

So, you have proved the initial statement

4 0

3 years ago

What is the area of the figure below?

svetlana [45]

24 cm sq is the answer

8 0

2 years ago

Read 2 more answers

On the farm, rather than use money, they exchange animals. The exchanges are as follows: 4 bunnies = 3 chickens, 5 chickens = 2

Radda [10]

We have to trade 5 bunnies for a donkey.

Solution:

To calculate how many bunnies could be exchanged for a donkey, we have to multiply the exchange rates of each animal/bird.

One bunny = 3/4 chickens (0.75 chicken),

One chicken = 2/5 pigs (0.4 pigs)

One pig = 2/3 donkeys (0.67 donkeys).

On multiplying all of the above rates we get,

0.75*0.4*0.67=0.2

Since we now know a bunnies worth is 0.2 donkey

Therefore, (1/0.2=5) 5 bunnies to trade for a donkey.

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3 years ago

Find the interquartile range of the data.

likoan [24]

The answer is 14. Here is how I work it out.

First we are going have to identify quartile 1 and quartile 3.

So after putting the numbers in order from least to greatest mark the number with a half way point.

This is optional but it will help us spot the sections better.

31,33,35,41,43,|46,48,49,49,50

Next put parentheses around the remaining groups of numbers.

(31,33,35,41,43,)|(46,48,49,49,50)

For the next step we have to find the median of each group.

(31,33,35,41,43,)|(46,48,49,49,50)

The median of each group, are called the quartiles, the median of the lower half is quartile 1, and the median of the upper half is quartile 3.

Now subtract quartile 1 from quartile 3.

49 – 35 = 14

So the interquartile range of this data set is 14.

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3 years ago

Write an equation for the relationship between time and distance for each horse

alexandr402 [8]

Idekxbbdbdbdbbddbbddbbdd

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2 years ago