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abruzzese [7]
3 years ago
5

Could some one help me in this?

Mathematics
2 answers:
Sunny_sXe [5.5K]3 years ago
7 0
3.14 x r^2 I belive this is right
bekas [8.4K]3 years ago
3 0

Answer:

A = 3.14 × r²

Step-by-step explanation:

A = πr²

A = 3.14 × r²

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Zoltan has a list of whole numbers, all larger than 0 but smaller than 1000. He notices that every number in his list is either
timama [110]

The largest number of different whole numbers that can be on Zoltan's list is 999

<h3>How to determine the largest number?</h3>

The condition is given as:

Number =  1/3 of another number

Or

Number = 3 times another number

This means that the list consists of multiples of 3

The largest multiple of 3 less than 1000 is 999

Hence, the largest number of different whole numbers that can be on Zoltan's list is 999

Read more about whole numbers at:

brainly.com/question/19161857

#SPJ1

4 0
2 years ago
The population of a colony of mosquitoes obeys the law of uninhibited growth. if there are 1000 mosquitoes initially and there a
inna [77]
<span>6554 Assuming that the growth the function will be of the form f(n) = P*R^n where P = Initial population R = Growth rate per period n = number of periods In this problem, the only period or interval I see are days. So we'll use n to represent the number of days since the start. The initial population is 1000, and the ratio would be 1600/1000 = 1.6. So the formula becomes f(n) = 1000*1.6^n So let's evaluate that function using n=4. f(n) = 1000*1.6^n f(4) = 1000*1.6^4 f(4) = 1000*6.5536 f(4) = 6553.6 Since we can't have a fractional mosquito, round to the nearest integer, giving 6554.</span>
7 0
3 years ago
How many five-digit integers (integers from 10,000 through 99,999) are divisible by 5?
lukranit [14]
Okay to solve this problem you would simply do this solution. It took me a minute, haven't had to do integers in a while. :D

From 10,000 to 99,999 the numbers divisible by 5 ends in 0 or 5.
The first (most significant) digit can be any of nine from 1 to 9.
The 2nd digit can be any of 10.
The 3rd digit can be any of 10.
The 4th digit can be any of 10.
The 5th digit can be either of 2.

<span>That's 9(10)(10)(10)2=18,000
</span>
I hope this helps. 
7 0
4 years ago
What is the square root of 216?
KATRIN_1 [288]
\sqrt{216}=\sqrt{36\cdot6}=\sqrt{36}\cdot\sqrt6=6\sqrt6


216|2\\108|2\\.\ 54|2\\.\ 27|3\\.\ \ 9|3\\.\ \ 3|3\\.\ \ 1\\\\216=2^2\cdot3^2\cdot2\cdot3\\\\\sqrt{216}=\sqrt{2^2\cdot3^2\cdot2\cdot3}=\sqrt{2^2}\cdot\sqrt{3^2}\cdot\sqrt6=2\cdot3\cdot\sqrt6=6\sqrt6
5 0
3 years ago
A rectangle has an area of 2020cm².
Nadya [2.5K]

Answer:

Length = 2020 units

Width = 1 unit

Step-by-step explanation:

We know that area of a rectangle = length x width

We also know  that perimeter = (2 x Length) + (2 x Width)

the goal is to to find Length and Width such that:

Condition 1: Length x width = 2020

Condition 2: (2 x Length) + (2 x Width)= maximum possible

We are also given that both Length and Width are whole numbers, hence we can start by finding the factors of 2020 into prime factors

Factor 2020: 2 x 2 x 5 x 101

By observation, we can see that the following combinations of the factors make up the required area of 2020:

Case 1: (1) x (2)(2)(5)(101) = 1 x 2020 = 2020

Perimeter = 2(1 + 2020) = 4042

Case 2: (2) x (2)(5)(101) = 2 x 1010 = 2020

Perimeter = 2(2+ 1010) = 2024

Case 3: (2)(2) x (5)(101) = 4 x 505 = 2020

Perimeter = 2(4 + 505) = 1018

Case 4: (2)(2)(5) x (101) = 20 x 101 = 2020

Perimeter = 2(20 + 101) = 242

Case 5: (5) x (2)(2)(101) = 5 x 404 = 2020

Perimeter = 2(5 + 404) = 818

Case 6: (2)(5) x (2)(101) = 10 x 202 = 2020

Perimeter = 2(10 + 202) = 424

From the above, it is clear that case 1 yields the largest perimeter

4 0
4 years ago
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