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Gwar [14]
3 years ago
15

Find the area of the regular figure below:​

Mathematics
1 answer:
Mice21 [21]3 years ago
6 0

Answer:

1,496.49 cm squared

Step-by-step explanation:

The regular figure below is a hexagon.

The formula for the area of a regular hexagon is A = \frac{3\sqrt{3} }{2} a^2. Here a = 24. Substitute and simplify for the area.

A = \frac{3\sqrt{3} }{2} a^2\\A = \frac{3\sqrt{3} }{2} 24^2\\A = \frac{3\sqrt{3} }{2} * 576\\A = 1496.49

You might be interested in
how many ways can 8 distinguishable books be placed in 5 distinguishable shelves so that each shelf contains one book
musickatia [10]

There are 6720 ways by 8 distinguishable books be placed in 5 shelves.

According to statement

The number of books (n) is 8

The number of shelves (r) is 5

Now, we find the ways by which the 8 books be placed in 5 distinguishable shelves

From Permutation formula

P(n,r) = n! / (n-r)!

Substitute the values then

P(n,r) = 8! / (8-5)!

P(n,r) = (8*7*6*5*4*3*2*1) / (3*2*1)

P(n,r) = 8*7*6*5*4

P(n,r) = 6720

So, there are 6720 ways by 8 distinguishable books be placed in 5 shelves.

Learn more about PERMUTATION here brainly.com/question/11732255

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5 0
2 years ago
Find the values of a.b,andc in the table a= b= c=.
Svetach [21]

Answer:

a. is 1 and b is 2 and c is 3 I did it in algebra pay attention in class

4 0
2 years ago
Find an integer x such that 0<=x<527 and x^37===3 mod 527
Greeley [361]
Since 527=17\times31, we have that

x^{37}\equiv3\mod{527}\implies\begin{cases}x^{37}\equiv3\mod{17}\\x^{37}\equiv3\mod{31}\end{cases}

By Fermat's little theorem, and the fact that 37=2(17)+3=1(31)+6, we know that

x^{37}\equiv(x^2)^{17}x^3\equiv x^5\mod{17}
x^{37}\equiv(x^1)^{31}x^6\equiv x^7\mod{31}

so we have

\begin{cases}x^5\equiv3\mod{17}\\x^7\equiv3\mod{31}\end{cases}

Consider the first case. By Fermat's little theorem, we know that

x^{17}\equiv x^{16}x\equiv x\mod{17}

so if we were to raise x^5 to the nth power such that

(x^5)^n\equiv x^{5n}\equiv x\mod{17}

we would need to choose n such that 5n\equiv1\mod{16} (because 16+1\equiv1\mod{16}). We can find such an n by applying the Euclidean algorithm:

16=3(5)+1
\implies1=16-3(5)
\implies16-3(5)\equiv-3(5)\equiv1\mod{16}

which makes -3\equiv13\mod{16} the inverse of 5 modulo 16, and so n=13.

Now,

x^5\equiv3\mod{17}
\implies (x^5)^{13}\equiv x^{65}\equiv x\equiv3^{13}\equiv(3^4)^2\times3^4\times3^1\mod{17}

3^1\equiv3\mod{17}
3^4\equiv81\equiv4(17)+13\equiv13\equiv-4\mod{17}
3^8\equiv(3^4)^2\equiv(-4)^2\mod{17}
\implies3^{13}\equiv(-4)^2\times(-4)\times3\equiv(-1)\times(-4)\times3\equiv12\mod{17}

Similarly, we can look for m such that 7m\equiv1\mod{30}. Apply the Euclidean algorithm:

30=4(7)+2
7=3(2)+1
\implies1=7-3(2)=7-3(30-4(7))=13(7)-3(30)
\implies13(7)-3(30)\equiv13(7)equiv1\mod{30}

so that m=13 is also the inverse of 7 modulo 30.

And similarly,

x^7\equiv3\mod{31}[/ex] [tex]\implies (x^7)^{13}\equiv3^{13}\mod{31}

Decomposing the power of 3 in a similar fashion, we have

3^{13}\equiv(3^3)^4\times3\mod{31}

3\equiv3\mod{31}
3^3\equiv27\equiv-4\mod{31}
\implies3^{13}\equiv(-4)^4\times3\equiv256\times3\equiv(8(31)+8)\times3\equiv24\mod{31}

So we have two linear congruences,

\begin{cases}x\equiv12\mod{17}\\x\equiv24\mod{31}\end{cases}

and because \mathrm{gcd}\,(17,31)=1, we can use the Chinese remainder theorem to solve for x.

Suppose x=31+17. Then modulo 17, we have

x\equiv31\equiv14\mod{17}

but we want to obtain x\equiv12\mod{17}. So let's assume x=31y+17, so that modulo 17 this reduces to

x\equiv31y+17\equiv14y\equiv1\mod{17}

Using the Euclidean algorithm:

17=1(14)+3
14=4(3)+2
3=1(2)+1
\implies1=3-2=5(3)-14=5(17)-6(14)
\implies-6(14)\equiv11(14)\equiv1\mod{17}

we find that y=11 is the inverse of 14 modulo 17, and so multiplying by 12, we guarantee that we are left with 12 modulo 17:

x\equiv31(11)(12)+17\equiv12\mod{17}

To satisfy the second condition that x\equiv24\mod{31}, taking x modulo 31 gives

x\equiv31(11)(12)+17\equiv17\mod{31}

To get this remainder to be 24, we first multiply by the inverse of 17 modulo 31, then multiply by 24. So let's find z such that 17z\equiv1\mod{31}. Euclidean algorithm:

31=1(17)+14
17=1(14)+3

and so on - we've already done this. So z=11 is the inverse of 17 modulo 31. Now, we take

x\equiv31(11)(12)+17(11)(24)\equiv24\mod{31}

as required. This means the congruence x^{37}\equiv3\mod{527} is satisfied by

x=31(11)(12)+17(11)(24)=8580

We want 0\le x, so just subtract as many multples of 527 from 8580 until this occurs.

8580=16(527)+148\implies x=148
3 0
3 years ago
Do y’all like Tyler the created
liq [111]

Answer:

nah bro thats demarcus

Step-by-step explanation:

5 0
3 years ago
Amalia and Alec are studying the growth of a plant over time. They measure the height of the plant at the end of each week for s
EleoNora [17]

After 7 weeks the plant will 11.25 units tall if the equation for the best-fit line to be y = 2.5 + 1.25x.

<h3>What is the line of best fit?</h3>

A mathematical notion called the line of the best fit connects points spread throughout a graph. It's a type of linear regression that uses scatter data to figure out the best way to define the dots' relationship.

\rm m = \dfrac{n\sum xy-\sum x \sum y}{n\sum x^2 - (\sum x)^2} \\\\\rm c = \dfrac{\sum y -m \sum x}{n}

We have a line of best fit:

y = 2.5 + 1.25x

Plug x = 7 weeks

y = 2.5 + 1.25(7)

y = 11.25 units

Thus, after 7 weeks the plant will 11.25 units tall if the equation for the best-fit line to be y = 2.5 + 1.25x.

Learn more about the line of best fit here:

brainly.com/question/14279419

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3 0
2 years ago
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