look for numbers in place of the letters such that the following mathematical expression hold PRQ/3=OV UVW/6=OV MNO/9=OV

ANEK [815]

Answer: Well if you were going to think of a circle you can divide it into 4 equal sizes in one of the is equal to two pentagons and a half. If you add all is equal to a loop full of pentagons [which makes 10 pentagons]. Because the ring is made out of regular pentagons, we can work out that each of the interior angles of each pentagon is 108* degrees. By extending the lines that two pentagons share, assuming they will all meet in the middle, it will create a triangle. As we know each angle of a [regular] pentagon is 108* degrees, we know the two base angles of the triangle would equal to 72** which leaves the top angle to be 36 degrees. As [the sum of] angles at a point is 360 degrees, and 36 is divisible by 360, it will make a complete ring. Also, as 360 Ã· 36 = 10, we know that the ring will be made out of 10 pentagons. as formula to calculate the the size of a interior angle of a polygon is (n×180−360)÷n (for n being the number of sides that the polygon has). because the triangle is made by extending the lines, and angles on a line is 180 degrees, 180−108 (an interior angle of a pentagon) =72.

4 0

3 years ago

Read 2 more answers

There is maybe be answer two

dolphi86 [110]

I believe it would be the first choice

Hope that helps!

4 0

2 years ago

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Find the nth term of the sequence 7,25,51,85,127

olya-2409 [2.1K]

Let a (n) denote the n-th term of the given sequence.

Check the forward differences, and denote the n-th difference by b (n). That is,

b (n) = a (n + 1) - a (n)

These so-called first differences are

b (1) = a (2) - a (1) = 25 - 7 = 18

b (2) = a (3) - a (2) = 51 - 25 = 26

b (3) = a (4) - a (3) = 85 - 51 = 34

b (4) = a (5) - a (4) = 127 - 85 = 42

Now consider this sequence of differences,

18, 26, 34, 42, …

and notice that the difference between consecutive terms in this sequence b is 8:

26 - 18 = 8

34 - 26 = 8

42 - 34 = 8

and so on. This means b is an arithmetic sequence, and in particular follows the rule

b (n) = 18 + 8 (n - 1) = 8n + 10

for n ≥ 1.

So we have

a (n + 1) - a (n) = 8n + 10

or, replacing n + 1 with n,

a (n) = a (n - 1) + 8 (n - 1) + 10

a (n) = a (n - 1) + 8n + 2

We can solve for a (n) by iteratively substituting:

a (n) = [a (n - 2) + 8 (n - 1) + 2] + 8n + 2

a (n) = a (n - 2) + 8 (n + (n - 1)) + 2×2

a (n) = [a (n - 3) + 8 (n - 2) + 2] + 8 (n + (n - 1)) + 2×2

a (n) = a (n - 3) + 8 (n + (n - 1) + (n - 2)) + 3×2

and so on. The pattern should be clear; we end up with

a (n) = a (1) + 8 (n + (n - 1) + … + 3 + 2) + (n - 1)×2

The middle group is the sum,

$\displaystyle 8\sum_{k=2}^nk=8\sum_{k=1}^nk-8=\frac{8n(n+1)}2-8=4n^2+4n-8$

so that

a (n) = a (1) + (4n ² + 4n - 8) + 2 (n - 1)

a (n) = 4n ² + 6n - 3

4 0

3 years ago

Help please, not sure how to do this...

olga nikolaevna [1]

Answer:

8.38

Step-by-step explanation:

The underlined number is in the hundredths place, so that means that we're going to round there. That means that you look at the number to its right and if it's lower than 5, it stays the same. However, if it's 5 or higher, you round it up. For example, if the number was 8.386, we would round it to 8.39.

5 0

3 years ago

1.

LenKa [72]

The answer would be C because they say if 3 new students arrived each year.

4 0

3 years ago

look for numbers in place of the letters such that the following mathematical expression hold PRQ/3=OV UVW/6=OV MNO/9=OV​

look for numbers in place of the letters such that the following mathematical expression hold PRQ/3=OV UVW/6=OV MNO/9=OV