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Artist 52 [7]
3 years ago
7

Can someone answer this please

Mathematics
1 answer:
bulgar [2K]3 years ago
7 0

Answer:

good luck man

Step-by-step explanation:

you will get answers after some time

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Find the measure of
adell [148]
I don’t know what the options are but I think it is 95 because 120-25 = 95
5 0
3 years ago
What is the area of a regular hexagon (6 sides) whose
Ganezh [65]

Step-by-step explanation:

With reference to the regular hexagon, from the image above we can see that it is formed by six triangles whose sides are two circle's radii and the hexagon's side. The angle of each of these triangles' vertex that is in the circle center is equal to 360∘6=60∘ and so must be the two other angles formed with the triangle's base to each one of the radii: so these triangles are equilateral.

The apothem divides equally each one of the equilateral triangles in two right triangles whose sides are circle's radius, apothem and half of the hexagon's side. Since the apothem forms a right angle with the hexagon's side and since the hexagon's side forms 60∘ with a circle's radius with an endpoint in common with the hexagon's side, we can determine the side in this fashion:

tan60∘=opposed cathetusadjacent cathetus => √3=Apothemside2 => side=(2√3)Apothem

As already mentioned the area of the regular hexagon is formed by the area of 6 equilateral triangles (for each of these triangle's the base is a hexagon's side and the apothem functions as height) or:

Shexagon=6⋅S△=6(base)(height)2=3(2√3)Apothem⋅Apothem=(6√3)(Apothem)2

=> Shexagon=6×62√3=216

8 0
3 years ago
I REALLY NEED HELP FAST, will mark you the brainliest if you answer this first :D
VLD [36.1K]

Step-by-step explanation:

required \: scale  \\ =  \frac{12 \frac{1}{2} }{1 \frac{9}{16} }  \\  \\  =  \frac{ \frac{25}{2} }{ \frac{25}{16} }  =  \frac{25}{2}  \times  \frac{16}{25}  =  \frac{1}{2}  \times  \frac{16}{1}   \\  \\ = 8 \:  :  \: 1

6 0
3 years ago
I need help with this
PIT_PIT [208]
2x+8y
2x+8times2
2x+16
18x
Your final answer is 18x!
7 0
3 years ago
Read 2 more answers
The first three terms in the binomial expansion of (a+b)^n in ascending powers of b, are p,q and r respectively. show that q^2/p
katen-ka-za [31]

First 3 terms are a^2 + n a^(n-)1 b + n(n-1)/2 * a^(n-2) b^2


So q^2 / pr = (n^2 * a^(2n-2) * b^2 ) / (1/2 * a^n * (n(n-1) * a^(n-2) * b^2 )


= n^2 * a^2n-2 * b^2

-----------------------------------

1/2 n(n-1) * a^(2n-2) * b^2


= 2n / n - 1 as required


given p = 4, q=32 and r = 96:-


32^2 / 4*96 = 2n / n-1

2n / n-1 = 8/3

6n = 8n - 8

2n = 8


n = 4 answer

7 0
3 years ago
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