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3 years ago

Find me area of the regular hexagon if the apothem is 4sqrt3 mm and a side is 8mm. Round the the nearest whole number.

Mathematics

1 answer:

lesya692 [45]3 years ago

8 0

Answer: THE LAST OPTION.

Step-by-step explanation:

To calculate the area of the regular hexagon you must apply the following formula:

$A=\frac{P*a}{2}$

Where P is the perimeter and a is the apothem.

The perimeter is:

$P=6*s$

Where s is the lenght of a side.

Then:

$P=6*8mm=48mm$

Then, the area is:

$A=\frac{48mm*4\sqrt{3}mm}{2}\\A=166mm^{2}$

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A bank contains 35 coins, all nickels and quarters. The total value of coins is $8.15. How many of each coin does the bank conta

Lina20 [59]

n + q = 35

n = 35-q

0.05n + 0.25q = 8.15

0.05(35-q) +0.25q = 8.15

1.75 -0.05q + 0.25q =8.15

1.75 + 0.20q = 8.15

0.20q = 6.40

q = 6.40 / 0.20 = 32

32 quarters and 3 nickels

5 0

4 years ago

Please help me to prove this!

Sophie [7]

Answer: see proof below

Step-by-step explanation:

Given: A + B = C → A = C - B

→ B = C - A

Use the Double Angle Identity: cos 2A = 2 cos² A - 1

→ (cos 2A + 1)/2 = cos² A

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · 2 cos [(A - B)/2]

Use Even/Odd Identity: cos (-A) = cos (A)

Proof LHS → RHS:

LHS: cos² A + cos² B + cos² C

$\text{Double Angle:}\qquad \dfrac{\cos 2A+1}{2}+\dfrac{\cos 2B+1}{2}+\cos^2 C\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}\bigg(2+\cos 2A+\cos 2B\bigg)+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\dfrac{1}{2}\bigg(\cos 2A+\cos 2B\bigg)+\cos^2 C$

$\text{Sum to Product:}\quad 1+\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\cos (A+B)\cdot \cos (A-B)+\cos^2 C$

$\text{Given:}\qquad \qquad 1+\cos C\cdot \cos (A-B)+\cos^2C$

$\text{Factor:}\qquad \qquad 1+\cos C[\cos (A-B)+\cos C]$

$\text{Sum to Product:}\quad 1+\cos C\bigg[2\cos \bigg(\dfrac{A-B+C}{2}\bigg)\cdot \cos \bigg(\dfrac{A-B-C}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+(C-B)}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-(C-A)}{2}\bigg)$

$\text{Given:}\qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+A}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-B}{2}\bigg)\\\\\\.\qquad \qquad \qquad =1+2\cos C \cdot \cos A\cdot \cos (-B)$

$\text{Even/Odd:}\qquad \qquad 1+2\cos C \cdot \cos A\cdot \cos B\\\\\\.\qquad \qquad \qquad \quad =1+2\cos A \cdot \cos B\cdot \cos C$

LHS = RHS: 1 + 2 cos A · cos B · cos C = 1 + 2 cos A · cos B · cos C $\checkmark$

5 0

3 years ago

David made a class banner out of a large rectangular piece of paper. He cut a triangular piece out of one side. what is the area

VikaD [51]

The height would be the same as the rectangular piece of paper as well as the length so what you would do is you do heights times Lanks divide that by two and you would get the Square inches of the banner hope it helps

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3 years ago

Please answer and explain #17 :)

IRINA_888 [86]

Answer:

Solution given:

we have

m<PTQ=m<RTS

x+20=3x+12.

20-12=3x-x

$\frac{8}{2}$

now

16.m<PTQ=x+20=4+20=24°

17.

again

m<PTR+m<PTQ=180°[supplementary]

m<PTR=180°-24°=156° is your answer

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3 years ago

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4 years ago