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3 years ago

What is the area of a hexagon inscribed in a circle with a diameter of 8

Mathematics

1 answer:

Goryan [66]3 years ago

8 0

Answer:

the side = the radius for a hexagon.

so, area = 6 times the area of one equilateral triangle of side 4.

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At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

3 years ago

? points Simplify completely : 5x - 20 9x - 36

pantera1 [17]

Answer:

5/9

Step-by-step explanation:

4 0

3 years ago

-3 (C - 10) + -4 = 2 C =

DENIUS [597]

-3(c-10) + -4= 2
-3c+ 30+ -4= 2
-3c + 26= 2
-3c= -24
C= 8

7 0

3 years ago

Read 2 more answers

Write the equation of a line in - form that contains the point (- 1, - 2) and is perpendicular to the line y = 5x - 10 degrees

galina1969 [7]

Answer:

y+2= -1/5(x+1)

Step-by-step explanation:

if lines are perpendicular their slopes are negative reciprocal

y=mx+b where m is the slope

y=5x-10 has the slope 5 so a perpendicular line will have slope -1/5

equation point slope form

(y-y1) = m(x-x1) where m is slope, and (x1,y1) any point that belongs to the line

y+2= -1/5(x+1)

3 0

3 years ago

Which ordered pair is a solution of the equation? y=−2x+5

RUDIKE [14]

Answer:

10 is your answer

4 0

3 years ago