You roll a fair 6-sided die.

The shape of the clock is a regular dodecagon with a radius of 14cm. Centered on the clock's face is a green circle of radius 9c

faust18 [17]

The area of a dodecagon is 1/2aP
where:
a-apothem
P-perimeter

In the figure you will draw a point to any side and make a line perpendicular from a center to that side. Then if u think u can see that with only one draw of a hypotenuse can form a triangle. Then after drawing the hypotenuse which has 2 triangles, make the top angle of 1 triangle there are 30 degrees. On the down part is perpendicular and the left or the right is the 60 degrees. After that using special right triangles like the 30, 60, 90,. Then at the base of the one it bisected dived them by 2 for the 2 triangle's base. After that solve using it since the radius is 14. The apothem is 14(sqrt of 3). So after solving them the perimeter of each dodecagon is 12 sided so 14x12 is 168

A=1/2aP
A=1/2(14)(sqrt of 3)(168)
A=1176(sqrt of 3)

Thats the area of the dodecagon now the circle

A=(pi)r^2
A=(pi)(9)^2
A=81pi

Therefore the color purple is greater than the color green

8 0

3 years ago

Given the function h(x) =1/3 |x-6| +4, evaluate the function when x = - 3, - 2, and 0

Masja [62]

|x| = x for x ≥ 0

examples:

|3| = 3; |0.56| = 0.56; |102| = 102

|x| = -x for x < 0

examples:

|-3| = -(-3) = 3; |-0.56| = -(-0.56) = 0.56; |-102| = 102

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Use PEMDAS:

P Parentheses first

E Exponents (ie Powers and Square Roots, etc.)

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

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$h(x)=\dfrac{1}{3}|x-6|+4$

Put the values of x to the equation of the function h(x):

$x=-3\to h(-3)=\dfrac{1}{3}|-3-6|+4=\dfrac{1}{3}|-9|+4=\dfrac{1}{3}(9)+4=3+4=7\\\\x=-2\to h(-2)=\dfrac{1}{3}|-2-6|+4=\dfrac{1}{3}|-8|+4=\dfrac{1}{3}(8)+4=\dfrac{8}{3}+\dfrac{12}{3}=\dfrac{20}{3}\\\\x=0\to h(0)=\dfrac{1}{3}|0-6|+4=\dfrac{1}{3}|-6|+4=\dfrac{1}{3}(6)+4=2+4=6$