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Mariulka [41]
3 years ago
9

29. Find the area of a regular hexagon with apothem of 6 inches.

Mathematics
1 answer:
victus00 [196]3 years ago
5 0

Answer:

124.7 in²

Step-by-step explanation:

A hexagon consists of six equilateral triangles, each of side a, and we can divide each of them into two right triangles.

So, we can calculate the area of one right triangle and multiply by 12.

The formula for the area of one triangle is

A = ½bh

Step 1. Calculate the length of the side a

Per the Pythagorean Theorem,

\begin{array}{rcl}h^{2} + \left(\dfrac{a}{2}\right)^{2} & = & a^{2} \\\\6^{2} + \dfrac{a^{2}}{4} & = & a^{2}\\\\36 & = &\dfrac{3a^{2}}{4}\\\\144 & = & 3a^{2}\\\\\end{array}\\\\

\begin{array}{rcl}a^{2} & = & \dfrac{144}{3}\\\\a & = & \dfrac{12}{\sqrt{3}}\\\\a & = & \dfrac{12\sqrt{3}}{3}\\\\a & = & 4\sqrt{3}\\\\\end{array}

2.Calculate the area of a small triangle

The base of a small triangle is  

b = ½a = ½ × 4√3 = 2√3

The area of one small triangle is

A = ½ bh = ½× 2√3 × 6 = 6√3 in²

3. Calculate the area of the hexagon

A = 12 × 6√3 = 72√3 = 124.7 in²

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melamori03 [73]

Answer:

a) C(2000)=1915

C(2014)=1915

b) C'(2000)=-\frac{275}{14}

C'(2014)=-\frac{275}{14}

c) C(t)=-\frac{275}{14}t+\frac{288405}{7}

d) t=2021

e) too early

Step-by-step explanation:

a)

Since C(t) is the concentration of CFCs in ppt in year t, all we need to do to solve this part is determine what the concentration of CFCs is in years 2000 and 2014. Luckily for us, the problem already gives us those values, so:

C(2000)=1915    concentration in year 2000

C(2014)=1915      concentration in year 2014

b)

By definition, the derivative of a function at a given point is interpreted as the slope of the tangent line to the point of interest, so in order to find this answer, we need to find the slope of the line. Since the problem specifies that the behavior is linear, this means that the slope will always be the same no matter the year, so we get:

m=C'(t)=\frac{C'(t_{2})-C'(t_{1})}{t_2-t_1}

so:

C'(t)=\frac{1640-1915}{2014-2000}=-\frac{275}{14}\approx -19.64

therefore:

C'(2000)=-\frac{275}{14}

C'(2014)=-\frac{275}{14}

c)

Since the behavior is linear, we can calculate it with the point-slope form of the line which is:

y-y_{1}=m(x-x_{1})

in this case:

C(t)-C(t_1)=C'(t)(t-t_{1})

so we get:

C(t)-1915=-\frac{275}{14}(t-2000)

and we can now solve for C(t) so we get:

C(t)=-\frac{275}{14}t+\frac{275000}{t}+1915

for a final answer of:

C(t)=-\frac{275}{14}t+\frac{288405}{7}

d)

So next we solve the equation for C(t)=1500 so we get:

1500=-\frac{275}{14}t+\frac{288405}{7}

1500-\frac{288405}{7}=-\frac{275}{14}t

-\frac{277905}{7}=-\frac{275}{14}t

t=2021 \frac{7}{55}

so we will reach that concentration level at the year 2021 approximately.

e)

Since the second derivative of the concentration function is greater than zero, this means that the original function might be a function in the form: .

This means that the decrease of the concentration levels is slower than that of a linear equation. So the projected date will be too early than the real date.

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3 years ago
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Answer:

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Step-by-step explanation:

Try using Symbolab, I use it all the time it gives the correct answer and it gives good explanations.

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5x-3x-2=13
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