hectictar

You can just type in " 1/6 " into the bar and it will turn that fraction into a decimal.

1/6 ≈ 0.166667

12 - 5   =   7

area of circle = π * (radius)²

radius = diameter / 2

radius = 28 yd / 2

radius = 14 yd

area of circle = π * (14 yd)²

area of circle = 196π    yd²

area of circle ≈ 196(3.14)  yd²   ≈   615  yd²

Assuming the party hat is a cone shape,

volume of cone = (1/3) * (π * radius² ) * (height)

volume of cone = (1/3) * (π * [18/2]²  ) * (25)

volume of cone = (1/3) * π * 81 * 25

volume of cone = 675π   cubic cm,

                            which is  ≈  2120.575 cubic cm

Every time x goes up by 1 unit, y goes up by 4 units....so, this looks linear!

\(x^{\frac23}(8x)^{\frac13} \\~\\ =x^{\frac23}\cdot 8^{\frac13} \cdot x^{\frac13} \\~\\ =x^{\frac23+\frac13}\cdot 2 \\~\\ =x^{\frac33}\cdot2 \\~\\ =x^1\cdot2 \\~\\ =2x\)

(4 - 5)^2   =   (-1)^2   =   (-1) * (-1)   =   1

Remember, a negative times a negative is a positive!

\((cd)^{\frac12}(d^{\frac13}) \\~\\ =c^{\frac12}\cdot d^{\frac12}\cdot d^{\frac13} \\~\\ =c^{\frac12} \cdot d^{\frac12+\frac13} \\~\\ =c^{\frac12}\cdot d^{\frac56} \\~\\ =\sqrt{c} \cdot \sqrt[6]{d^5}\)

We need an equation that tells us the speed of the car at any time.

To do that, we can take the derivative of the position equation.

Let's call that function v(t) .

v(t)  \(=\frac{d}{dt} (\frac14t^2+1) \\~\\ =\frac{d}{dt} \frac14t^2 +\frac{d}{dt}1 \\~\\ =(\frac14)(2)(t)+0 \\~\\ =\frac12t\)

And the average speed is given by...

average speed  \(=\frac{\text{distance}}{\text{time}} \\~\\ =\frac{s(10)-s(0)}{10} \\~\\ =\frac{(\frac14(10)^2+1)-(\frac14(0)^2+1)}{10} \\~\\ =\frac{5}{2}\)

Now we just need to know what t equals when v(t) = \( \frac52 \)

\(\frac52 = \frac12t \\~\\ \frac{5}{2}\div\frac{1}{2} =t \\~\\ 5=t\)

Here I made a triangle where A is the first sighting, B is the light, and C is the second sighting.

The question is, what is the length of BC ?

On this first one, I just focused on the angles.

m∠BAC = 107º - 60º   =   47º

measure of the green angle = 180º - 107º   =   73º

Since the two vertical lines are parallel and line AC crosses them, we have two parallel lines cut by a transversal. So.. the two green angles are the same.

Now let's just look at the second triangle.

We can use law of sines to find BC.

\(\frac{1.5}{\sin 26}=\frac{BC}{\sin 47} \\~\\ \frac{1.5}{\sin 26}*\sin 47 =BC \\~\\ 2.503\text{ km} \approx BC\)