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igor_vitrenko [27]
2 years ago
11

ut%20out%20exactly%20%7D" id="TexFormula1" title="\text{I was eating cookies and had some thoughts. If I wanted to cut out exactly }" alt="\text{I was eating cookies and had some thoughts. If I wanted to cut out exactly }" align="absmiddle" class="latex-formula">\frac{1}{3}of the cookie to share with someone, how far from one side would I have to make a straight cut to get that exact amount? How far would I have to cut if I wanted to cut off\frac{1}{n}\text{ of the cookie?}
\text{Basically, the question is, find the value of }a\text{ given only n, and r}

\text{One way of finding this, is by finding the area of the shaded reigon, Q in terms of}
\text{r, a, and b, and equating it to the area of the fraction of the cookie then solving for a.}

\text{In math, this means solving } \frac{1}{n}\pi r^2=Q \text{ for }f(r,n)=a.

\text{From the diagram, we can see that }r=a+b

\text{Eventually, by 2 different means, I found 2 equations that, if solved, would give the}\text{ relationship between r, n, and a.}\text{They are as follows:}

\text{1. }\frac{1}{n}\pi r=r\theta-bsin(\theta) \text{ where }\theta=cos^{-1}(\frac{b}{r})

\text{2. }\frac{1}{n}\pi=\theta-sin(2\theta)\text{ where }\theta=cos^{-1}(\frac{b}{r})

\text{These 2 equations are equivalent, but annoying to solve.}

\text{To claim these points, please solve for a in terms of r and n, showing all work.}
\text{I would like an analytic solution if possible.}[\tex][tex]\text{All incorrect, spam, or no-work solutions will be reported.}

Mathematics
1 answer:
andrew-mc [135]2 years ago
5 0

In the attachement, there is what I came up with so far. I think that finding 'a' is non-trivial, if possible at all.

A_c - the area of a circle

A_{cs} - the area of a circular segment

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A bulletin board has an area of 1/ 12 yd and a length of 1/3 yard the bulletin board is divided into six equal sections right vi
gulaghasi [49]

Answer: The 6 sections has an area of 1/72 square yards

Step-by-step explanation:

Length= area / length

Length = 1/12 ÷ 1/3 = 1/12 × 3/1 = 1/4 yards

Since the width and length are different and the 6 squares must be arranged in a 2 × 3 pattern.

The long side is 1/3 yard

1/4 ÷ 2/1 = 1/4 × 1/2 = 1/8 yard

And the width is 1/3 ÷ 3/1 = 1/3 × 1/3 = 1/9 yard

The 6 sections are 1/8 × 1/9 and each one has an area of 1/8×1/9 = 1/72 square yard

8 0
3 years ago
Read 2 more answers
A toy cannon ball is launched from a cannon on top of a platform. The equation h(t) =- 5<img src="https://tex.z-dn.net/?f=t%5E%7
DanielleElmas [232]

Answer:

Part A)

No

Part B)

About 2.9362 seconds.

Step-by-step explanation:

The equation  \displaystyle h(t)=-5t^2+14t+2  models the height h in meters of the ball t seconds after its launch.

Part A)

To determine whether or not the ball reaches a height of 14 meters, we can find the vertex of our function.

Remember that the vertex marks the maximum value of the quadratic (since our quadratic curves down).

If our vertex is greater than 14, then, at some time t, the ball will definitely reach a height of 14 meters.

However, if our vertex is less than 14, then the ball doesn’t reach a height of 14 meters since it can’t go higher than the vertex.

So, let’s find our vertex. The formula for vertex is given by:

\displaystyle (-\frac{b}{2a},h(-\frac{b}{2a}))

Our quadratic is:

\displaystyle h(t)=-5t^2+14t+2

Hence: a=-5, b=14, and c=2.

Therefore, the x-coordinate of our vertex is:

\displaystyle x=-\frac{14}{2(-5)}=\frac{14}{10}=\frac{7}{5}

To find the y-coordinate and the maximum height, we will substitute this value back in for x and evaluate. Hence:

\displaystyle h(\frac{7}{5})=-5(\frac{7}{5})^2+14(\frac{7}{5})+2

Evaluate:

\displaystyle \begin{aligned} h(\frac{7}{2})&=-5(\frac{49}{25})+\frac{98}{5}+2 \\ &=\frac{-245}{25}+\frac{98}{5}+2\\ &=\frac{-245}{25}+\frac{490}{25}+\frac{50}{25}\\&=\frac{-245+490+50}{25}\\&=\frac{295}{25}=\frac{59}{5}=11.8\end{aligned}

So, our maximum value is 11.8 meters.

Therefore, the ball doesn’t reach a height of 14 meters.

Part B)

To find out how long the ball is in the air, we can simply solve for our t when h=0.

When the ball stops being in the air, this will be the point at which it is at the ground. So, h=0. Therefore:

0=-5t^2+14t+2

A quick check of factors will reveal that is it not factorable. Hence, we can use the quadratic formula:

\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Again, a=-5, b=14, and c=2. Substitute appropriately:

\displaystyle x=\frac{-(14)\pm\sqrt{(14)^2-4(-5)(2)}}{2(-5)}

Evaluate:

\displaystyle x=\frac{-14\pm\sqrt{236}}{-10}

We can factor the square root:

\sqrt{236}=\sqrt{4}\cdot\sqrt{59}=2\sqrt{59}

Hence:

\displaystyle x=\frac{-14\pm2\sqrt{59}}{-10}

Divide everything by -2:

\displaystyle x=\frac{7\pm\sqrt{59}}{5}

Hence, our two solutions are:

\displaystyle x=\frac{7+\sqrt{59}}{5}\approx2.9362\text{ or } x=\frac{7-\sqrt{59}}{5}\approx-0.1362

Since our variable indicates time, we can reject the negative solution since time cannot be negative.

Hence, our zero is approximately 2.9362.

Therefore, the ball is in the air for approximately 2.9362 seconds.

5 0
3 years ago
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Which equation results from adding the equations in this system? x + 6 y = 9. Negative x + 2 y = negative 15.
Lunna [17]

Answer:

Step-by-step explanation:

x + 6y = 9

-x + 2y = -15

-------------------add

8y = - 6 <=== u see that the x terms cancel each other out

5 0
2 years ago
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NEED ANSWER ASAP!!! PLEASE HELP
lys-0071 [83]

Answer:

Answer "C"

Step-by-step explanation:

I believe this because it would not make sense to compare the already known equation to the unknown equation and does make sense to compare the unknown equation to the known equation

(pls dont delete this moderators)

3 0
2 years ago
If Hector is 8 years old and Mary is 3 years old how old will Mary be when Hector is 16
IRINA_888 [86]
Okay so 8 (hector) - 3(Mary) = 5 years apart
So, 16 (hector) - 5 (years apart) = 11 (Mary’s age)
7 0
3 years ago
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