@sophisticatedwolf Typically, when posed with these problems, you will consider the positive number only, unless otherwise specified. The reason is because who actually cares about negative integers

@bulba_bulbasaur said in No one finished Module 6!:

@bulba_bulbasaur said in No one finished Module 6!:

@rz923 said in No one finished Module 6!:

@bulba_bulbasaur said in No one finished Module 6!:

how do you quote?

Here:
E7B11E1E-7DF7-49C0-8EFB-C6BDF897A97C.jpeg

Green in the honour of the green shirt! 👕

ok thx!

Ze quote chain
XD

very likely

@the-blade-dancer I find that the correct answer has another answer that is very similar to it. Whenever there are 2 similar answers it is usually one of those and then there is another answer that is completely off.

@aaronhma You can try some of the other modules first. I started at module 2 and then decided that I didn't need module 0 or 1.

@debbie yeah I read about those a year ago. Depressing that they're on hold.

@debbie This is still not fixed yet. Would you please prioritize it?

https://forum.poshenloh.com/topic/318/typo

@divinedolphin Fixed 🔧 🔧

👍 yeah good idea

@The-Blade-Dancer You are really observant! This new feature only got added a week ago. 🙂 Now during business hours (EST), there will be a live help chat with one of our marketing specialists.

@divinedolphin Thanks for asking! It's actually the exact same mock test as the one that Prof. Loh wrote last year, so if you did that one already, maybe you'll find the livestream review tonight (9/19/20 at 6:30 p.m. EST) redundant. On the other hand, we might be getting 450+ viewers, so it might be a lot of fun to watch anyway! 🙂

okay. thx! 🙂

@spaceblastxy1428 Thanks so much for updating this question with more info!

I tried it, and I'm sorry that I don't have an elegant way to do it, just a grungy way 🔨 🔨 ⚒

Here is the grungy way:

Angle-chasing (which we learned about in these lessons: M2W2D5-Y-2, M2W4D13-Ch-2) tells us
that \( \angle BYZ = 50^{\circ} \) and \(\angle YAB = 70^{\circ},\) so since \( \angle BZY = 50^{\circ}\) and \(\angle AYZ = 70^{\circ}\) (which is given), we know that triangle \(\bigtriangleup BYZ\) and \(\bigtriangleup AYZ\) are both isosceles.

I'll let \( BZ = BY = 1,\) since all lengths will be relative and only the angles matter, and normalize all other lengths around this length. Then triangle \(\bigtriangleup BYZ\) has lengths of \(1, 1,\) and a third leg, the length of which can be found using the Law of Sines:

$$ \frac{\sin 50^{\circ}}{1} = \frac{\sin 80^{\circ}}{YZ}$$

$$ YZ = \frac{\sin 80^{\circ}}{\sin 50^{\circ}} $$

From angle-chasing, we know that \( \overline{AZ} \) and \(\overline{YB}\) intersect at right angles. Let's call this intersection point \(D.\) Then we get four right triangles: \( \bigtriangleup DYZ, \bigtriangleup DAY, \bigtriangleup DBZ, \text{ and } \bigtriangleup DAB.\)

Then \(BD + DY = 1\) and \(AD + DZ = \frac{\sin 80^{\circ}}{\sin 50^{\circ}}.\)

From the Pythagorean Theorem on \(\bigtriangleup DBZ,\) we know \(BD^2 + DZ^2 = 1,\) so \(DZ^2 = 1 - DB^2.\)
From the Pythagorean Theorem on \(\bigtriangleup DYZ,\) we know that \(DY^2 + DZ^2 = \left( \frac{\sin 80^{\circ}}{\sin 50^{\circ}} \right)^2 .\)
Substitute for \(DZ^2\) into the first equation to solve for \(DB,\) which equals \( 1 - \frac{1}{2} \left( \frac{\sin 80^{\circ}}{\sin 50^{\circ}} \right)^2.\)

Then solve for \(DZ\) using the Pythagorean Theorem on \(\bigtriangleup DBZ,\) which says \( DZ^2 = 1 - DB^2.\)

Once you have found \(DZ,\) then use it to find \(AD,\) since \(AD + DZ = \frac{\sin 80^{\circ}}{\sin 50^{\circ}}.\)

Once you have found \(AD,\) then using the value for \(DB\) from earlier, we can solve for \(\angle ABY:\)

$$ \tan ABY = \frac{AD}{DB} $$

$$ \angle ABY = \arctan \frac{AD}{DB} $$

Sorry I didn't do the calculations... they were just so grungy! But the logic of the steps is all here.

🙂

@spaceblastxy1428 Interesting question! This is actually similar to a question that Prof. Loh covers in the Day 16 Your Turn question of Module 1! Can we assume that the acceleration is constant in this question?

I wasn't sure how long the car drove at \(1\) mph for, so I just assumed that it immediately starts accelerating.

We can graph the speed in mph vs. time (in minutes) like this:

 
20200628-forum-speed-intervals-50-percent.png
 

The acceleration is constant at \( \pm 1 \text{ } \frac{\text{mph}}{\text{min}}\) for each straight segment of the zigzag. From physics, we learn that

$$ \text{distance} = \text{speed} \times \text{time} $$

(Though actually in physics, they use velocity instead of time, since velocity takes into account the direction of where an object is moving.)

Thus the distance can be interpreted as the area under a graph of speed versus time, like in the following example:

20200628-forum-speed-intervals-averages-area-example-25-percent.png

Here the distance traveled is equal to

$$ \text{ distance } = 4 \frac{\text{miles}}{\cancel{\text{hour}}} \times \frac{1}{12} \text{ } \cancel{\text{hour}} = \textcolor{red}{\frac{1}{3} \text{ miles }}$$

Even though our graph doesn't look as nice and simple as the one above, the distance traveled is also the same: it's the area under the speed versus time curve.

20200628-forum-speed-intervals-area-25-percent.png

The area above is equivalent to the area of the shape below, found by averaging the speed for each interval:

20200628-forum-speed-intervals-averages-area-25-percent.png

Remember that we must convert the time duration of each interval from minutes to hour, to match the units of the speed (miles per hour). Thus we go at

$$\begin{aligned} 2 & \text{ mph for } \frac{1}{30} \text{ hour } \\ 2.5 & \text{ mph for } \frac{1}{60} \text{ hour } \\ 3 & \text{ mph for } \frac{1}{30} \text{ hour } \\ 3.5 & \text{ mph for } \frac{1}{60} \text{ hour } \\ 4 & \text{ mph for } \frac{1}{30} \text{ hour } \\ 4.5 & \text{ mph for } \frac{1}{60} \text{ hour } \\ 5 & \text{ mph for } \frac{1}{30} \text{ hour } \\ 5.5 & \text{ mph for } \frac{1}{60} \text{ hour } \\ 6 & \text{ mph for } \frac{1}{30} \text{ hour } \\ 6.5 & \text{ mph for } \frac{1}{60} \text{ hour } \\ \dots \\ 59 & \text{ mph for } \frac{1}{30} \text{ hour } \\ 59.5 & \text{ mph for } \frac{1}{60} \text{ hour } \\ \end{aligned} $$

In the question, it's not clear how long the car spends driving at \(60 \text{ mph},\) so I'll just assume that once it reaches \(60 \text{ mph} \) it no longer continues driving.

Thus the total distance, in miles, is

$$\begin{aligned} & 2 \times \frac{1}{30} + 2.5 \times \frac{1}{60} + 2 \times \frac{1}{30} + 2.5 \times \frac{1}{60} \\ & + 3 \times \frac{1}{30} + 3.5 \times \frac{1}{60} + 4 \times \frac{1}{30} + 4.5 \times \frac{1}{60} \\ \ldots \end{aligned} $$

I'll leave it to you to figure out how to simplify this expression! 🙂