NotThatSmart

I mean, it is credits to Melody. Her answer helped me understand it!

If none of the square are consecutive, then that means the yellow squares must occupy squares 1, 3, and 5. 

This means only squares 2 and 4 are left to fill, with either red or blue. 

We have 

Blue, Blue

Red, Blue

Blue, Red

Red, Red

There sre 4 ways we can do it!

Thanks! :)

Oh wait! That's so smart. 

I totally forgot that the children could row the boat back!

I was confused since only one adult could fit on the boat, meaning it would be "impossible" to succeed! I totally missed out on that LOL. 

Well, thanks a lot CPhill!

~NotThatSmart

Let's draw a diagram. 

The Green part is where the points would satisfy the conditions. 

There are 18 triangles congruent to each other in this graph. If we count, 12 of them are in the green area. 

Thus, we have \(\frac{12}{18} = \frac{2}{3}\)

Thanks! :)

~Oh, and thank you Melody for the graph

First, let's start by calculating the distance Willl covers before Grace even can start. We have:

\((45 min)(50 m/min) = 2250 m\)

This means that when Grace starts to row, she and Will have \((2800 m) – (2250 m) = 550 m\)to cover before meeting up.

We have them rowing at each other at the same exact time, so we have \((50)(t) + (30)(t) = 550\) where t is the time it takes. 

Now, we solve for t, 

\(80t = 550 \\ t = 550/80 = 6.875\)

This is approximately 6 minutes and 52 seconds. 

We add this on to the 45 minutes it took Will to row, and add that time onto 2 P.M. 

We get that Will and Grace meet at \(2:51:52\) P.M.

Thanks! :) 

First, let's note that distance = rate*time. 

From the problem, we know that 

\(d = 40 t \\ d = 55(t - 25/60)\)

\(40 t = 55 ( t-25/60) \\ -15t = -275/12 \\ t = 55/36\)

Multiplying this by 40 miles per hour, we get 

\( 40 \cdot 55/36 = 61 + 1/9 miles \)

61 1/9 miles is hour answerr!

Thanks! :)

Thanks! That's nice of you to say!

First, let's cross multiply and simplfy. We get 

\(8 (a + 2) = b ( a + 1) \\ 8a + 16 = ab + b\)

Now, in order to progress, we have to be able to factor it, 

\(ab + b - 8a = 16 \\ ab + b - 8a - 8 = 16 - 8 \\ (a + 1) ( b - 8) = 8\)

That means the number of order pairs there are is the number of factors 8 has, including negatives. 

There are 8 factors, meaning the answer is 8. 

Here are the ordered pairs that work. 

\((3, 10), (1, 12), (0, 16), (7, 9), (-5, 6), (-3, 4), (-2, 0), (-9, 7)\)

Thanks! :)

First off, let's combine some like terms and move all terms to one side. 

We get \(6x^{2}-21x-31=0\). Unfortunately, this can't be factored without radicals, so let's use the Qudratic Formula. 

We get \(x=\frac{21\pm \sqrt{(-21)^{2}-4\cdot 6(-31)}}{2\cdot 6}\). Simplifying, we get 

\(x=\frac{\sqrt{1185}+21}{12}\\ x=\frac{-\sqrt{1185}+21}{12}\)

We want to take the biggest value of x we could, which would be \(x=\frac{\sqrt{1185}+21}{12}\). 

Thanks! :)