NotThatSmart

First, let's convert 1/288 into base 25 form. 

We get that $(BD)_{25} = (11 × 25^1) + (13 × 25^0) = (288)_{10}$

This means that $\frac{1}{288}_{10} = \frac{1}{BD}_{25}$

Doing base division, we have that

$\frac{1}{BD}_{25} = 0.\overline{0.02468ACEGIKN}$

So when  $1/288$ is in base 25, the decimal is repeating. 

So repeating is our answer. 

Thanks! :)

First, let's calculate the distance Will travels before grace even leaves. 

Since he traveled for 45 minutes, we have

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

So when Grace starts, she and Will have

$(2800 m) – (2250 m) = 550 m$ to cover. 

Let's let t be the amount of time it takes for the two to catch up to each other after Grace starts rowing. 

We have the equation

$(50)(t) + (30)(t) = 550 \\ 80t = 550 \\ t = 550/80 = 6.875$

This is approximately 6 minutes and 52 seconds. 

So the clock time they meet is 6 min 52 sec after Grace starts.

Grace started at 2:45, so add 6 min 52 sec and the clock will read 2:51:52 pm when they meet  

Thus, our final answer is $2:51:52$

Thanks! :)

If angle XWZ = 60°, angle WXY = 180° - 60° = 120° because XY and WZ are parallels.

However, this means that Angle ZXW must be less than 120°.

In the problem though, it clearly defines ZXW as 120, meaning this problem is invalid. 

Thanks! :)

Let's note that 1/17 is a repeating decimal. 

$1/17 = 0.\overline{ 0588235294117647}$

There are 16 repeating digits in the decimal. 

Dividing 4000 by 16, we get

$4000/16 = 250$

Since it is divisble, then we can conclude that the last digit of the repeating decimal is the correct answer. 

Thus, 7 is the final answer. 

Thanks! :)

When Dirk finishes, Edith has run 14km  and Foley has run 9 km

This means that Foley runs 9/14 as fast as Edith

When Edith has run 24km, Foley has run $24 (9/14) ≈ 15.43 km$

So Foley is $24 - 15.43 ≈ 8.57$ km behind!

Thanks! :)

We can use the information we already know to solve this problem!

First off, let's note that $(x+y)^3 = x^{3}+3x^{2}y+3xy^{2}+y^{3}$, which we can simplify to $8=5+6xy$. Solving for xy, we get $xy=1/2$. 

This may seem useless, but it will come into play later. 

Now, we know that $(x+y)^2 = x^2+2xy+y^2$, which contains x^2+y^2. Isolating this, we get $x^2+y^2 = (x+y)^2 - 2xy$. 

We already know all the terms! We can plug in 1/2 and 2 which we know from above! We get $x^2 + y^2 = 4 - 1 = 3$. 

So 3 is our answer!

Thanks! :)

I defintely didn't do this in a smart way, but a more complicated manner. 

First, let's put s in terms of t so we can do some subsitutions. We get

$s=-\frac{2\left(-4t+1\right)}{5}$ from the first equation. 

Subsituting this value of s into the second equation, we get

$3t-6\left(-\frac{2\left(-4t+1\right)}{5}\right)=9-2t$

Now, we solve for t. We get

$\frac{-23t+12}{5}=9\\-23t+12=45\\t=-\frac{33}{23}$

Now, we subsitute t back into the first equation to find s. We get

$s=-\frac{62}{23}$

Thus, we finally find the ordered pair as

$s=-\frac{62}{23},\:t=-\frac{33}{23}$

So our final answer is $(-62/23, -33/23)$

Thanks! :)