NotThatSmart

The length of $\overline{AB}$ is $4,$...Find the length $AB.$

The answer is 4. :)

\(244224 \pmod {64} \equiv 0 \)

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! :)

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! :)