NotThatSmart

This problem actually has a very simple answer. 

Let's note the two sidelengths given, We have \(WX = 2\) and \(XY=3\)

In order to find the area of WXYZ, we must find the height and base. 

The base of the rectange is \(WX\) and the height of rectangle is \(XY\)

The problem gave all the dimensions we need to already solve this problem. We have

\(WX \cdot XY = 2 \cdot 3 = 6\)

So our final answer is 6. 

Thanks! :)

First, let's find how far Will gets before Grace even starts to row. 

We have the handy equation \(d=rt\) where d is distance, r is rate, and t is time. 

We have \( (45 min)(50 m/min) = 2250 m \), so will paddled 2250m before Grace started. 

When Grace starts to canoe, the two have \((2800 m) – (2250 m) = 550 m \) before they meet up. 

Thus, we can write the equation

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

This rounds to approximately 6 minutes and 52 seconds. 

This means the two spent about \(51:52\) minutes to meet eachother. 

This means they met each other at \(2:51:52 pm\)

So 2:51:52 is our final answer. 

Thanks! :)

First, let's simplfy the equation a bit. Distributing and expanding, we get

\(6c+3=84c-9+{135}c-18\)

Isolating c and combining even more like terms, we get

\(-213c=-30\)

Now, we simply isolate c, and we get

\(\frac{-213c}{-213}=\frac{-30}{-213}\)

\(c=\frac{10}{71}\)

So our final answer is 10/71. 

Thanks! :)

We can define variables to solve this problem. 

Let's let the common ratio of this geometric sequence be r.

We have

\(16r = 1/4 \\ r = (1/4) / 16 = 1/64\)

So the common ratio is 1/64. 

Now, we calculate for the 30th term. We have'

\( (1/4) r^6 = (1/4)(1/64)^6\\ = (1/4) ( 1/4^3) ^6 \\ =(1/ 4)^19\\ =0.000000000003638\)

So our final answer is 0.000000000003638.

Thanks! :)

We can set variables to solve this problem. 

Let's let x be the same constant added to every single number.

The 3 terms in the geometric series must be \(60+x, 120+x, 140+x\)

Since these 3 numbers form a geoemtric series, we can write the equation

\(\frac{60+x}{120+x}=\frac{120+x}{140+x}\)

Now, we simply solve for x. 

We have \((60+x)(140+x)=(120+x)^2\)

Expanding out everything and combining all like terms, we get

\(-40x-6000=0\)

\(x=-150\)

Our geometric series is now \(-90, -30, -10\)

The common ratio is \(-30/-90 = 1/3\)

So our answer is 1/3. 

Thanks! :)

We can use casework to solve this problem. 

First, le'ts consider cases with \(4-0-0\), where all 4 balls are in one box. 

Since each box is indistinguiashable, there is only 1 way to complete this. 

Second, let's consider cases with \(3-1-0\)

We can calculate this by using \(4 \choose 1 \), so there are 4 ways. 

Third, let's consider cases with \(2-2-0\)

There are \(4 \choose 2\) ways to complete this, so there are 6 cases. 

Lastly, we have \(2-1-1\)

We have \(4 \choose 2 \) ways, to do this, so 6 additional cases. 

Now, we add all of these up. We get \(1 + 4 + 6 + 6 = 17\)

So our final answer is 17. 

Thanks! :)

First, it's important to note that 1/17 is a repeating decimal. We have

\(1/17=0.\overline{0588235294117647}\)

Now, we must find which digit is the 4000th. 

We have \(4000/17=235R5\)

So, our answer is the fifth digit in the 16 repeating digits. 

The number we are looking for is 2. 

So our final answer is 2. 

Thanks! :)

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

We get

\(x^{2}+10x+9=0\)

We can put this into factored form to find x. We get

\((x+9)(x+1)=0\)

From this, we know that 

\(x=-1\\ x=-9\)

So our final answer is -1 and -9. 

Thanks! :)

Let's find how far everyone ran when Dirk finished. 

When Dirk finishes, Edith has ran 14km while Foley has only ran 9km. 

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

We can now find how far Foley ran when Edith finished the race. 

Plugging in 24, we find that Foley has only ran \(24 (9/14) ≈ 15.43 km\) when Edith ran 24. 

This means Foley was \(24 - 15.43 ≈ 8.57\)km behind Edith. 

So our final answer is 8.57km. 

Thanks! :)