All Answers 354446 Answers

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! :)

Sorry, that is not correct. Thanks for trying.

The correct solution is this:

For the 4, 0, 0 distribution we have only 1 way.

For the 3, 1, 0 distribution we have 4 ways.

The 2, 2, 0 distribtution is where NotThatSmart makes a mistake. It would appear to be \(4 \choose 2\) = 6 ways, but there is double counting. For example, choosing balls 1 and 3 leads to the same split as choosing balls 2 and 4. So, for the 2, 2, 0 distribution there are only 3 ways, making the total amount 14 ways.

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! :)

In order to solve this problem, we must find the common ratio of the numbers. 

Since it's a geometric series, setting r as the common ratio, we get

\(r = .036 / .4 = 36 / 400 = 9 /100 = .09 \)

We can calculate the sum using a really handy formula. 

\(\frac{a}{1-r}\) where a is the first number and r is the common ratio. 

Plugging in the values we already now, we get that the sum is equal to

\( .4 / [ 1 -.09] = .4 / .91 = 40 /91 \)

So our final answer is 40/91

Thanks! :)

The smallest possible prime divisor a number can have is of course, \(2\)

If we check the multiplication, we can see that 3 odd numbers can't possibly mulitply to be an even. 

So, \(5^{19} * 7^{13} * 3^{31}\) is not divisble by 2. 

The next smallest is 3. 

The third term of the multiplication is \(3^{31}\)

This means, no matter what, this number has to be divisible by 3. 

So 3 is our final answer. 

Thanks! :)

The fraction is terminating.

The answer is 3/4.

a^3 - 1/a^3 = 27.

 {   an  =  n* a1  +  (n) (n-1) / 2 * d   }

23a1 + 253d = 1/4   mult through by  33   →   759x + 8349y = 33/4 

33a1 + 528d = - 1/5 mult through by  -23  →    759x - 12144y = 23/5

Add these

-3795d  = 257 / 20

d = -257 / 75900

23a1 +  253 (-257/75900)  = 1/4

a1 = 83  /1725

a43  = 43(83/1725)  +  903(-257/75900)  =   -15007 / 15180

1000 (-8/9)^(n -1)  > 1

(-8/9)^(n -1)   is  only positive when  n  is odd

So we can  write

1000 (8/9)^(n -1)  >  1

(8/9)^(n - 1) > 1/1000     take the log of both sides  and by a log property we can write

(n -1) log (8/9) > log (1/1000)        

{ divide both sides by log (8/9)....this is negative so reverse the inequality sign }

n - 1  <   log (1/1000) / (log (8/9)

n <  1 + log (1/1000) /log (8/9)

n < 59.6

So 

n = 59

The number of terms > 1 =   [ 59 + 1 ] / 2  =  30