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We can do some rearrangements to do the following:

\(\frac{a + 2}{a + 1} = \frac{b}{3}\), 

\(\frac{3a + 6}{a + 1} = b\), 

\(\frac{3a + 3}{a + 1} + \frac{3}{a + 1} = b\)
\(3 + \frac{3}{a + 1} = b\).

From this, we have to find values of a for which \(\frac{3}{a + 1}\) is an integer. Since there's no limitation on positive or negative integers, we find that if
\(a = 2 \rightarrow b = 4\)
\(a = -4 \rightarrow b = 2\)

Hey, NotThatSmart has the right method, but accidentally used 100 instead of 120 as the second term :

(120+n)/(60+n)  =  (160+n)/(120+n)    cross multiply to get 

n^2 + 240n + 14400 = n^2 +220n + 9600 

20n = -4800

n = -240        would be the constant to add to the numbers ....

SO the geometric sequence would be 

-180        -120         -80      and the common ratio would be   -120/-180 =   2/3 

The answer is TP = 7*sqrt(3).

We're converting the power of a microwave from Watts (W) to a new unit, Zaps (z). We know the conversion rates for both Watts and Zaps to SI units (kg, m, s, min).

Watts to SI:

1 W = 1 kg * m^2 / s^3

Zaps to SI:

1 z = 1 kg * m^2 / min^3

We want to find the power of a 900 W microwave in Zaps. To do this, we can write an equality where the power is the same but expressed in different units:

900 W = P z

Now we can manipulate the equation to solve for P (power in Zaps). We can do this by introducing a conversion factor that equates Watts and Zaps.

This factor will cancel out the desired units (kg and m^2) and leave us with a factor that relates Watts and Zaps through time units (seconds and minutes).

a. We know from the definitions of Watts and Zaps that:

- 1 W / (1 kg * m^2 / s^3) = 1 z / (1 kg * m^2 / min^3)

b. This simplifies to:

- 1 W * (min^3 / s^3) = 1 z

c. This conversion factor is equal to 1 because we're converting between equivalent units that express the same fundamental quantities (mass, length, time) but in different time scales (seconds vs minutes).

Apply the conversion factor to the original equation:

900 W * (min^3 / s^3) = P z

Since the conversion factor is 1 (as derived previously), we have:

900 W = P z

Therefore, the power of the 900 W microwave in Zaps is also 900 Zaps. However, to express the answer in scientific notation, we should recognize that 900 can be written as 9.00 x 10^2.

Answer: The power of the 900 W microwave in Zaps is 9.00 x 10^2 Zaps.

We're converting the power of a microwave from Watts (W) to a new unit, Zaps (z). We know the conversion rates for both Watts and Zaps to SI units (kg, m, s, min).

Watts to SI:

1 W = 1 kg * m^2 / s^3

Zaps to SI:

1 z = 1 kg * m^2 / min^3

We want to find the power of a 900 W microwave in Zaps. To do this, we can write an equality where the power is the same but expressed in different units:

900 W = P z

Now we can manipulate the equation to solve for P (power in Zaps). We can do this by introducing a conversion factor that equates Watts and Zaps.

This factor will cancel out the desired units (kg and m^2) and leave us with a factor that relates Watts and Zaps through time units (seconds and minutes).

a. We know from the definitions of Watts and Zaps that:

- 1 W / (1 kg * m^2 / s^3) = 1 z / (1 kg * m^2 / min^3)

b. This simplifies to:

- 1 W * (min^3 / s^3) = 1 z

c. This conversion factor is equal to 1 because we're converting between equivalent units that express the same fundamental quantities (mass, length, time) but in different time scales (seconds vs minutes).

Apply the conversion factor to the original equation:

900 W * (min^3 / s^3) = P z

Since the conversion factor is 1 (as derived previously), we have:

900 W = P z

Therefore, the power of the 900 W microwave in Zaps is also 900 Zaps. However, to express the answer in scientific notation, we should recognize that 900 can be written as 9.00 x 10^2.

Answer: The power of the 900 W microwave in Zaps is 9.00 x 10^2 Zaps.

Let's write an equation for this problem. 

Since they form a geometric series, let's let n be the number that is added. 

We have the equation

\(\frac{100+n}{60+n}=\frac{160+n}{100+n}\\ (100+n)^2=(60+n)(160+n)\\ 10000+n^2+200n=9600+n^2+220n\\ 400=20n\\ n=20\\\)

Thus, adding 20 to each term, we get the sequence

80, 140, 180. 

Since 

\(120/80 = 3/2 \\ 180/120=3/2\)

The common ratio is just 3/2. 

Thanks! :)

Since the number of members stayed the same and the amount of wall tripled, it takes triple the time

\(40*3=120\) minutes

So 120 minutes it our final answerr. 

Thanks! :)

You could also look at it this way:

You can see that this will always end with 10/18, i.e. there will always be a remainder of 10.

Draw AM , BN perp to CD

Let DN = x

CN = 12 - ( AB + DN)  =  12 - (6 + x)  =  6-x

AM = BN

So....By the Pythagorean Theorem

AD^2 - DM^2  = BC^2 - CN^2

5^2 - x^2  = 7^2 - ( 6-x)^2

25 - x^2 = 49 - x^2 + 12x - 36

25 =  12x + 13

12 = 12x

x =1 = DM

Note that we have two right triangles ADM  and PDX

Angles AMD and PXD are equal = 90°

And angle ADM = angle PDX

So  triangle ADM is similar to triangle PDX

PD / AD = 6/5

So

DX / DM = 6/5

DX = (6/5)DM = (6/5) (1) = 6/5

So

sqrt [DP^2 - DX^2] = PX  

sqrt [ 6^2 - (6/5)^2 ] =  sqrt [ 36*25 - 36 ] / 5 = sqrt [ 900 - 36] / 5  = sqrt [ 864] / 5  =

sqrt [ 144 * 6 ] / 5  =  12sqrt (6) / 5 = PX

Similarly

Right Triangles BCN and PCY are similar

PC/BC  = 6/7

So

CY / CN  = 6/7

CY / 5 = 6/7

CY = 30/7

So

sqrt [ PC^2 - CY^2 ] = PY

sqrt [ 6^2 - (30/7)^2] = sqrt [ 36 * 49 - 900 ] / 7  =  sqrt [ 1764 - 900] / 7 = sqrt [ 864]/7  =

sqrt [ 144 *6 ] / 7 = 12sqrt(6) / 7 = PY

\(100^{100} = (10^2)^{100} = 10^{200}.\)
Now we can check the remainder of the powers of 10 being divided by 18.

\(10^2 \equiv 10 mod(18) \because 10^2-(18*5) = 10\)

\(10^3 \equiv 10 mod(18) \because 10^3-(18*55) = 10\)

\(10^4 \equiv 10 mod(18) \because 10^4-(18*555) = 10\)

\(10^5 \equiv 10 mod(18) \because 10^5-(18*5555) = 10\)

\(10^6 \equiv 10 mod(18) \because 10^6-(18*55555) = 10\)

...

Therefore we can conclude that the remainder of \(100^{100}\) divided by 18 is 10

The answer is 224.

To determine how many points \((x, y)\) where both \(x\) and \(y\) are positive integers lie below the hyperbola \(xy = 16\), we need to find the integer pairs \((x, y)\) such that \(xy < 16\).

### Step-by-Step Solution:

1. **Consider values of \(x\) and find corresponding \(y\) values**:

   For each \(x\), \(y\) must satisfy \(1 \leq y < \frac{16}{x}\).

2. **Calculate pairs for each \(x\)**:

   - \(x = 1\):

     \[
     xy < 16 \implies y < \frac{16}{1} \implies y < 16 \implies y = 1, 2, 3, \ldots, 15 \quad (\text{15 values})
     \]

   - \(x= 2\):
     \[
     xy < 16 \implies y < \frac{16}{2} \implies y < 8 \implies y = 1, 2, 3, \ldots, 7 \quad (\text{7 values})
     \]

   - \(x = 3\):
     \[
     xy < 16 \implies y < \frac{16}{3} \implies y < 5.33 \implies y = 1, 2, 3, 4, 5 \quad (\text{5 values})
     \]

   - \(x = 4\):
     \[
     xy < 16 \implies y < \frac{16}{4} \implies y < 4 \implies y = 1, 2, 3 \quad (\text{3 values})
     \]

   - \(x = 5\):
     \[
     xy < 16 \implies y < \frac{16}{5} \implies y < 3.2 \implies y = 1, 2, 3 \quad (\text{3 values})
     \]

   - \(x = 6\):
     \[
     xy < 16 \implies y < \frac{16}{6} \implies y < 2.67 \implies y = 1, 2 \quad (\text{2 values})
     \]

   - \(x = 7\):
     \[
     xy < 16 \implies y < \frac{16}{7} \implies y < 2.29 \implies y = 1, 2 \quad (\text{2 values})
     \]

   - \(x = 8\):
     \[
     xy < 16 \implies y < \frac{16}{8} \implies y < 2 \implies y = 1 \quad (\text{1 value})
     \]

   - \(x = 9\) and higher:
     \[
     xy < 16 \implies y < \frac{16}{x} \implies y < \frac{16}{x} \implies y = 1 \quad (\text{1 value if } x \leq 15 \text{ else no values})
     \]

3. **Count the total number of pairs**:

   Summing all the valid \(y\) values for each \(x\):
   \[
   15 + 7 + 5 + 3 + 3 + 2 + 2 + 1 + 1 = 39
   \]

Therefore, there are \( \boxed{39} \) points of the form \((x, y)\) where both coordinates are positive integers and lie below the hyperbola \(xy = 16\).

To determine the number of points of the form \((x,y)\), where both coordinates are positive integers, that lie below the graph of the hyperbola \(xy = 16\), we need to find the integer pairs \((x, y)\) such that \(xy < 16\).

### Step-by-Step Solution:

1. **Identify the Range for \(x\)**:

   - For each \(x\), we need \(y\) to be an integer such that \(xy < 16\).

   - Since \(x\) must be a positive integer, consider possible values of \(x\) starting from 1 up to the point where \(x \cdot 1 = 16\), so \(x\) ranges from 1 to 15.

2. **Count \(y\) Values for Each \(x\)**:

   - For each \(x\), find the largest integer \(y\) such that \(y < \frac{16}{x}\).

   Here’s how this works for each \(x\) from 1 to 15:

   - \(x = 1\): \(xy < 16 \implies y < \frac{16}{1} = 16\). So, \(y\) can be 1 to 15 (15 values).

   - \(x = 2\): \(xy < 16 \implies y < \frac{16}{2} = 8\). So, \(y\) can be 1 to 7 (7 values).

   - \(x = 3\): \(xy < 16 \implies y < \frac{16}{3} \approx 5.33\). So, \(y\) can be 1 to 5 (5 values).

   - \(x = 4\): \(xy < 16 \implies y < \frac{16}{4} = 4\). So, \(y\) can be 1 to 3 (3 values).

   - \(x = 5\): \(xy < 16 \implies y < \frac{16}{5} = 3.2\). So, \(y\) can be 1 to 3 (3 values).

   - \(x = 6\): \(xy < 16 \implies y < \frac{16}{6} \approx 2.67\). So, \(y\) can be 1 to 2 (2 values).

   - \(x = 7\): \(xy < 16 \implies y < \frac{16}{7} \approx 2.29\). So, \(y\) can be 1 to 2 (2 values).

   - \(x = 8\): \(xy < 16 \implies y < \frac{16}{8} = 2\). So, \(y\) can be 1 (1 value).

   - \(x = 9\) to \(x = 15\): For these values, \(y < \frac{16}{x}\) will always be less than 2, so \(y\) can only be 1 (1 value each).

3. **Summarize the Counts**:

   \[
   \begin{aligned}

   &15 \text{ values for } x = 1, \\

   &7 \text{ values for } x = 2, \\

   &5 \text{ values for } x = 3, \\

   &3 \text{ values for } x = 4, \\

   &3 \text{ values for } x = 5, \\

   &2 \text{ values for } x = 6, \\

   &2 \text{ values for } x = 7, \\

   &1 \text{ value for } x = 8, \\

   &1 \text{ value each for } x = 9 \text{ to } 15.

   \end{aligned}

   \]

   Adding these up:

   \[
   15 + 7 + 5 + 3 + 3 + 2 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 44
   \]

Therefore, the number of points \((x, y)\), where both coordinates are positive integers, that lie below the graph of the hyperbola \(xy = 16\) is:
\[
\boxed{44}
\]

I'm getting an answer of 42.  I will post the solution later.

Thanks for the guidance, CPhill. 

Sorry for the original problem. :)

Seems OK, NTS  !!!!

This is defintely not a smart tactice, but let's expand everything. 

I won't show all the steps, but we eventually get

\(x^{7}-7x^{6}+10x^{5}-13x^{4}+8x^{2}-9x+14\)

Thus, our answer is -9. 

Thanks! :)