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Rather than using brute force Bosco you could have used this division fact

Divisibility rule of 11: This particular rule states that the given number can only be completely divided by 11 if the difference of the sum of digits at odd position and sum of digits at even position in a number is 0 or 11.

The number of diagonals of a   n-gon equals :    n (n-3)/2  

so: 

n (n-3)/2 = 2n     Solve for n 

n^2 - 3n = 4n 

n^2 - 7n = 0 

n(n-7) = 0              shows  n = 7    sides 

To determine which integers \( n \) in the range \( 70 \leq n \leq 90 \) can be expressed in the form

\[
n = ab + 2a + 3b,
\]

we can rearrange this equation:

\[
n = ab + 2a + 3b = ab + 2a + 3b = a(b + 2) + 3b.
\]

Thus, we can factor the expression:

\[
n = a(b + 2) + 3b = (b + 2)a + 3b.
\]

Rearranging further, we can consider the expression more carefully:

\[
n = ab + 2a + 3b = a(b + 2) + 3b.
\]

To find how \( n \) varies based on \( a \) and \( b \), we can think of expressions:

\[
n = ab + 2a + 3b = a(b + 2) + 3b.
\]

To explore values of \( n \), we can set specific values for \( b \) and investigate how \( n \) changes. We start by expressing \( a \) in terms of \( n \) for different values of \( b \):

1. Rearranging gives us:

\[
n - 3b = a(b + 2).
\]

This implies

\[
a = \frac{n - 3b}{b + 2}.
\]

For \( a \) to be a positive integer, \( n - 3b \) must be divisible by \( b + 2 \) and \( n - 3b > 0 \). This leads to key constraints:

- \( n > 3b \) (or equivalently, \( b < \frac{n}{3} \)),

- \( n - 3b \) should be divisible by \( b + 2 \).

Now, let’s analyze integers within the specified range \( 70 \leq n \leq 90 \).

**For \( b = 1 \):**

\[
n = a(1 + 2) + 3 \cdot 1 = 3a + 3 \Rightarrow n - 3 = 3a \Rightarrow n = 3a + 3 \Rightarrow n - 3 \equiv 0 \, (\text{mod } 3).
\]

Thus, \( n \equiv 0 \, (\text{mod } 3) \) which gives us possible \( n \): \( 72, 75, 78, 81, 84, 87, 90 \).

**For \( b = 2 \):**

\[
n = a(2 + 2) + 3 \cdot 2 = 4a + 6 \Rightarrow n - 6 = 4a \Rightarrow n = 4a + 6 \Rightarrow n - 6 \equiv 0 \, (\text{mod } 4).
\]

Thus, \( n \equiv 2 \, (\text{mod } 4) \) which gives us \( n \): \( 70, 74, 78, 82, 86, 90 \).

**For \( b = 3 \):**

\[
n = a(3 + 2) + 3 \cdot 3 = 5a + 9 \Rightarrow n - 9 = 5a \Rightarrow n = 5a + 9 \Rightarrow n - 9 \equiv 0 \, (\text{mod } 5).
\]

Thus, \( n \equiv 4 \, (\text{mod } 5) \) which gives us \( n \): \( 74, 79, 84, 89 \).

**For \( b = 4 \):**

\[
n = a(4 + 2) + 3 \cdot 4 = 6a + 12 \Rightarrow n - 12 = 6a \Rightarrow n = 6a + 12 \Rightarrow n - 12 \equiv 0 \, (\text{mod } 6).
\]

Thus, \( n \equiv 0 \, (\text{mod } 6) \) which gives us \( n \): \( 72, 78, 84, 90 \).

Continuing this process for \( b = 5 \) and \( b = 6 \) eventually provides specific integers. We can compile our results and find the integers \( n \):

By synthesizing:

- All candidate results from \( b = 1, 2, 3, 4, 5, 6 \):

The integers derived include:

- \( n = 70, 72, 74, 75, 78, 79, 81, 82, 84, 86, 87, 89, 90 \).

Finally, we count unique integers:

\[
\{70, 72, 74, 75, 78, 79, 81, 82, 84, 86, 87, 89, 90\}
\]

Counting these gives:

\[
\text{Total count} = 13.
\]

Thus, the answer is

\[
\boxed{13}.
\]

Let \(BD=x\). By angle bisector theorem, \(\frac{AB}{BD} = \frac{AC}{CD}\) which is equivalent to \(\frac{4}{x}=\frac{5}{3-x} \Rightarrow 12-4x=5x \Rightarrow 9x=12 \Rightarrow x=\frac{4}{3}\). The area of \([\triangle{ADC}]=\frac{DC\cdot AB}{2}=\frac{(3-\frac{4}{3})\cdot4}{2}=\frac{\frac{5}{3}\cdot4}{2}=\frac{10}{3}\approx3\)

\(\triangle{AEM}\) and \(\triangle{B EM}\) share the same base and the same height, so \([\triangle{AEM}]\) = \([\triangle{BEM}]\) and therefore,  \([\triangle{BEM}]\) = \(\frac{1}{2}[\triangle{BAM}]\). \([\triangle{BAM}]\) and \([\triangle{CAM}]\) also share the same base and the same height, so \([\triangle{BAM}]\) = \([\triangle{CAM}]\) and therefore,  \([\triangle{BAM}]\) = \(\frac{1}{2}\triangle{ABC}\). Draw line \(EC\).  \(\triangle{BEM}\) and \(\triangle{CEM}\) also share the same base and the same height, so \([\triangle{BEM}]\) = \([\triangle{CEM}]\) and therefore,  \([\triangle{BEM}]\) = \(\frac{1}{2}[\triangle{BEC}]\). In the end, you get \([\triangle{BEC}]=2[\triangle{CEM}]=2\cdot\frac{1}{2}[\triangle{BAM}]=2\cdot\frac{1}{2}\cdot\frac{1}{2}[\triangle{ABC}]=\frac{1}{2}[\triangle{ABC}]=\frac{1}{2}\cdot14=7\).

 Point E is on the midpoint of AB and [ABC] is 14. Draw line EC. SInce ∆BEC has the same base and the same height has ∆AEC. [BCE] + [ACE] = [ABC]. So [BCE]= 7

What you did was take the long way around to    

say that two times zero equals one times zero.    

And then you divided by zero to obtain 1 = 2.     

_.   

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! :)

tip: if you want to actually make your LaTeX show, don't put dollar signs around it, there's a specific tool for it

let the cubic be ax^3 + bx^2 + cx + d, where a, b, c, and d are the coefficients

f(1) = 0 gives a + b + c + d = 0

so the sum of the coefficients is 0

please tell me if there are any errors

We can let a hot dog be H and a roll be R.

8H + 7R = $13.60

7H + 8R = $13.40

If we add the equations, 

we get

15H + 15R = $27

that means that 

H+R=$27/15

so the answer is 27/15.

Each gallon covers 400 square ft, which is 20ft x 20ft, which is around6x6 meters which is around 36 sq meters. 900/36 will be 25 gallons.