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H is a function, so in order to find h(-67), we just need to plug in -67 into h(x):

h(-67) = -49*-67 -125 = 3158

You have a total supply of $1000$ pieces of candy, and an empty vat.  You also have a machine that can add exactly $7$ pieces of candy per scoop to the vat, and another machine that can remove exactly $6$ pieces of candy with a different scoop from the vat.  When these two machines are done, there is only one piece of candy left in the vat.  What is the smallest possible number of times the the first machine added candy to the vat?    

     One      

     Let the 1st machine cycle one time .....  7 pieces added    

     Let the 2nd machind cycle one time ....  6 pieces removed     ==>> 1 piece remaining in vat    

     _.         

The expected length of the longest run is 4/3.

We are tasked with finding the area of a regular hexagon \(IJKLMN\) where the interior angles are all equal. However, the given side lengths of the hexagon vary, suggesting that this is not a regular hexagon, but rather an **equiangular hexagon** (a hexagon with equal angles but varying side lengths).

### **Solution By Steps**

**Step 1: Properties of an equiangular hexagon**

For any equiangular hexagon, the sum of the interior angles is \(720^\circ\), and since all the angles are equal, each interior angle is:

\[
\frac{720^\circ}{6} = 120^\circ
\]

This is characteristic of any equiangular hexagon. Additionally, an important property of equiangular hexagons is that opposite sides of the hexagon are parallel. We can utilize this property to find relationships between the sides and the area.

**Step 2: Application of the area formula for an equiangular hexagon**

There is a known formula for the area of an equiangular hexagon with side lengths \(a\), \(b\), and \(c\) such that opposite sides are equal:
\[
\text{Area} = \frac{3\sqrt{3}}{2} \times (a^2 + b^2 + c^2)
\]

where \(a\), \(b\), and \(c\) are the three distinct side lengths of the hexagon.

From the problem, we know:

- \(IJ = LM = 3\) (opposite sides are equal),

- \(JK = MN = 4\) (opposite sides are equal),

- \(KL = IN = 6\) (opposite sides are equal).

Thus, we have:

- \(a = 3\),

- \(b = 4\),

- \(c = 6\).

**Step 3: Calculate the area using the formula**

Now, we substitute the values of \(a\), \(b\), and \(c\) into the area formula:

\[
\text{Area} = \frac{3\sqrt{3}}{2} \times (3^2 + 4^2 + 6^2)
\]

First, compute the squares of the side lengths:

\[
3^2 = 9, \quad 4^2 = 16, \quad 6^2 = 36
\]

Now, sum these values:

\[
9 + 16 + 36 = 61
\]

Substitute this into the formula:

\[
\text{Area} = \frac{3\sqrt{3}}{2} \times 61
\]

Simplify:

\[
\text{Area} = \frac{3\sqrt{3} \times 61}{2} = \frac{183\sqrt{3}}{2}
\]

### **Final Answer**

The area of the hexagon \(IJKLMN\) is:

\[
\frac{183\sqrt{3}}{2} \text{ square units}.
\]

Let  \(BC = a, AC = b\) just to simplify our eyes a bit. 

Let's note that through the Pythagorean Theorem, we can find AM and BN in terms of a and b. 

We get \(\overline{AM} = \sqrt{b^2 + \left(\dfrac a2\right)^2} = \dfrac12 \sqrt{4b^2 + a^2}\)

and \(\overline{BN} = \sqrt{a^2+ \left(\dfrac{b}{2}\right)^2} = \dfrac12 \sqrt{4a^2 + b^2}\)

Because AM and BN both equal 1, we get the equations

\(4b^2 + a^2 = 1^2 \\4a^2 + b^2 = 1^2\)

Adding the two equations together and dividing, we get

\(a^2 + b^2 = \dfrac{1+ 1}{5} = 2/5\)

Since the hypotenuse is just the square root of a^2 + b^2, we get

\(\sqrt{a^2+ b^2} = \sqrt{2/5}\)

Thanks! :)

https://web2.0calc.com/questions/geometry_2526

A positive integer is called terrific if it has exactly 5 positive divisors. What is the smallest number of primes that could divide a terrific positive integer?  

The smallest number of prime divisors is one.     

As an example, 16 has 5 positive divisors:  1, 2, 4, 8, 16.  2 is a prime number.    

_.   

On second thought.....I would say 24      1   2    3    4     6    8    12   24 

I believe that would be '30'

  divisors  1  2   3   5   6   10   15   30 

Answer C    f(x) = 3 (x+10)      then    y = 3 (x+10) 

              one output value.

The   DOMAIN

of a function is the set of all input values, or x-values, for which the function is defined.

The   RANGE

of a function is the set of all output values, or y-values, for which the function is defined.

To write the equation y = ax + b in function notation, substitute   f(x ) 

for y.

16 has factors    1  2  4  8  16 

Variable time :)

x = Chinese students

y = Spanish students

x = y + 5

Now its time for a ven diagram (sort of):

Chinese students: x

Spanish students: y (x - 5)

both: z

Using the principle of inclusion-exclusion: we can express this into:

(x - z) + (x - 5 - z) + z = 12

x - z + x - 5 = 12

2x - z - 5 = 12

2x - z = 17

Only 15 take spanish, so:

y = 15, x = 15 +5  = 20

Plugging x = 20 into 2x - z = 17, we can get that there are 23 students taking both. 

First, If a positive integer is divisible by 6, it must be divisible by three and 2, meaning n must be divisible by 3 and 2.

Second, n has exactly 8 positive integers, giving us hints about the prime factorization and their powers.

Here is a pretty neat trick:

for any integer x, you can tell the number of factors by taking each of the prime's exponent adding one to it, and multiplying the whole thing.

ex: 12: 2^2 * 3^1, factor amount: 3 * 2 = 6

8 can be expressed as 1*8, 2*4, 4 *2, and 1*8. 

Remember, n has had 3 and 2, so using the clue above, n could be 3^7 * 2^0 ( not possible), 3^1 * 2^3, 3^3 * 2^1, 2^7 * 3^0 ( not possible)

Some are not possible, leaving us with only 3^3 * 2^1 and 3^3 * 2^1, and both of them have only two prime divisors, so the answer is two.  

However, I have not proven if we can add more prime factors. 

The answer is no, and let's express that as x. If we do add x (x^1 basically), we would end up with:  3^1 * 2^3 * x^1 and 3^3 * 2^1 * x^1

Both are not possible, as if we add 1 to all the exponents that would simply not multiply up to 8, as we added an x^1 term, and if we add one to that, we would get an EXTRA *2, resulting in 16 factors in both cases. 

We could multiply these numbers together, but we could also do this. 

\(40 \cdot 6 \cdot75\cdot 12 = \\ 40 \cdot6 \cdot[3 \cdot25] \cdot[ 3\cdot4 ] = \\ 40 \cdot 6 \cdot [ 3 \cdot 3 ] \cdot [ 25 \cdot4] = \\ 6 \cdot 40 \cdot 9 \cdot 100 = \\ 54 \cdot 4000 = \\ 54 \cdot 4 \cdot1000 = \\ 216 \cdot10^3\)

Thus, there are 3 terminal zeroes. 

Thanks! :)

A terminal zero is a $0$ that appears at the end of a number. For example, the number $3,800$ has two terminal zeros.    
How many terminal zeroes does $40 \cdot 6 \cdot 75 \cdot 12$ have?    

     40 • 6 • 75 • 12  =  216,000 ... so. three terminal zeroes.      

_.   

Let's know that XZW will form a triangle

\(\angle ZXW  = 90^\circ \\ \angle XWZ = 60^\circ\)

Thus, \( \angle XZW =  180 - 90 - 60  =   30\)

30 degrees is our answer. 

Thanks! :)

The number $100$ has four perfect square divisors, namely $1,$ $4,$ $25,$ and $100.$     
What is the smallest positive integer that has exactly $2$ perfect square divisors?    

That would be four ... it has two perfect square divisors, namely one and four.    

_.