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In order to find the x- intercept, we need to find the function. Using the point slope formula:

y = 2x/5

The x intercept is when y = 0, so the x intercept is (0,0). (also the y intercept)

h is a function, so in order to find h(x), plug 45 into the equation :

h(45) = (1/9) * (90 - 45)

h(45) = 45/9 = 5

So the answer is B

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

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. 

Lets express this as normal notation:

(sorry I don't use Latex)

1/x + 1/y + 1/z = 13/12

Wait! We know what 1/x + 1/y + 1/z is ! We just express both as a common fraction (by expressing them as a common denominator)! 

(1/x + 1/y + 1/z) = (x + y) + xyz/xyz

So now:

(x + y + xyz/)xyz = 13/12

x + y +z= 13, and xyz = 12

Two equations, three variables isn't possible to solve, so we can only test numbers :(, luckily, the problem says only three distinct positive integers!

After some simple testing, we should get that only 1, 4, and 8 work. 

I hope it helps, and of course, your welcome! 

1.The class size must be between 10 and 15, and when arranged in a line, the only way to do so it in one line.

Well, how do we do that? Only if the number is prime! Why?

2. If the students are arranged in a grid the number of students is L x W, but its a single fill line, so either the length or width is one! So the number of students must be prime, since if it had other factors, there would be more available gride formation of students!  

3. The biggest prime between 10 and 15 is 13!

So 13 in our answer

:))

Since the numbers of workers doesnt change, we can times the time by 480/100, so 50 * 4.8 = 240 minutes.

Here for better visualization:

40 members. 100 sq wall.    50 min

40 members.  480 sq wall.  ?

we just times it by 4.8 to get the time. 

I hope it helps!!!