nitsuno said: Hi Mike :D I have a data management & statistics math question if you can answer it ~ Consider the word SARCASM: a) How many different 4-letter combinations exist using the letters in this word? and b) How many different 5-letter permutations exist using the letters in this word? Note: SARC and SRAC are considered the same combination but different permutations. I'm totally stuck D: thank you for your help!

Hi there!

In SARCASM, there are 5 unique letters- SARCM.

a) This would be 5 choose 4, or 5C4. If you know some higher math, this is a binomial coefficient. Use a calculator to find 5C4, and you’ll get 5 unique 4-letter combinations.

b) Permutations mean that order matters. We can have any letter as first, then we have 4 left to choose from, then 3, then 2, then 1. So that’s 5*4*3*2*1 = 5! = 120 permutations.

Matt and Josh high five when the four digits in the digital clock are an even number (ex. 12:34pm =1234) and when the total of the numbers are divisible by three. What is themaximum number of times Matt and Josh can high five in a span of 24 hours?

I’m assuming that we use 12-hour time (ex: 8:00 pm = 8:00am, not 20:00), so then we only have to work with 12 hours instead of 24. There are 1440 minutes in a day, so 720 even minutes per day. Therefore 720 is an upper bound.

We want even numbers- so that’s 30 per hour.

We need the sum of the digits to be divisible by 3- so let’s go through each possible “minute” and find hours that make the sum of the digits divisible by 3.

00: 3,6,9,12

02: 1, 4, 7, 10

04: 2, 5, 8, 11

06: 3, 6, 9, 12

08: 1, 4, 7, 10

10: 2, 5, 8, 11

12: 3, 6, 9, 12

I’m skipping the rest and assuming the pattern continues. It looks like that for each minute there are four hours that make the sum of the digits divisible by 3.

So 30 even minutes per hour * 4 possible hours per even minute *2 since this is per 12 hour period= 240 times.

240 </ 720 so this makes sense. But I’m not sure if it’s right.  This is my guess.

If you could help me answer question 10 & 13, that would be FANTASTIC :))))

10) First, let’s look at the y-intercept part. By definition, the y-intercept is the value of y when x=0. Since we want y=1, we just plug this into the function and see what we get.

1 = 0 + 0 + c, therefore c=1.

Now, let’s think about relative max. The derivative =0 at a relative max by definition.

So therefore y’ = 2ax +b. Since x=3 and y’=0, let’s plug in and see what we get.

0 = 2a(3) +b

0 = 6a +b (call this equation 1 for now)

This isn’t too helpful, so let’s try to come up with some other equation to relate a and b to solve a system of equations.

We also know that y(3) = 12, since it says that (3,12) is a point on the curve.  Also, we just found that c=1.

12 = a(3^2) +b(3) + 1

11 = 9a +3b. (call this equation 2)

So now we use (equation 1) and (equation 2) to solve for a and b.

We know b= -6a, so we can plug that into the second equation: 11 =  9a + 3(-6a), therefore a= -9/11 (which also tells us that the parabola opens downwards), and then b = 45/11 (from b= -6a).

13) cubic, local max at (2,4)  local min at (0,0)

This is kind of like the problem I just did, we need to find a system of equations relating the coefficients. (Since we are restricting the function to have certain properties simultaneously, the values of the coefficients will be unique.)

By definition, the derivative =0 at these points. So y’ = 3ax^2 +2bx + c = 0.

If we want a min at x=0: 0 = 0 + 0 + c. Therefore c=0.

If we want a max at x=2: 0 = 3a(2)^2 + 2b(2) + c. Then 0= 12a + 4b + 0, and to simplify we get 0 = 3a + b (equation 1)

We also know that (2,4) and (0,0) are on the curve: 4 = a(2)^3 + b(2^2) + c(2) + d, and 0 = 0 + 0 + 0 + d. So d=0.

So 4 = 8a + 4b +2c. We know already that c= 0. So 4 = 8a + b. (equation 2)

Again we have two equations and two unknowns so we can solve. b = 4- 8a from equation 2. Plugging into equation 1, 0 = 3a + (4-8a), therefore a= 4/5. Then b= 3- 8(4/5), so b= -32/5.

superwholoctopus said: Hey I was wondering if you could explain how to answer this question for me if you get the time - what is the average rate of change of y=(-2x^2-3x)-1 with respect to x over the interval [-9,9]

Sure! Average rate of change is kind of like finding slope.

We’re already given our change in x.

Next, find y(-9) and y(9), which is the y-values at those points. (-136 and -190 respectively.)

So now we apply a slope formula: (y2 -y1)/ (x2- x1) = (-190 - (-136)) / (9- (-9)) = -54/18 = -3.

Another way to picture the average rate of change is if you draw a line between the two points, what is the slope of that line? That’s kind of what you’re doing here.

Check out this page for more info. http://cims.nyu.edu/~kiryl/Precalculus/Section_2.4-Average%20Rate%20of%20Change/Average%20Rate%20of%20Change.pdf It’s probably geared more towards a calculus class (while I tried to explain it without calculus) but it’s not too bad to understand. Let me know if you need anything else!

Anonymous said: can you solve this and explain how?? :( Teresa drove to a resort 319 miles from her house. her speed one was 50mph while the second part was 48 mph? if her total driving time was 6.5 hours, how long did she travel at each rate

Let x = the time spent travelling 50mph (number of hours)

Let y = the time spent traveling 48 mph (number  of hours)

From the problem, the total distance was 319, so 50x + 48y = 319.

The total time was 6.5, so 6.5 = x +y.

The easiest way to solve this is to do y = 6.5- x, and substitute that into the first equation. 50x + 48(6.5 -x) = 319 and solve. Or, you could solve for x in terms of y and substitute that instead. Either way you should get the same answer, but I’m too lazy right now to actually solve it :P

gorgeouslytangledtears said: hi im in algebra! and I'm trying to get this: 2x-3y=-1 into y=mx+b but I KEEP GETTING A FRACTION! (i have to graph it) so it wouldnt make sense! help?!?

Yes, you will get a fraction as an answer.

Subtract 2x from each side and then divide both sides by -3.

That gives slope-intercept form and you can more easily graph it.

jhartal said: I'm a bit buffled by derivatives... like I know how to take the limit and use the various rules to calculate them, and also that they represent the slope of the tangent line at a particular point but I don't get how they are connected to the actual "y" values? If the deriv. at point x = 3 of a function evaluates to 15 for example, does that mean that f(4) will be 15 units higher? Or that f(2) was 15 units smaller? Ugh.

The derivative at a point has nothing to do with the y-values of the original function. The behavior at x-values always match up across derivatives though (for example, if there’s a change in concavity [inflection point] at a certain x-value, then you will able to see that that point actually has a change in concavity by looking at the graph).

I’ll use your example but make it a little more clear.  If we have f(x) and f’(x), and we know that f’(3)=15, it means that at x=3 on the graph of f(x), the tangent line has a slope of 15. On f’(x), the y-value when x=3 is 15.

Here’s another example: consider y=x^2. The derivative is 2x.

Let’s say we pick a point x=3. f(3) = 9.

So the point (3,9) is on the graph of f(x).

Let’s say we want to find the derivative at that point. We know f’(x)= 2x. Plug in 3, since we want the derivative when x=3 (has nothing to do with the y-values). We get f’(3) = 2*3 = 6

So notice here how we can pick any point on the derivative function and the y-value tells us the derivative at that x-value on f(x).

Looking at the graph of y=x^2, at the point (3,9). At this point, the tangent line at  has a slope of 6.

So the y-value of the derivative (function) at a point x=a is equal to the slope of the tangent line at x=a on the graph of f(x).

Also, a side note, this is why derivatives of functions that only differ by a constant have the same derivatives. For example, y= x^2 +6 and y=x^2 -100000 have the same derivative, so the derivative only tells us about the slope of tangent lines at a certain x-value on f(x), and it doesn’t matter what the y-value is at that point on f(x).

forgetwhatyouveheard said: PLEASE HELP ME!!! Joice deposits \$350 every 6 months into an account which pays 2.8%/a compunded semi- annually. What is the total amount of this annuity after 5 years?

I’m not sure, I’ve never done a problem where you have recurring deposits :( If I had to guess, I’d say to calculate the interest on the first \$650, and then use that total amount to find the next interest, etc. and repeat until you have 5 years worth. But there has to be a batter way to do it since that’s tedious and takes a while D:

Anonymous said: The formula I = 3v3/2 relates the cathode current I in milliamperes and the voltage V is in volts for a certain picture tube. Find the cathode current when the voltage is 64 volts. A.48 ; B.288 ; C.1536 ; D.2660. Thanks :)

Ugh this is why I hate physics.

we have  i = 64, so plug in and solve for v. I don’t know what formula that is because the syntax is incorrect, but just substitute and solve.

Anonymous said: write a quadratic in y=ax2+bx+c form with solutions of -3/7 and 1/2 with a y intercept of 6. Thanks!

We know the roots/solutions are -3/7 and 1/2. Think about when we solve a quadratic, we end up with (x+3/7)(x-1/2) = 0.

so we have y = (x+3/7)(x-1/2)

We will have to add some number to get an intercept of 6. On the y-axis, x = 0, and we want y =6. Substitute and solve for the constant:

6 = (0+3/7)(0-1/2) +c

c = 6- (3/14)

And to write it in the ax^2 form, foil and add the c value we just got.

winterparadox said: It is always great when no one in the class seems to know what is going on either. F'(x), derivative, and basically all the problems say "limit definition" which when a kid asked today what that was the prof said we didn't, don't, wont be going over that because it is too much. You want to help/teach?

Okay so quick lesson on a derivative: Derivative is the rate of change at a point on a graph; or pictorially  the slope of the tangent line to the curve at a given point.

There exists a shortcut called the Power Rule that is useful for finding derivatives of polynomial functions easily.  For polynomials, the derivative is always one degree less than the original function. (*)

So the power rule says that you take the exponent of the variable and multiply it by the coefficient of that term, and that is the new coefficient of the derivative term. Then you subtract one from the original coefficient, and that is a term of the derivative.

Example: 2x^3 + 4x^2 -6x -5

f’(x) = 2*3 x^(3-1) + 2*4 x^(2-1) - 6(1) x^(1-1) - 5*0

= 6x^2 +8x -6 (see * above- the derivative’s degree is one less than the original function)

The derivative of a constant is 0 (since it’s really like c(x^0) = c ) The power rule works for all powers of x (even fractional).

This only works for polynomials, and there are so many more things to know about derivatives that I can’t possibly write them out here. Anything besides a polynomial (trig, logs, exponentials) use formulas that you can only memorize, there’s no way to figure them out.

So back to your question: That’s the “shortcut” of a polynomial derivative.

The long way is the limit definition.  I’m assuming you know what limits are for this.  Any derivative, regardless of what the function is, can be found using this definition.

AHHHH there’s so much material here that it’s hard to write it all out. First, read this for the meaning of h: http://www.math.hmc.edu/calculus/tutorials/limit_definition/- just replace delta x with h (since h is more often the abbreviated form), and this link provides the background for the pictorial representation of a derivative, which is EXTREMELY helpful.

There are two definitions; either work but sometimes one is easier than the other.

(1) lim h—>0 of [f(x+h)- f(x)] /h (I like this one, not sure why, but the trick here is to factor out an h from the top terms so it cancels with the h on the bottom- because you can’t have 0 in the denominator)

f’(a) = lim(x—>a) of [f(x) - f(a)] / (x-a) is the next one- this one is best used for when you have to find the derivative at a point without finding the actual derivative first.

The link I said provides a few examples to show. You shouldn’t get anything too complicated because the long way takes a long time (that’s why we have shortcuts!)\

Ahhhh I hope that helped, it’s like trying to explain the core of calculus in a short post. D:

invisibull said: Hey, I was wondering if you could help me(: I'm working with z-scores and the question states, Given the mean and a variance of 73.1 find the z-score for each. I was wondering how to change the variance into Standard Deviation? I know it's something to do with the square root..? Thanks!

standard deviation = sqrt(variance). Also, in other words, (standard deviation)^2 = variance

Also, z = (mean - given value) / (standard dev)

jadenhatesyou-deactivated201312 said: Could you help with a precalc question? The pop of a town in 1890 was 6250. If it increased at a rate of 3.75% per year, what was the pop in 1915? Also, what year was the pop expected to double? We were given the answers, which are 15689 and1908, but I don't know how to do the work

Ooh yay I like these questions! :P

Remember the PERT formula: it’s great because it’s one formula that applies to a few types of problems. P = e^(rt), where P = “new amount”, e = e (the weird number, 2.7….), r = rate as a decimal, and t = time.

So for the first one, we have P = e^(3.75*25) [25 since we want the change from 1890 to 1915] = _____ (use your calculator)

The next part is the trickier one, where we have to use logs to solve. We get 2*6250=e^(3.75x), where x is the year that has a population that is double that of the original year.  So to solve for x, take the ln of both sides, which should get rid of the e, and then it should be an easy equation from there.

40ozfree said: 1. A racetrack is 440 yards long. Approximately what percent on a mile is a racetracks length? 2. You have a photo that measures 3 in wide and 4 in long. You want to proportionally enlarge it as much as possible so that it fits on a greeting card that is 5 in by 6 in with a 0.25 in margin on all sides. A. Find the possible length of the enlarged design. B. what percent of the original length will the enlarged length be? C. What will be the width of tbe enlarged design? Can you please help me? :/

1- A mile is 1760 yds, so 440/1760 = .25—> 25%.

2- a- Since we want a margin of .25 on all sides, subtract .5 from the desired length and width: so our new dimensions are 4.5 x 5.5.

b- Let x be the constant of proportionality. Since the length is 5.5, 5.5 = 4x. x = 138%

c- 4.5 (from part a)

Anonymous said: for the problem (-14x+2x)/1 and (3x+1) what would you change the window to so you could find the solution? thanks!

Try -3 </ x,y </ 3