CPhill

If f(x) = 5x - 6, what is the equation for f^-1(x)

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So, we want to find the inverse of this......there are  couple of ways to do that......let's look at one

First.....let's write this as

y = 5x - 6      ..........now let's get "x" by itself......add 6 to both sides

y + 6 = 5x     ..........divide both sides by 5

(y + 6)/ 5  = x      ....now, just "switch" x and y

(x + 6)/ 5  = y      

If we have an inverse - and we do - it simply"reverses" the x and y coordinates - in "math" terms, if (a,b) is on the original graph, then (b,a) is on the inverse graph. Let's pick a point on the original graph, say, (6,24). So (24,6) should be on the on the inverse graph - and it is!! This is always a good way to check to see if the "inverse" is correct.

One more thing...... replace the "y" on the inverse with f^-1(x) 

So we have ..... f^-1(x) = (x + 6)/ 5

And that's it....

If I invest 10 000$ and average an interest return of 2% every month, how many years will it take before I get to 1 000 000 $ ?

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We have

A_^t = A₀(1 + r)^12t

Where:

A_t = amount accumulated at the end of some time  (in our case, $1 000 000)

A₀ = Amount invested ($10 000)

r = interest rate expressed as a decimal  (.02)

12 = compoundings per year

t = time in years (what we're looking for)

So we have

1000000 = 10000(1 +.02)^12t         .........let's divide both sides by 10000

100 = (1.02)^12t                             .........now, take the the "log" of both  sides

log (100) = log (1.02)^12t                ........by a "log" property, we can bring the 12t "out front" on the right

log (100) = (12t) log (1.02)            ........divide both sides by log (1.02)

log (100) / log (1.02) = 12t             ........divide both sides by 12

log (100) /(12 log (1.02)) = t     =      about 19.38 years

y=-x+4 ? how you find this?

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This is just the graph of a line with a y-intercept of 4 and an x-intercept of 4. Thus, the points (0,4) and (4,0) lie on the line. Graph it for yourself, and you'll see that it has a "negative" slope - it "falls" from "left" to "right."

3=(21.25-t)/t

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Let's start by multiplying both sides by "t"......note, we are explicitly assuming that "t" isn't = 0

(and it's not!!!)  ...   so we have

3t = 21.25 - t       .....now add t to both sides......we have

4t = 21.25            .....divide both sides by 4

t = 21.25 / 4   =    5.3125

6y^2-60y+126

y^2-6y-7

simplify please

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Here's the first one.....first, take out the common factor, 6,......so we have

6 [ (y² - 10y + 21) ]

Now, we need to factor inside the brackets. We're looking for two numbers that multiply to 21 and sum to -10.

The first two things I think of are (-7) and (-3). I think those will work. So we have...

6(y - 7)(y - 3)

For the second one, we're looking for two numbers that multiply to -7 and sum to -6. (-1) and (+7) multiply to -7, but their sum is 6. Let's just change the signs. (+1) and (-7) multiply to -7 and sum to -6. Looks like those are what we need. So we have...

(y + 1)(y - 7)

how do you find miles per hour?

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Given what???    Feet per second??   Kilometers per hour??

Re: if 6 in the divisor and 1.2 is the quotient, what is the dividend?

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(dividend) / (divisor) = quotient

So...........(dividend) / 6  = 1.2

So...... (dividend) = (1.2) X 6 = 7.2

Re: why is 5 to the power of 0 one?

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In general, any real number raised to the "0" power is = 1

The only exception is  0⁰.........this is known as an "indeterminate" form - we can't define what this might be!!!

(It also involves division by 0......that's a no-no!!)

Here's another way to look at it.......just subtract one side from the other.......if you get "0," both things must be equal !!!

Is it possible to do (-1)^4/3 

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No, because we would have to use a negative real number as a log base and that creates problems.

Also, note that the normal exponential curve is "unbroken" and non-periodic.

But (-1)² = 1  and (-1)³ = -1 and (-1)⁴ = 1........notice how the graph would "oscillate" back and forth between 1 and -1 as x increases in integer value (like a periodic function) and it would also be "undefined" between integer values (a "broken" graph). (That's why you get an "error" on the calculator when you try to evaluate (-1)^(4/3) - your exponent of 4/3 lies "between" integers.)