Daewei

While this does get you there eventually, it's much longer and harder than is really necessary. You could also use matrices, but it also is much more time-consuming and more complex than is really necessary. If you have a feel for about how high you can get with those kinds of numbers, you can easily poke around that area until you find the peak and get the answer much quicker. 

I'm sorry, your photo didn't attach. Maybe try again?

Hello, Friend!

For this equation, the easiest way to do it is to plug in different times. Let's start!

First off, it'll probably take a second for the football to get near it's peak, so let's start with t=1.

Ok, not quite 25. Let's try t=1.25!

Great! We found the answer to A - the ball reaches 25 feet after 1.25 seconds. But does it keep rising enough to hit 30 feet, or will it plateau too soon? Let's try t=1.5.

As it turns out, the ball has reached it's plateau and rises no higher! In the end, we find that the answer is yes, just barely, for A but a resounding no for B. 
 

I hope this helped you get a firmer understanding of this kind of equation! Physics seems hard now, but you'll look back soon enough and say, "That was actually kinda fun!" Good luck on your journey, my friend!

I have completed around 60% of the lessons on Khan academy, mainly focusing on the programming and computer science courses. Is there anything in particular I can help you with?

I'm not sure where you learned this, but there is a lot of extraneous math in your solution. All the "1^ -1" is completely unnecessary and confusing, although it does somehow give the correct solution. Here's a simpler way to do it:

256^x-2 = 2^-x-7 | 256 is equal to 2⁸, so you can make the bases the same

2^8x-16=2^-x-7 | Now, you can take away the bases and solve algebraically. If you perform log₂ on both sides, you're left with just the exponents, so I'm just not showing a step.

8x-16=-x-7 | Next, you solve algebraically.

8x+x-16+16=-x+x-7+16

9x=9 | Divide both sides by nine.

x=1 | You have your answer!

Now, let's plug it back in to check if it works.

256^x-2 = 2^-x-7 | Replace x with our answer, 1

256^1-2=2^-1-7 | Change 256 to 2⁸

2^8-16=2^-1-7 | Simplify

2^-8=2^-8

It works! It may look hard at first, but once you try it out it's much easier and saves you the extra steps.

TLDR; Match your bases, take the log and solve algebraically.

50%. Since the outcomes of the previous flips don't affect the current flip, it merely depends on the number of flips and the odds of getting heads. Unless you want to count the miniscule chance of getting neither heads nor tails, where the coin lands on it's side and you slowly descend into madness as all math, physics, and probability start to fall apart, it's a pure 50% chance that you'll get heads. There is also an equal number of odd numbers between 1 and 2016, so it's not skewed odds. Thus, your answer is 50%.

Since you're adding each number, n_next=99:

n_x=n_x-1+x | This is our basic formula

n_x=n_x-1+100 | We know that we're adding 100 to the current x, and the current x is equal to n_x-1+x, so the next x must be 100

n₁₀₀=n_100-1+100

n₁₀₀=n₉₉+100 | We subtract 1, giving us the previous x

So, the easiest way to solve this is to work backwards. Ed ends up with a grand total of $150. Before that, she earned an extra $50, so we subtract that from $150 leaving us with $100. If she spent half her paycheck and ended up with $100, then we can multiply that $100 by 2 giving us her initial weekly paycheck: $200

Here's the work in normal form:

150

 -50

100

  x2
200

I'm not sure what you're trying to ask. Are you looking for a graph of the function?

Hello, Charley!

Web2.0Calc is a site dedicated to helping people understand and appreciate all forms of mathematics. The forums are a place for people to ask questions about parts of math they don't understand, whether it's a specific question or a short explanation of a whole part of math. If you have any other questions, feel free to ask!

Happy to welcome you to the family,

Daewei

It depends on the original format of the question. If it was originally written as 2*log(x+2) then no, you can't treat the two like a normal interger until you get rid of the logarithm. If, however, it was written 2*log(x)+2, then the two is not a part of the logarithm and can be treated as a normal interger because it is one. Once you determine the right format, it's merely a matter of stacking the logs and solving algebraically.

Hope this helped!

-Daewei