shuttle

Shuttle

AsadRehman...

What happened to your Jose's and Patricia's question that I just answered? Did you delete it and why?

There is no simplification possible for the expression $x+5b+1$, unfortunately.

Shuffle - switch the f into a t

yes!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

answer is correct!

Doodle?

This is the formula you would use to compare the 2 investments:

FV = PV x [1 + R]^N

FV =$50,000 x [1 + 0.04]^2

FV =$50,000 x      1.0816

FV =$54,080 - Jose's investment.

FV =$50,000 x [1 + 0.04/4]^(2*4)

FV =$50,000 x       1.01^8

FV =$50,000 x        1.0828567056280801

FV =$54,142.84 - Patricia's investment.

$54,142.84 - $54,080 =~$63 - extra that Patricia's investment earned.

1, 2, 4... ( x2)

1, -2, 4 ... (x -2)

there could be more but the definate answer is that there is more than one way to solve this question. 

No problem! Don't worry about it.

Solve for x over the real numbers:
(66 - 2^x)/(2^x + 3) = (4 - 2^x)/(2^(x + 1) + 6)

Hint: | Expand out terms of the left hand side.
(66 - 2^x)/(2^x + 3) = 66/(2^x + 3) - 2^x/(2^x + 3):
66/(2^x + 3) - 2^x/(2^x + 3) = (4 - 2^x)/(2^(x + 1) + 6)

Hint: | Look for an expression to multiply both sides by in order to clear fractions.
Multiply both sides by 2^(x + 1) + 6:
132 - 2^(x + 1) = 4 - 2^x

Hint: | Move everything to the left hand side.
Subtract 4 - 2^x from both sides:
128 + 2^x - 2^(x + 1) = 0

Hint: | Simplify 128 + 2^x - 2^(x + 1) = 0 by making a substitution.
Simplify and substitute y = 2^x.
128 + 2^x - 2^(x + 1) = 128 - 2^x
 = 128 - y:
128 - y = 0

Hint: | Divide both sides by the sign of the leading coefficient of 128 - y.
Multiply both sides by -1:
y - 128 = 0

Hint: | Solve for y.
Add 128 to both sides:
y = 128

Hint: | Perform back substitution on y = 128.
Substitute back for y = 2^x:
2^x = 128

Hint: | Perform the prime factorization of the right hand side.
128 = 2^7:
2^x = 2^7

Hint: | Equate exponents.
Equate exponents of 2 on both sides:
x = 7

Sorry Alan.

In half an hour, your friend would have run:
3.5 x 1/2 hour=1.75 miles. And at the same time, you would have run:
4 x 1/3 hour =1 1/3 miles. So, you could not have caught him at the finish line.
Let the time when you caught up with him =t minutes. So you have:
3.5t = 4(t - 10), solve for t
t = 80 minutes - your friend's running time when both of you will meet. And: 80 - 10 = 70 minutes - your running time.

Like so:

$(\sqrt{n^3+7n^2}-\sqrt{n^3})/\sqrt n \rightarrow \sqrt{n^2+7n}-\sqrt{n^2}\\ \rightarrow (n^2+7n)^{1/2}-n \rightarrow n(1+7/n)^{1/2}-n \\ \rightarrow n(1+(1/2)(7/n)+(1/2)(-1/2)(1/2!)(7/n)^2+...)-n\\ \rightarrow n + 7/2 - 49/(8n) + ... - n \rightarrow 7/2 - 49/(8n) + ...$

As all the higher order terms involve a power of n in the denominator, they go to zero as n goes to infinity, so we are left with just 7/2 as the limit.

Hey...I was wrong. (Again.)

Sorry Boseok.

If your friend runs at a rate of 3.5 mph and has a 10 minute - 1/6 hr -  head start......he will be

3.5 * (10/ 60)  =  3.5 * (1/6)  = 7/12 of a mile ahead of you when you start

The distance he covers in one minute  = 3.5 / 60  =  7/ 120 miles

The distance the you cover in one minute  = 4/60   =  8/120 miles

So....the distance that you close on him 1/2 hour after you start running [ 30 minutes ]  =

7/12 - [8 - 7 ] / 120 * 30  =  7/12 - 1/4 = 1/3 mile   which is less than the 7/12 mile distance that you have to close

Here's a graph that shows the closing distance between you and your friend x minutes after you start running : https://www.desmos.com/calculator/pavi3luxdt

As the graph shows....you will catch you friend after 70 minutes when both of you have run 14/3 miles  [ 4 + 2/3 ] miles......at that point the closing distance you have to make up = 0 miles

thats really nice melody thank you, my dads no here , he's in another state, and my mom well, idk why but i just want to tell her, im tired of doing it , and ive got through pretty much the whole day pretending that today is like any other day for me, im having a really bad head ache right now , and all i want is this day to end, and me be normal and nice and not be mean or rude or angry at my mom for not remembering my b'day...and ofcourse you wishes have made me feel better, i'd like to know when is your birthday?i would really love to remeber it,,,seeing that noone remebered mine....

Here's a graph:

I wanted to send you that beautiful post that the troll did for you but I don't know where it is.   

I bet Ginger can find it and she will probably read this post.

Maybe she can post the address for us.

That would cheer you up a little I think!

Please Ginger could you do that for us please ???    Pretty please !!!

You could tell your Mum and Dad.

Maybe they will feel so bad that you get even more thant you would have in the first place.  

I am real;ly sorry that you birthday has not been a success.

I hope you enjoy my cake anyway. It won't make you fat  !!         

$\sqrt[3]{27x^6y^3}$ can be simplified further.

Let's write all of the variables as numbers to the power of the 3.

$\sqrt[3]{3^3\left(x^2\right)^3y^3}$

To make this even easier, i'll distribute the cubed root to every term.

$\sqrt[3]{3^3}*\sqrt{\left(x^2\right)^3}*\sqrt[3]{y^3}$

Now, simplify.

$3x^2y$

Happy Birthday ROSALA!!!

Can I have a piece of your cake please ?   

Maybe this will help you :)