All Answers

+2354

ppppppp:
2/3 x 1/8 =

I am guessing you want your answer as a fraction.

When multiplying fractions, you simply need to multiply nominators and denominators.

Therefore; 2/3 * 1/8 = (2*1)/(3*8) = 2/24 = 1/12

Hence, your answer is 1/12

For more information: http://www.mathsisfun.com/fractions.html

reinout-gApr 9, 2014

+130561

Have you tried the on-site calculator????......I think it should do the job just fine!!!..............

CPhillApr 9, 2014

its 2thirds multiplied by 1eigth please answer

GuestApr 9, 2014

+130561

find the general solution of du/dt=e^(u+2t)

Refreshing my memory of differential equations, I think we have a separable equation here
We can write
du/dt = e^(u) * e^(2t)......and separating the variables, we have

e^(-u) du = e^(2t) dt....and integrating both sides gives us

∫ e^(-u) du = ∫ e^(2t) dt.....now, perform the integration, and we get either

-e^(-u) = e^(2t) / 2 + C or -e^(-u) = e^(2t) / 2 - C

Note that we can't use the first one, because we would have to take a negative LN in the next step (which isn't possible)........i.e., we have to "fudge" the problem a bit and take the negative sign in front of the constant, C..........multiplying the second one through by - 1, we have

e^(-u) = C - e^(2t) / 2 Now take the LN of both sides

LN (e^(-u)) = LN (C - e^(2t) / 2) ........ =

-u = LN (C - e^(2t) / 2)........... =

u(t) = - LN (C - e^(2t) / 2)

(Notice that we're assuming that C > e^(2t) / 2)

I believe that's the answer........I checked this with an online solver and it arrived at the same solution, but if any of the other forum members spot a flaw in my reasoning....I'll stand corrected!!

CPhillApr 9, 2014

+33666

reinout-g:
Alan:
hiii:
which is greater: .06 or 60%. and how do you show that on a number line?

60% is 60/100 which is 0.06, so they are the same thing.

Actually 60/100 = 0.6, which is ten times as big as 0.06.

therefore, 60% is bigger

Oops!! You are right of course. I must stop doing mental arithmetic late at night!

AlanApr 9, 2014

+118725

Ethan Bates:
The problem is: (3t + 4)^5
Instead of simply expanding, my instructor wants me to use Pascal's Triangle.

You have alreay had 2 good answers, I am offering a third.

140409 - Pascal's triangle.JPG

This is pascal's triangle - you are going to use the row where the 5 is the second number.

(3t + 4) ⁵= 1*(3t) ⁵*4 ⁰ + 5* (3t) ⁴*4 ¹ + 10*(3t) ³*4 ² + 10*(3t) ²*4 ³ + 5*(3t) ¹*4 ⁴ + 1*(3t) ⁰*4 ⁵

=(3t) ⁵ + 5* (3t) ⁴*4 + 10*(3t) ³*4 ² + 10*(3t) ²*4 ³ + 10*(3t)*4 ⁴ + 4 ⁵

MelodyApr 9, 2014

Thanks for the quick responses! First time using this site and I'll certainly come back.

GuestApr 9, 2014

This is what I think it is although I learned this 2 months ago
(a+b)^x

1(3t)^5+5(3t)^4(4)^1+10(3t)^3(4)^2+10(3t)^2(4)^3+5(a)^1(b)^4+1(b)^5

the numbers 1,5,10,10,5,1 before each set of a's and/or b's are part of the triangle. If the equation were to the ^4th then it would be 1,4,6,4,1

your final answer though is:

243t^5+1604t^4+4320t^3+5760t^2+3840t+1024

GuestApr 9, 2014

+6252

Quote:
The problem is: (3t + 4)^5
Instead of simply expanding, my instructor wants me to use Pascal's Triangle.

Neat question.

(a+b) ⁿ can be written as

Sum[k={0,n} Choose(n,k) a ^kb ^n-k]

i.e. Choose(n,0) b ⁿ+Choose(n,1) a b ^n-1 + ... Choose(n,n-1) a ^n-1b + Choose(n,n) a ⁿ

Now you're probably thinking ok wth is "Choose"?

This is the binomial coefficient, "n choose k" and they are the values of a given row in Pascal's triangle. Come down n rows and go over k elements in the triangle and you get the Binomial coefficient corresponding to "n choose k" = Choose(n,k) = Binomial(n,k) (different names for the same thing).

So in this case if you look at the n=5 row (really the 6th row, the first row is n=0) of Pascal's triangle you see it's

1 5 10 10 5 1

in other words

Choose(5,0)=Choose(5,5) = 1
Choose(5,1)=Choose(5,4) = 5
Choose(5,2)=Choose(5,3) = 10

so (a+b) ⁵ = 1*a ⁵ + 5 a ⁴ b + 10 a ³ b ² + 10 a ² b ³ + 5 a b ⁴ + 1 b ⁵

now let a=3t and b=4 and substitute those into the formula above and simplify things and you're done

RomApr 9, 2014

+118725

helpneeded105:
What two numbers multiply to 600, and add to 17?
Everything Alan said is correct - There are no (real) solutions. I just want to talk about it a little more.

Original problem:
30+17a-20a^2

I am going to look at your original problems some more.

30+17a-20a ² = -20a ² + 17a +30

If you want to factorise this you need to look for 2 numbers that add to 17 and multiply to -20*30= -600
These numbers are not easy to find because thay are not rational. I could talk about this a bit more but I suspect Alan is correct when he said maybe the original question was
30+17a-20a^2 =0
in which case we have
-20a ² + 17a +30 =0
You can solve this using the quadratic formula. This will help you remember it.
http://www.youtube.com/watch?v=O8ezDEk3qCg
For you substitute in, a=-20, b=17 and c=30.
The unknown in your question is an a - usually it is an x.
Please do not get confused, the a in the formula is the number in front of the squared unknown. I mean the a in the formula is -20. do you understand what I am trying to say?

There are 2 solutions. They will be

(-17+sqrt(17^2-4*-20*30))/(2*-20))=
and

(-17-sqrt(17^2-4*-20*30))/(2*-20))=

These answers should be identical to Alan's answers. Mine are approximations, Alan's are exact.

MelodyApr 9, 2014

+118725

Chris, you have made a small mistake. It is -3 not 3.

5s - 28 = -3
Add 28 to both sides
5s = 25
Dividing by 5 on each side gives us

s = 5

Once you fix your entry I will delete this one.

MelodyApr 9, 2014

+118725

3_29_14:
andre surveyed a random sample of 30 students to find the heights of seventh grade students in his school. the mean of the data in his sample was 6103 inches. which of these is most likely the mean height of all the seventh grade students?

I think that some of your question is missing. Was it a multiple choice question?

MelodyApr 9, 2014

+118725

HelP:
:o log 2x = log 10

Assuming that they are both the same base, 2x=10

MelodyApr 9, 2014

+118725

Bre.20:
-x+2=4
x=

Subtract 2 from both sides and then think about it.
If you still don't get it, post again.

MelodyApr 9, 2014

Apr 8, 2014

+118725

Hi Zahcarry123,

Thank you for your answer Zahcarry123, it is great when people get involved and answer (or ask) questions.

I just wanted to mention these things.

1) Your last answer is better but it is an approximation. Why would you approximate an answer when it can easily be written exactly? (either as a fraction or as a recurring decimal)

2) Where is your working - you are supposed to be helping jumper2332 to understand.

3) If you were a member you would be able to edit/delete incorrect posts so that you would not have to leave incorrect answers on tha page. This is just one of the reasons that you may like to join up.
Here are some more reasons;
http://web2.0calc.com/questions/become-a-member-why-would-you-want-to Just highlight the address - left click - choose 'goto'. It will open in a new page.

MelodyApr 8, 2014

+111

: if i was faster than sonic and sonic was faster than a car and the car was faster than me who will be faster .tony
thats not possible. If you are faster than sonic, sonic cannot be faster than something faster than you. lets say you run 10 mi hr (for example) sonic must go less than 10 mi an hr. the car is faster than you, so it must be faster 10 mi an hr. this is impossible, as you can see, so that is the answer

unknownApr 8, 2014

+2354

60/100 = 0.6, which is ten times as big as 0.06.

therefore, 60% is bigger

reinout-gApr 8, 2014

+130561

The "formula" for computing compound interest is given by:

A = P(1 + r /n)^(n * t)

Where A = total amount accumulated after some specified period of time
P = the amount invested (18000 in our case)
r = annual interest rate expressed as a decimal (.09 here)
n = number of compoundings per year (semiannually = 2)
t = number of years = (6)

So we have

A =

18000( 1 + (.09 / 2))^(6 * 2)

And there you are......

CPhillApr 8, 2014

.002

GuestApr 8, 2014

+33666

helpneeded105:
What two numbers multiply to 600, and add to 17?

The way to approach these problems is to let the two numbers be called a and b (or any other two letters you like!). Then you write
a + b = 17
a*b = 600

Rearrange the first to get b = 17 - a by subtracting a from both sides. Substitute this into the second to get a*(17 - a) = 600.
This can be written as 17a - a² = 600, which can be rearranged as:
a² - 17a + 600 = 0.
This is a quadratic equation that can be solved to find two values for a.
However, neither value exists on the real number line; they are what are called complex numbers.
Unless you've dealt with complex numbers it's best not to take this any further here!
If you made a mistake with the 600, and meant 60, then the quadratic factorises nicely into two real positive integers.

Original problem:
30+17a-20a^2