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Which is equivalent to (5^2)^3

=5^(2x3)=5^6. The answer is "A", but don't write it "5^2×^3". Copy it the way I wrote it.

Here's a graph to illustrate the result:

The symmetry of the parabola dictates that if the vertex is at (5, 3) and it goes through (2, 0) then it must also go through (8, 0).

So there are three conditions from which to find the three unknowns:

3 = 25a + 5b + c.   (1)

0 = 4a + 2b + c.     (2)

0 = 64a + 8b + c.   (3)

From (2) we get.  c = -4a - 2b.  (4)

Put this into (1) and (2).  

3 = 21a + 3b.     (5)

0 = 60a + 6b.     (6)

Divide (5) by 3 and (6) by 6:

1 = 7a + b.    (7)

0 = 10a + b.  (8)

From (8) we get. b = -10a.   (9)

Put this into (7)

1 = -3a.  or:               a = -1/3

Put this into (9):        b = 10/3

Put these into (4):     c = -16/3

See if this image helps

x = 9

y = 4

Here is how Wolfram/Alpha sums it up:

http://www.wolframalpha.com/input/?i=1.06+*%E2%88%91%5B5000*1.1236%5En%5D,+for+n%3D0+to+14

Possible derivation:
d/dx(tan^(-1)(2 x) sin(2 x))
Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = tan^(-1)(2 x) and v = sin(2 x):
  =  tan^(-1)(2 x) (d/dx(sin(2 x)))+(d/dx(tan^(-1)(2 x))) sin(2 x)
Using the chain rule, d/dx(sin(2 x)) = ( dsin(u))/( du) 0, where u = 2 x and ( d)/( du)(sin(u)) = cos(u):
  =  (d/dx(tan^(-1)(2 x))) sin(2 x)+cos(2 x) d/dx(2 x) tan^(-1)(2 x)
Factor out constants:
  =  (d/dx(tan^(-1)(2 x))) sin(2 x)+2 d/dx(x) tan^(-1)(2 x) cos(2 x)
The derivative of x is 1:
  =  (d/dx(tan^(-1)(2 x))) sin(2 x)+1 2 tan^(-1)(2 x) cos(2 x)
Using the chain rule, d/dx(tan^(-1)(2 x)) = ( dtan^(-1)(u))/( du) 0, where u = 2 x and ( d)/( du)(tan^(-1)(u)) = 1/(1+u^2):
  =  2 tan^(-1)(2 x) cos(2 x)+(d/dx(2 x))/(4 x^2+1) sin(2 x)
Factor out constants:
  =  2 tan^(-1)(2 x) cos(2 x)+2 d/dx(x) (sin(2 x))/(1+4 x^2)
The derivative of x is 1:
Answer: |  =  2 tan^(-1)(2 x) cos(2 x)+1 (2 sin(2 x))/(1+4 x^2)

Looks like you made this one up! No solution in sight!!!!

Did you make this one up yourself? Looks like it!!.

x=0

Very good work Melody. It is basically an ordinary FV of 15 payments @ 6% compounded annually, or 12.36% compounded biennially. The FV will be to the end of the 29th year and you would simply add 6% to the total for the last year. Or:

FV=P{[1 + R]^N - 1/ R}=FV OF $1 PER PERIOD.
FV=5,000{[1.1236]^15 -1 /0.1236}

FV=$191,888.80 X 1.06 =$203,402.13

200 + 20 000 000 + 100 = 20 000 300

Hi Kreyn,

I like the way you worded your question.

Let's see.

\(y+1=\frac{-4}{3}(x-5)\\ \mbox{Multiply both sides by 3}\\ 3(y+1)=-4(x-5)\\ 3y+3=-4x+20\\ 3y=-4x+20-3\\ 3y=-4x+17\)

Maybe you would like to try the second one yourself?

Ask if you do not understand any of my answer.  :)

If the two roots of the quadratic 3x^2+5x+k are \(\frac{-5\pm i\sqrt{11}}{6}\) , what is k?

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ \frac{-5\pm i\sqrt{11}}{6} = {-5 \pm \sqrt{5^2-4*3*k} \over 6}\\ \frac{-5\pm \sqrt{-11}}{6} = {-5 \pm \sqrt{25-12k} \over 6}\\ -11=25-12k\\ -36=-12k\\ k=3 \)

Good one Maths Yea !  :D

An accountant decided to open an IRA for her retirement. She planned to deposit $5,000 at the end of the first year. Then she planned to skip the payment for the 2nd year and make the $5,000 deposit every other year. In other words, she deposits her $5,000 every  2nd. year, so that in 30 years she will have made a total of 15 deposit. If she is able to earn 6% compounded annually, what would the balance in her IRA be in 30 years?
Thanks for any help.

deposits end of 1{,3,5,7,9,11,13,15,17,19,21,23,25,27, 29}.    yes that is 15 deposits

6% annually = is what in 24 months?

\((1+0.06)^2=(1+x)^1\\ 1.06^2-1=x\\ x=0.1236=12.36\% \;\;every \;\;2\;\; years \)

I think I will add in the first payment later.  So I am starting at the end of year 1 but not including the first payment.

So I am going to start just after  the 1st payment is mad at the end of year 1 and finish just after the 15th deposit is made at the end of year 29.

So I will have to add the last years interest later too.

So that is (14 deposits made at the end of each 2 years + the first year will be invested for 28 years,) then the amount that totals will be invested for 1 more year.

14 deposits made at the end of each 2 years=

\(5000*\frac{(1.1236^{14}-1)}{0.1236}\)

1 deposit is in the bank for 28 years will grow to     5000*0.06²⁸

^{So I think the answer is       ((5000*(1.1236^14-1)/0.1236)+5000*1.06^28)*1.06 = $203,402.13}

^{That seems like an awful lot but it could be correct.}

^{If you used a formua for future value of an anuity due formula it would  be easier but not everyone is supposed to use that formula.}

  joke

3(-10)/8-2/(-10)

\(\frac{3\times (-10)}{8}-\frac{2}{-10}\\ =\frac{-30}{8}+\frac{2}{10}\\ =\frac{-15}{4}+\frac{2}{10}\\ =\frac{-15}{4}\times \frac{5}{5}+\frac{2}{10}\times \frac{2}{2}\\ =\frac{-75}{20}+\frac{4}{20}\\ =\frac{-75+4}{20}\\ =\frac{-71}{20}\\ =-3\frac{11}{20}\\\)

check with the calc :)

3(-10)/8-2/(-10) = -3.55

Yes thanks!

Can you please explain how this simplification process works?

\(\sqrt{\frac{13}{4}}=\frac{\sqrt{13}}{\sqrt{4}}=\frac{\sqrt{13}}{2}\)

Does that help? 

I got the same answer

You multiply each fraction by the others denominator to produce 2 fractions with an equivilent denominator. You can then just add the 2 numerators over the same denominator.

\(\frac{A}{B}+\frac{C}{D}=\frac{AD}{BD}+\frac{CB}{BD}=\frac{AD+BC}{BD}\)

As follows:

.

It is best to ask just one question at a time.

Learn from the answer or ask questions about it.

THEN

move to the next question if you still need help with it.

\(\begin{align*} y &= 2x^2 + kx + 6 \\ y &= -x + 4? \end{align*}\\~\\ y=2x^2+kx+6\\ -x+4=2x^2+kx+6\\ 0=2x^2+kx+x+6-4\\ 0=2x^2+(k+1)x+2\\ \mbox{If there is to be only one solution, the discriminant must equal 0}\\ \triangle=(k+1)^2-16=0\\ (k+1)^2=16\\ k+1=\pm4\\ \mbox{The only negative solution for k is }\\ k=-5 \)

Here is the graph.  

Thanks to the three of you.  :)

I rather like Mathyeah's answer.  :)

Or        13*1