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sqrt(1 + x)^ -1 =1 - 1/2x + [1.3/2.4]x^2 - [1.3.5/2.4.6]x^3 + [1.3.5.7/2.4.6.8]x^4.........etc.

N(t) = N(0)*e^-kt  where  k = ln(2)/tau   tau is half-life.

k = ln(2)/432 ≈ 1.605*10^-3 per year

N(20) = 200*e^{-1.605*10^(-3)*20} ug ≈ 193.7 ug = 190 ug to the nearest 10 ug.

R =200 x 2^-(20/432)

R =200 x 2^-(1/21.6)

R =200 x 0.96841927.....

R =193.684 ug remains after 20 years.

 (1 + x)^ -1/2  at x = 0

Using the Maclaurin Series, we have

f(0) + f ' (0)x  + f ''(0) x^2 / 2!  +   f  '''(0) x^3 / 3! + ......  + f⁽ⁿ⁾(0) x^n / n!

f(0)  =  1

f ' (x) =  ( -1/2) (1 + x)^( -3/2)   and   f ' (0)  =  (-1.2)  ( 1 + 0)^-(3/2)=   (-1/2)

f '' (x)  =  (3/4) (1 + x) ^(-5/2)  and   f '' (0)  = (3/4)  (1 + 0)^(-5/2)  =   (3/4)

f ''' (x)  =   (-15 / 8 ) ( 1 + x)^(-7/2)   and  f''' (0)  = (-15/8) (1 + 0)^(-7/2)  = ( - 15 / 8)

So we have the first four terms

1 - (1/2)x + (3/4)x^2/2   - (15/8)x^3 / 6 + ..... +  f⁽ⁿ⁾(0) x^n / n!  =

1 - x/2  + 3x^2/8  - 15x^3/48 + .....+  f⁽ⁿ⁾(0) x^n / n!  =

1 - x/2  + 3x^2/8  - 5x^3/16 + ...... +  f⁽ⁿ⁾(0) x^n / n!

Note that a = b/2 so:

.

Factor the following:
6 m^2 n - 12 m n^2 - 10 m n + 20 n^2

Factor 2 n out of 6 m^2 n - 12 m n^2 - 10 m n + 20 n^2:
2 n (3 m^2 - 6 m n - 5 m + 10 n)

Factor terms by grouping. 3 m^2 - 6 m n - 5 m + 10 n = (10 n - 6 m n) + (3 m^2 - 5 m) = m (3 m - 5) - 2 n (3 m - 5):
2 n m (3 m - 5) - 2 n (3 m - 5)

Factor 3 m - 5 from m (3 m - 5) - 2 n (3 m - 5):
Answer: | 2n (3m - 5) (m - 2n)

a- ∑[1/(2n), n, 1, 120] =2.6844341436766............etc.............(1)

b-∑[1/(n), n, 1, 60] =4.679870412951737.............etc. ...........(2)

a + b =2.6844341436766..  +   4.679870412951737........ =7.364304556628337 ????????!!!.

Factor the following:
6 q^2 - 7 q r - 24 r^2

6 q^2 - 7 q r - 24 r^2 = 6 q^2 - 7 q r - 24 r^2:
6 q^2 - 7 q r - 24 r^2

The coefficient of q^2 is 6 and the coefficient of r^2 is -24. The product of 6 and -24 is -144. The factors of -144 which sum to -7 are 9 and -16. So 6 q^2 - 7 q r - 24 r^2 = 6 q^2 - 16 q r + 9 q r - 24 r^2 = 3 q (2 q + 3 r) - 8 r (2 q + 3 r):
3 q (2 q + 3 r) - 8 r (2 q + 3 r)

Factor 2 q + 3 r from 3 q (2 q + 3 r) - 8 r (2 q + 3 r):
Answer: | (2q + 3r ) (3q - 8r)

How did you do it?

Find the FV of the $27,500 @ a given rate such as 12% compounded monthly...................(1)
Then find the FV of the 36 monthly payments of $516.60 @ the same given rate of 12% C.M..........(2)
Subtract (2) from (1) above and the result should be as close to $15,500 as possible. Iterate (1) and (2) above to get as close as possible to $15,500. That interest rate, compounded monthly, would be the return to the car dealership. This is how it would be calculated manually to get the answer.
However, we are going to use this fine online calculator to find the rate directly. We enter -$27,500 under PV. Notice that we enter it as a negative amount because it is money "owed" from the point of view of the car dealership. Then we enter $15,500 under FV, $516.60 under PMT, and 36 under NP, or number of payments. We choose the "beginning" mode since the lease payments are paid in advance or at the beginning of the month. Then we simply press "IR", or interest rate. It will give us the interest rate "per period", or in this case "per month". Then, we simply multiply it by 12 to give us annual rate compounded monthly.
We find that the calculator gives a monthly rate of 0.854169 x 12 =10.25% compounded monthly. Here is the financial calculator:
https://arachnoid.com/finance/index.html?pv=0&fv=100000&np=200&ir=1&pb=false

6q^2 - 7qr - 24r^2

\( 6q^2 - 7qr - 24r^2 = 6(q-\frac83r) (q+\frac32r) \)

you need $66132 at the end of twenty years. yo know that the best you can do is to make equal payments into a bank account on which you can earn 5% of interest annually. what amount must you plan to pay annually to achieve your objective?

FV = P x {[1 + R]^N - 1 / R}

66,132 = P x {[1 + 5%]^20 - 1 / 5%}

66,132 = P x {1.05^20 - 1 / 0.05}

66,132 = P x        33.0659541.......

P = 66,132 / 33.0659541......

p = $2000.00 - annual payment required to save $66,132 in 20 years @ 5% annual rate.

Like so:

.

completely factor this expression?

6m^2n - 12mn^2 - 10mn + 20n^2

\(\begin{array}{|rcll|} \hline && 6m^2n - 12mn^2 - 10mn + 20n^2 \\ &=& 2n ( 3m^2 - 6mn - 5m + 10n) \\ &=& 2n (m-2n) (3m-5) \\ \hline \end{array}\)

So that you have: (22.2222...)^2 =-493.827 / (2* -7.5)=~32.92m     

This is a wonder of slop to behold! This makes it very clear that (22.2222...)^2 = ~32.92m. I could guess who the bone brain (BB) is that presented this lazy slop.     

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Sir Alan assumes because you are doing grade 11 physics you should be able to do grade 8 algebra.

To answer your question for “+” becoming “/” in the solution:  It is not the same equation. Alan’s solves for “s” in the second equation.
Here’s a predigested  presentation:

\(v^2=u^2+2as\\ \text {Solve for s }\\ s=(v^2-u^2)/(2a)\\ s=\dfrac{(0^2-22.22^2)}{(2(-7.5))} \leftarrow \small \text {(-7.5 because scooter is decelerating) }\\ s=\dfrac{(-(22.22^2)}{-15} \leftarrow \small \text {(Notice the negative divided by a negative. This will give a positive solution.)}\\ s=\dfrac{(-493.82)}{-15}\\ s=32.92m \\\)

Because the "-" sign cancels when you divide 80,000/3,600=22.2222......., which becomes negative when you move it to the left side of the equation. So that you have: (22.2222...)^2 =-493.827 / (2* -7.5)=~32.92m. Or, you could cancel the negative sign by multiplying both sides by -1, which is the same thing.

And why did + in your formula become / in your solution? Thanks. :]

Can I ask why -7.5 suddenly became 7.5 in your solution?

Thanks, heureka......your answer is more elegant.....but hectictar and I had more fun folding paper........!!!!!!

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Albert is riding his scooter at a velocity of 80km/h, when he sees an old woman crossing the road 45m away. He immediately steps hard on the brakes to get the maximum deceleration of 7.5m/s^2. How far will he go before stopping? 

Do you need to do this with claculus or with physics formulas?

FORMULA METHOD

Here are the motion formulas that you should learn.

\(80km/hour =80000/(60*60) m/sec=800/36 m/sec=800/36 m/sec=22.\dot2 m/s\\\)

u=22.2 m/s

v=0

t=?

a=-7.5

s=?

I want stopping distance, s and i do not have t so that means I can use the last formula

\(v^2=u^2+2as\\ 0=22.\dot2^2+2*-7.5*s\\ s\approx 32.92m\)

The scooter can stop in 33m so Albert should stop in time.    

NOW USING CALCULUS

\( \ddot x=-7.5\\ \dot x = -7.5t+22.\dot2\\ x=\frac{-7.5t^2}{2}+22.\dot2t\\ x=-3.75t^2+22.\dot2t \)

When \(\dot x = 0\)

\(0=-7.5t+22.\dot2\\ t\approx2.963sec\\ x\approx-3.75*2.963^2+22.\dot 2*2.963\\ x\approx 32.92m \)

Fortunately the answers are the same :)

Three sides of triangle ABC has measure 3, 4, 6 .O is point such that angle angle OBC=angle OAB then find angle OAB.

It goes on.

Coordinate system x, y on the triangle.
A (0,0)
B (6,0)
O (x_o,y_o)

angle OAB = OBC = δ

Equation A

\(m=tan\ (δ)\)

\(tan\ (δ)=\color{blue}\frac{y-y_o}{x-x_o}\)                                Point-direction form

Equation B

\(m=tan\ (180°-\beta\ +\ δ)\)  

\(\beta=52.831°\)

\(tan\ (180°-52.831°\ +\ δ)=\frac{y-y_o}{x-x_o}=\frac{y-y_B}{x-x_B}\)   Two-point form

\(tan\ (127.169°\ +\ δ)={\color{blue}\frac{y-y_o}{x-x_o}}=\frac{y}{x-6}\) 

I have to think. Sorry. 

Guest...if you're still on....check my answer again...I made a slight mistake the first time....!!!