Alan

It's a quadratic in t.  

Rewrite it as  1/2at² +v₀t+x₀-x=0   

Multiply by 2/a:    t² +(2v₀/a)t +2(x₀-x)/a = 0

Use the quadratic formula:

t = { -(2v₀/a) + sqrt[ (2v₀/a)² - 4*2(x₀-x)/a) ] }/2    → { -v₀ + sqrt[ v₀² - 2a(x0-x) ] }/a

(Because t is almost certainly time here, we don't need to look at the other solution)

"Hi Can you show steps in solving this problem:

solve(1/x+1/y=3/4, x^2/y+y^2/x=9)"

As follows:

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Let m = mother's age.

m = 3k + 1      (1)

m + k = 57      (2)

from (2) we have m = 57 - k.    (3)

Replace the m in (1) using (3)

57 - k = 3k + 1

Subtract 1 from both sides

56 - k = 3k

Add k to both sides

56 = 4k

Divide both sides by 4

14 = k     (or just k = 14)

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Here's a visual representation:

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ABPA:  \(p_1=4+4+4\sqrt2 \)

BCDB: \(p_2=8+8+8\sqrt2 \)

BP:   \(p_3=4\)

ABCDPA: \(p=p_1+p_2-p_3\rightarrow20+12\sqrt2\)

Couple of examples of commutative matrices:

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No, it is not true in general that P^-1AP = APP^-1.  Here is a simple counter example:

It is always true that APP^-1 = A

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29 + 92 = 121

38 + 83 = 121

47 + 74 = 121

56 + 65 = 121

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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.

Note that a = b/2 so:

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