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## Masters of Education Graduate with Mathematics Expertise

### Brookline, MA (02446)

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# Roman's Responses in WyzAnt Answers

#### Find the area of the parallelogram?

Find the area of the parallelogram spanned by the vectors <1, 0, -1> and <-2, 2, 0>.

I have troubles with dot products.

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Asked by Sun from Los Angeles, CA
20

This one is actually a classic problem for cross products.

<1, 0, -1> × <-2, 2, 0>

|  i     j    k  |

=    |  1    0   -1 |

| -2    2    0 |

= <0*0 - (-1)*2, (-1)*(-2) - 1*0, 1*2 - 0*(-2)>

= <2, 2, 2>

Area = ||<2,2,2>|| = √(22 + 22 + 22) = 2√3.

#### Limit questions

Evaluate the limit..

Lim Δx-->0   sin ( x + Δx ) - sin x / Δx

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Asked by Aqib from Carrollton, TX
10

By the definition of derivatives, you get the following

limΔx→0 [ sin( x + Δx ) - sin x ] / Δx = d/dx sin x = cos x, assuming that the angle x is measured in radians.

If you want to derive it, you will need to use the angle sum formula which is

sin(a+b) = sin a cos b + cos a sin b.

You get

limΔx→0 [ sin( x + Δx ) - sin x ] / Δx

= limΔx→0 [ sin x cos Δx + sin Δx cos x - sin x ] / Δx

= limΔx→0 (sin x)(cos Δ - 1) / Δx + lim Δx→0 (cos x)(sin Δx) / Δx

= (sin x) limΔx→0 (cos Δx - 1) / Δx + (cos x) lim Δx→0 (sin Δx) / Δx

Now all we need is to calculate the two limits

limΔx→0 (sin Δx) / Δx can be computed by drawing a picture. I won't draw it here, but from the correct picture you get that

sin Δx < Δx < tan Δx for 0 < Δx < Π/2

and since sin Δx, Δx, and tan Δx are odd functions,

sin Δx > Δx > tan Δx for -Π/2 < Δx < 0.

Dividing either inequality by Δx and taking reciprocal gives.

1 > (sin Δx) / Δx > cos Δx for -Π/2 < Δx < Π/2 and Δx ≠ 0.

Since limΔx→0 cos Δx = 1, the squeezing theorem implies that

limΔx→0 (sin Δx) / Δx = 1.

We can use this limit to evaluate the other limit as follows.

limΔx→0 (cos Δx - 1) / Δx

= limΔx→0 (cos Δx - 1)(cos Δx + 1) / [Δx (cos Δx + 1)]

= limΔx→0 (cos2 Δx - 1) / [Δx (cos Δx + 1)]

= limΔx→0 (-sin2 Δx) / [Δx (cos Δx + 1)]

= -(limΔx→0 (sin Δx) / Δx)(limΔx→0 (sin Δx) / (cos Δx + 1))

= 1*(sin 0) / (cos 0 + 1)

= 0 / 2

= 0

Plugging these limits in gives

limΔx→0 [ sin( x + Δx ) - sin x ] / Δx = 0 sin x + 1 cos x = cos x.

#### Limits.....

Lim θ-->Π/2     cot θ - cos θ / cos3θ

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Asked by Aqib from Carrollton, TX
00

limθ→Π/2 (cot θ - cos θ) / cos3θ

= limθ→Π/2 (csc θ - 1) cos θ / cos3 θ

= limθ→Π/2 (csc θ - 1) / cos2 θ

= limθ→Π/2 (1 - sin θ) csc θ / (1 - sin2 θ)

= limθ→Π/2 (1 - sin θ) csc θ / [(1 - sin θ)(1 + sin θ)]

= limθ→Π/2 csc θ / (1 + sin θ)

= csc(Π/2) / [1 + sin(Π/2)]

= 1 / (1+1)

Two diseases A and B have spread within a certain population. We
know that 15% of this population have the disease A, 20% of this
population have disease B and 25% have at least one of the two diseases.
We choose a random person among this population.

(a) What is the probability that this person contracted both diseases?
(b) Given that the chosen individual has at least one of the diseases,
what is the probability that this person has contracted both dis-
eases?

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Asked by Mathew from Williamston, NC
20

Start by making a table like so.

B:Yes         B:No

A:Yes       W              X

A:No         Y               Z

We know that 15% have disease A, so 1: W+X = 15%.

We know that 20% have disease B, so 2: W+Y = 20%.

We know that 25% have at least one of the two diseases, so 3: W+X+Y = 25%.

The four entries in the table total the entire population, so 4: W+X+Y+Z = 100%

Subtracting equation 2 from equation 3 gives us X = 5%.

Subtracting equation 1 from equation 3 gives us Y = 10%.

Plugging these in to equation 3 gives W + 5% + 10%= 25%, so W = 10%.

Finally, subtracting equation 3 from equation 4 gives us Z = 75%.

We can now fill in the table:

B:Yes         B:No

A:Yes        10%          5%

A:No         10%          75%

You should now be able to answer both questions.

#### equation of a locus

Find the equation of a locus of point whose distance from the point (2,-2) is equal to its distance from the line      x-y=0

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Asked by Aqib from Carrollton, TX
20

First of all, note that it will be a parabola, and that (2,-2) is it's focus and x - y = 0 is it's directrix.

Let's work with the squares of the distances instead since then there won't be any radicals.

Given a point (x,y) on this parabola, we have the squared distance from the focus as

D2focus = (x - 2)2 + (y + 2)2.

To get the squared distance from the directrix, D2directricx , we we can use vectors.

Pick a vector perpendicular to the directrix.

Recall that vector < a , b > is perpendicular to line ax + by = c

For example, v = < 1 , -1 > is perpendicular to our directrix x - y = 0.

Let u = < x , y >.

Then the desired distance is Ddirectricx = u · v / ||v||.

Thus D2directricx = (<x , y> · <1 , -1> / || <1 , -1> ||)2

= (x - y)2 / 2

The locus you are looking for is the same as D2focus = D2directrix which is

2[(x - 2)2 + (y + 2)2] = (x - y)2

2x2 - 8x + 8 + 2y2 + 8y + 8 = x2 - 2xy + y2

x2 + 2xy + y2 - 8x + 8y + 16 = 0

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