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thank you triggsy! just had a look at Partial Fraction Decomposition. Turns out they haven't taught us that yet.

thanks!

I don't know what you mean by 'standard form'.

Assuming that the p inside the bracket is the same as the P at the front,  the equation can be written as

\(\displaystyle \frac{dP}{dt}=\frac{rP}{K}(K-P),\)

from which we have, (assuming that the r and K are constants),

\(\displaystyle \int\frac{dP}{P(K-P)}=\frac{r}{K}\int dt.\)

Now it's a partial fractions job on the first integral.

Tiggsy

ultimately I have to solve it, but I should be right once it's in standard form.

I was thinking i'm being too picky and try to solve it with the separation methord. But I don't have an expression with t...

I cannot open this.

sin(2pi+pi/6)

this must equal    sin (pi/6)  Becasue it is just sine (pi/6 + 1 rotation)

sin(pi/6) = 0.5

I really do not know what this means "Note that identically sized groups are indistinguishable."

BUT

In how many different ways can 3 men and 4 women be placed into two groups of two people and one group of three people if there must be at least one man and one woman in each group?

The groups will be      MW,     MW,     MWW

First, How many ways can MWW be chosen     3*4*3 

Now how many ways can a MW be chosen for the next group.    2*2

The left overs make the last group.

So I have     3*4*3*2*2 = 144 ways

Lucas numbers question

Find  \((a,b)\) such that

\(L_n = a\phi^n + b\widehat{\phi}^n\)

(a,b) = (1,1)

2.

The mean average temperature in Buffalo NY is 47.5 degrees. The temperature fluctuates 23.5 degrees above and below the mean temperature. If t =1 represents January, the phase shift of the sine function is 4. 

a. write a model for the average monthly temperature in Buffalo.

let x=month where December is 0 and 12 etc

let y= temperature

let a= amplitude

let h=average temperature

The wavelength will be 2pi/n

\(y=asin[n(x-k)]+h \)

h=47.5

a=23.5

k=4

The wavelength is 12 months and wavelength = 2pi/n

so

\(12=\frac{2\pi}{n}\\ n=\frac{2\pi}{12}=\frac{\pi}{6}\)

So we have:

\(y=asin[n(x-k)]+h\\ y=23.5sin[\frac{\pi}{6}(x-4)]+47.5\\ \)

b. according to the model, what is the average temperature in March?    x=3

23.5*sin(pi/6(3-4))+47.5 = 35.75

So, according to the modal the average March temp is 36 degrees

c. according to the model, what is the average temperature in August?     x=8

23.5*sin(pi/6(8-4))+47.5 = 67.851596988924

So, according to the modal the average August temp is   68 degrees

Here is the graph

Let ABCD be a parallelogram. We have that M is the midpoint of AB and N is the midpoint of BC. 

The segments DM and DN intersect AC at P and Q, respectively.

If AC = 15, what is QA?

\(\text{Let $NC = \dfrac{AD}{2} $ } \)

intercept theorem:

\(\begin{array}{|rcll|} \hline \dfrac {QC}{NC} &=& \dfrac{QA}{AD} \quad & | \quad QC = AC - QA \\\\ \dfrac {AC - QA }{NC} &=& \dfrac{QA}{AD} \\\\ (AC - QA )\cdot AD &=& NC \cdot QA \quad & | \quad NC = \dfrac{AD}{2} \\\\ (AC - QA )\cdot AD &=& \dfrac{AD}{2} \cdot QA \quad & | \quad : AD \\\\ AC - QA &=& \dfrac{ QA}{2} \\\\ AC &=& QA + \dfrac{ QA}{2} \\\\ AC &=& \dfrac{3}{2}QA \\\\ \dfrac{3}{2}QA &=& AC \\\\ QA &=& \dfrac{2}{3}AC \quad & | \quad AC = 15 \\\\ QA &=& \dfrac{2}{3}\cdot 15 \\\\ QA &=& 2\cdot 5 \\\\ \mathbf{QA} & \mathbf{=} & \mathbf{10} \\ \hline \end{array}\)

Using the quotient rule, what is the derivative of...

What is the derivative of this function?

\(g(t) = (t^3+3t^2+t)/(t^3)\)

g(t) = (t^3+3t^2+t)/(t^3)

I'm not getting how the extended power rule works

Formula power rule:

\(\begin{array}{|rcll|} \hline f(x) &=& x^n,\quad n \ne 0 \\ f'(x) &=& n\cdot x^{n-1} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline g(t) &=& \dfrac{t^3+3t^2+t} {t^3} \\\\ &=& \dfrac{t^3} {t^3} +\dfrac{3t^2} {t^3}+\dfrac{t} {t^3} \\\\ &=& 1 + 3\cdot t^{2-3} + t^{1-3} \\\\ &=& 1 + 3\cdot t^{-1} + t^{-2} \quad & | \quad \text{power rule} \\\\ g'(x) &=& 0 + 3\cdot \left((-1)\cdot t^{-1-1} \right)+ (-2) \cdot t^{-2-1} \\\\ &=& -3\cdot t^{-2} - 2 \cdot t^{-3} \\\\ &=& - \dfrac{3}{t^2} - \dfrac{2}{t^3} \\\\ \hline \end{array} \)

1. A leaf floats on the water bobbing up and down. The distance between its highest and lowest point is 4 centimeters. It moves from its highest point down to its lowest point and back to its highest point every 10 seconds. Write a cosine function that models the movement of the leaf in relationship to the equilibrium point.

**I got y = 4 cos(π/5)m as my answer - where y is the movement of the leaf and m is the relationship of the leaf to the equilibrium point, but I'm not sure if it is right. 

\(y=acos(nt)\)

a is the ampitude and that is 2

It is 2 cm from the centre of the wave to the crest (or trough)

The angle is in radians. if n=1 the period is 2pi   

in general though the period is 2pi/n

The question requires that the period is 20     

\(20=\frac{2\pi}{n}\\ 20n=2\pi\\ 10n=\pi\\ n=\frac{\pi}{10}\)

\(y=2cos(\frac{\pi t}{10})\qquad \text{where t is in seconds}\\\)

What is the derivative of this function?

2.

1.  Choose values for  x  and then find  f(x)  and  g(x)  using that  x  value.

For the first  x  value I choose  0 . Then...

f(x)  =  f(0)   =   0² - 2(0) + 2  =  2

g(x)  =  g(0)  =  3(0) - 0²  =  0

For the second  x  value I choose  1 . Then...

f(x)  =  f(1)  =  1² - 2(1) + 2  =  1 - 2 + 2  =  1

g(x)  =  g(1)  =  3(1) - 1²  =  3 - 1  =  2

For the third  x  value I choose  1/2 . Then...

f(x)  =  f(1/2)  =  (1/2)² - 2(1/2) + 2  =  1/4 - 1 + 2  =  1/4 + 1  =  5/4

g(x)  =  g(1/2)  =  3(1/2) - (1/2)²  =  3/2 - 1/4  =  6/4 - 1/4  =  5/4

For the fourth  x  value I choose  2 . Then...

f(x)  =  f(2)  =  2² - 2(2) + 2  =  4 - 4 + 2  =  2

g(x)  =  g(2)  =  3(2) - 2²  =  6 - 4  =  2

Do you see how to do this? I'll let you pick the fifth  x  value.

Notice that when  x = 1/2 ,  f(x) = 5/4  and  g(x) = 5/4

So when  x = 1/2 ,  f(x) = g(x)

And when  x = 2 ,  f(x) = 2  and  g(x) = 2

So when  x  = 2 ,  f(x) = g(x)

The  x  values that make  f(x) = g(x)  are  1/2  and  2 .

So the  x  values that are solutions to the system are  1/2  and  2 .

Expand and we have that

x^2 - (p + q)x + pq  = r^2 - (p+ q)r  + pq

x^2  - (p + q)k - r^2 + (p + q)r   = 0

The sum of the roots  =  (p + q) / 1   = p + q

The product of the roots  is [ -r^2  + (p + q)r] / 1  =  -r^2  + (p + q)r

Since  r  is one solution

Let  s   be the other   and we have

So

r + s  = p + q

s  = p + q  - r

Verify  that

r * s  =    r ( p + q - r)   =  -r^2  + (p + q)r

Angles A and C are supplementary, or add up to 180 degrees:

(x - 5) + (x + 5) = 180

2x = 180        divide both sides by 2

x = 90 degrees.

Angle D = 90 - 8 =82 degrees.

Angle B = 180 - 82 = 98 degrees.

area of sector   =   (1/2) * (radius)² * measure of central angle in radians

area of sector   =   (1/2) * 12² * 2.3      sq. meters

area of sector   =   (1/2) * 144 * 2.3      sq. meters

area of sector   =   72 * 2.3     sq meters

area of sector   =   165.6        sq meters

I think I see what you did......remember  12² = 12 * 12   ....NOT....  12 * 2     

A)   ∠A  intercepts arc CD.

B)   ∠BED  intercepts arc BCD.

C)   The measure of an inscribed angle is half the measure of its intercepted arc. So...

       The measure of ∠BED is half the measure of arc BCD

       The measure of ∠BED   =   (1/2)(the measure of arc BCD)

       85°   =   (1/2)(the measure of arc BCD)

       2(85°)   =   the measure of arc BCD

       170°   =   the measure of arc BCD

D)   The measure of ∠A   =   (1/2)(the measure of arc CD)

       The measure of ∠A   =   (1/2)(128°)

       The measure of ∠A   =   64°

E)   The sum of all the arcs in a circle is 360° , so...

       The measure of the rest of the circle   =   360° - 82°

       The measure of the rest of the circle   =   278°

Let R be the intersection of the medians

DR  =1/3  of DP  = 6

ER  =1/3  of EQ  = 8

So....since the medians are perpendicular....then  DE  is the hypotenuse of right triangle DRE  

So....call S  the midpoint  of DE...so  DS  = 5

So.... SF  is another median

The sine of RDE  = 8/10  = 4/5

cos RDE   = 3/5

So the distance  from  R to S  can be found using the Law of Cosines  as

RS^2  = DS^2 + DR^2 -2(DS * DR)cos(RDE)

RS^2  = 5*2 + 6^2  - 2(30)cosRDE

RS^2  = 61 - 60(3/5)

RS^2  = 61  - 36

RS^2  = 25

RS  = 5

But  RS  = 1/3  of SF....so SF  = 15

We need to find  sin DSR

Using the Law of Sines we have

sinDSR / DR  = sinRDE/ RS

sinDSR/6  = sinRDE / 5

sinDSR / 6   = (4/5) / 5

sinDSR / 6  = 4/25

sinDSR  = 24/25

And the cos DSR   = √ [1  - (24/25)^2 ]  = √ [625 - 576 ] / 25  = √49 / 25 =  7/25

So...we can find DF with the Law of Cosines as

DF^2  = DS^2  + SF^2  - 2(DS*SF)cos(DSR)

DF^2  = 5^2  + 15^2  - 2(5*15)(7/25)

DF^2  = 250  - 2*3*7

DF  = 250 - 42

DF^2  = 208

DF = √208  = 4√13

Thanks so much, CPhill!

OK

We need to  find the difference between the x coordinates of A and B= (5 -2)  =3

Since  AC  = 3AB

3 times this difference plus the x  coordinate of A  will give us the x coordinate of C

So    3(5-2)  + 2  =   3*3  + 2  = 11 

Similarly.......3 times the difference between the y coordinates of A  and B  plus the y coordinate of A  will give us the y coordinate of C....so we have

3 ( 9 - 3)  +  3   =  3*6  + 3  =  21

So....the coordinates of C   are   (11, 21)

The  volume  of the 8 steel balls  =

8 * (4/3)pi    =    32/3 pi

So....to find the radius, r, of the large ball, we have

32/3 pi  =  (4/3)pi (r)^3

(32/ 3) (3/4)  = r^3

32 / 4  =  r^3

8  =  r^3

r  = 2