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We are in vacation now , but those questions for UOS exam.

so think befor answering next time.

No not your physics homework

Thanks Rom: Here is my attempt at this:

(x + 2009)^2 - (2009 - x)^2 = 2009^2

Expand out terms of the left hand side:
8036 x = 4036081

Divide both sides by 8036:
x = 2009/4 = 502.25

∑[k*(k - 1) /2 *cos (k*(k - 1) *pi /2), k, 19, 98] = - 40
A1 - A18 = (0, -1, -3, 6, 10, -15, -21, 28, 36, -45, -55, 66, 78, -91, -105, 120, 136, -153)
A19 - A98 =(-171, 190, 210, -231, -253, 276, 300, -325, -351, 378, 406, -435, -465, 496, 528, -561, -595, 630, 666, -703, -741, 780, 820, -861, -903, 946, 990, -1035, -1081, 1128, 1176, -1225, -1275, 1326, 1378, -1431, -1485, 1540, 1596, -1653, -1711, 1770, 1830, -1891, -1953, 2016, 2080, -2145, -2211, 2278, 2346, -2415, -2485, 2556, 2628, -2701, -2775, 2850, 2926, -3003, -3081, 3160, 3240, -3321, -3403, 3486, 3570, -3655, -3741, 3828, 3916, -4005, -4095, 4186, 4278, -4371, -4465, 4560, 4656, -4753)>>Total = - 40

Note: somebody should ckeck this. Thanks.

reposted for readability

\(\text{Square A and Square B are both $2009$ by $2009$ squares.$\\$ Square A has both its length and width increased by an amount $x,\\$ while Square B has its length and width decreased by the same amount $x.\\~\\$ What is the minimum value of $x$ such that the difference in area between$\\$ the two new squares is at least as great as the area of a $2009$ by $2009$ square? }\)

All right, thanks! 

A lattice point in the  -plane is a point both of whose coordinates are integers (not necessarily positive). How many lattice points lie on the hyperbola ?       \(x^2-y^2=17\)

\(x^2-y^2=17\\ y=\pm\sqrt{x^2-17}\)

\(x\in \mathbb Z\ |\ \{[x^2-17]\}\subset \{squares\}\)

There are only four grid points P.

\(P_1(-9,-8)\\ P_2(-9,\ 8)\\ P_3(9,-8)\\ P_4(9,\ 8)\\\)

  !

\(\text{This is straightforward application of the binomial distribution}\\ p[\text{k answers correct}] = \dbinom{10}{k} \left(\dfrac 1 5\right)^k \left(\dfrac 4 5\right)^{10-k}\\~\\ p[0] = \left(\dfrac 4 5\right)^{10}\\ p[0\leq k \leq 3] = \sum \limits_{k=0}^3 p[k]\\ p[2\leq k \leq 10] = 1 - p[0\leq k \leq 1]= 1- p[0]-p[1]\\~\\ \text{I leave the plug and chug to you}\)

The force needed to loosen a bolt varies inversely with the length of the handle of the wrench used. A wrench with a handle length of 9 inches requires 375 pounds of force to loosen a certain bolt. A wrench of 15 inches will require how many pounds of force to loosen the same bolt?

The answer is 225pounds. 

PLEASE DO NOT ASK FOR ANY EXPLANATIONS BECAUSE I ALREADY KNOW IT WORKS.(I TESTED IT OUT ON MY MATH PROGRAM)

If it is a multiple of six, and you are losing two of the cars, then the remaining number of cars will be divisible by two but not by six since it will not be divisible by three.

The first few values are 6, 12, 18, 24, 30, 36 which means 4, 10, 16, 22, 28, 34. The numbers four and eight work, so add two more.

Thus, there should be three values that will work.

Thank you so much!!!

Hi Melody,

Thank you so much..okay, yes, a very logic approach....thank you..

Hi Juriemagic,

I do not like the wording of this problem one little bit but this is what it means.

You should be able to see now   

Thank you very much!! :)

Let \(f(x) = 5x+3\) and \(g(x)=x^2-2\). What is \(g(f(-1))\)?

\(\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{5x+3} \quad | \quad x=-1 \\ f(-1) &=& 5(-1)+3 \\ f(-1) &=& -5+3 \\ \mathbf{f(-1)} &=& \mathbf{-2} \\ \hline && \mathbf{g(f(-1))} \quad | \quad f(-1)=-2 \\ &=& g(-2) \quad | \quad \mathbf{g(x)=x^2-2},\ x=-2 \\ &=& (-2)^2-2 \\ &=& 4-2 \\ &=& \mathbf{2} \\ \hline \end{array}\)

\(\mathbf{g(f(-1))} = \mathbf{2} \)

SOLVE SECOND ORDER INITIAL VALUE PROBLEM.
\(y'' +6y' +5y = 0 ,\ y(0)=1 \text{ and } y' (0) = 1\)

\(\text{Let $y=e^{rx}$} \\ \text{Let $y'=re^{rx}$} \\ \text{Let $y''=r^2e^{rx}$} \)

\(\begin{array}{|rcll|} \hline \mathbf{y'' +6y' +5y} &=& \mathbf{0} \\\\ r^2e^{rx} +6re^{rx} +5e^{rx} &=& 0 \\ \underbrace{e^{rx}}_{\neq 0}\cdot \underbrace{ \left(r^2 +6r +5\right) }_{=0} &=& 0 \\ && \begin{array}{|rcll|} \hline r^2 +6r +5 &=& 0 \\ r &=& \dfrac{-6\pm \sqrt{36-4\cdot 5}}{2} \\ r &=& \dfrac{-6\pm \sqrt{16}}{2} \\ r &=& \dfrac{-6\pm 4}{2} \\\\ r_1 &=& \frac{-6+4}{2} \\ \mathbf{r_1} &=& \mathbf{-1} \\\\ r_2 &=& \frac{-6-4}{2} \\ \mathbf{r_2} &=& \mathbf{-5} \\ \\ \hline \end{array} \\\\ y &=& c_1\cdot e^{r_1\cdot x} + c_2\cdot e^{r_2\cdot x} \\ \mathbf{y} &=& \mathbf{c_1\cdot e^{-1\cdot x} + c_2\cdot e^{-5\cdot x} } & (1) \\ \mathbf{y'} &=& \mathbf{-c_1\cdot e^{-1\cdot x} -5\cdot c_2\cdot e^{-5\cdot x} } & (2) \\ \hline y &=& c_1\cdot e^{-x} + c_2\cdot e^{-5\cdot x} \quad | \quad y(0)=1,\ x=0,\ y=1\\ 1 &=& c_1\cdot e^{-0} + c_2\cdot e^{-5\cdot 0} \\ \mathbf{1} &=& \mathbf{c_1 + c_2} & (3) \\\\ y' &=& -c_1\cdot e^{-x} -5\cdot c_2\cdot e^{-5\cdot x} \quad | \quad y'(0)=1, \ x=0,\ y'=1\\ 1 &=& -c_1\cdot e^{-0} -5\cdot c_2\cdot e^{-5\cdot 0} \\ \mathbf{1} &=& \mathbf{-c_1 - 5c_2} & (4) \\\\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (3) & \mathbf{1} &=& \mathbf{c_1 + c_2} \\ (4) & \mathbf{1} &=& \mathbf{-c_1 - 5c_2} \\ \hline (3)+(4): & 1+1 &=& c_1 + c_2+-c_1 - 5c_2 \\ & 2 &=& c_2 - 5c_2 \\ &-4c_2 &=& 2 \quad | \quad :(-4) \\ & \mathbf{c_2} &=& \mathbf{-\dfrac{1}{2}} \\\\ & 1 &=& c_1 + c_2 \\ & 1 &=& c_1 -\dfrac{1}{2}\\ & c_1 &=& 1+\dfrac{1}{2} \\ & \mathbf{c_1} &=& \mathbf{ \dfrac{3}{2}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{y} &=& \mathbf{ c_1\cdot e^{-1\cdot x} + c_2\cdot e^{-5\cdot x} } \\ \mathbf{y} &=& \mathbf{ \dfrac{3}{2}\cdot e^{-x} -\dfrac{1}{2}\cdot e^{-5\cdot x} } \\ \hline \end{array}\)

x mod 4 = 3,    x mod 3 = 2,     x mod 5 =2, solve for x

The smallest x that will satisfy the 3 congruences is:

x = 47  and  52  - 47 = 5 cards were lost.

Hi tommarvoloriddle,

nope, that's how it is....I cannot upload pictures so I'll  try do describe it. The line goes from P to Q. Then from Q extends to T. An arrow pointing from P to Q indicates 20cm. An arrow REVERSE from T to Q indicates K2. 

Okay, they give a hint and say remember to set PT as 3K, and TQ(QT) AS 2K..so PQ is then K....Then they give the answers...no explanation...

Hi juriemagic!

A quick question about something that I don't think matters either way-

You mean

"T bisects PQ in the ratio 3:2" not

"T bisects PQ in the ratio 3:-2"

right?

THANK YOU!

I didn't know there was a formula...

Thank you for your answers!

But I think your way is the better way, as my math textbook is just teaching you steps you can use but are crazily bulky.

I learned it using steps like above but you would continue it like this(found my math textbook yay)

:

Then you have to find the mean of the last column

(1+4+0+9)/4

14/4

7/2

then square root that to get 

sqrt(7/2)

and the answers sqrt(7/2)

which is around 1.87

(Textbook way, no formula...)

I think This problem was categorized terribly as it accepts the population answer.

wait- there's a formula?

3,2,4,7

Find the standard deviation of the data set.
Round your answer to the nearest hundredth.

Standard deviation of the sample

\(s^2=\frac{\sum (x_i-\overline x)^2}{n-1}\)

\(n=4\\ \overline x=\frac{3+2+4+7}{4}\\ \overline x=4\)

\(s^2=\frac{(3-4)^2+(2-4)^2+(4-4)^2+(7-4)^2}{4-1}\\ s^2=\frac{1+4+0+9}{3}\\ s^2=\frac{14}{3}\\ s=\sqrt{\frac{14}{3}}=2.1602469\)

\(s\approx 2.16\)

Standard deviation of the population

\(s^2=\frac{\sum (x_i-\overline x)^2}{n}\)

\(s^2=\frac{(3-4)^2+(2-4)^2+(4-4)^2+(7-4)^2}{4}\\ s^2=\frac{1+4+0+9}{4}\\ s^2=\frac{14}{4}\\ s=\sqrt{\frac{14}{4}}=1.87083\)

\(s\approx 1.87\)

  !

3,2,4,7

Find the standard deviation of the data set.
 Round your answer to the nearest hundredth.

Standard deviation of the sample

\(s^2=\frac{\sum(x_i-\overline x)^2}{n-1}\)

\(n=4\\ \overline x=4\)

\(s^2=\frac{(3-4)^2+(2-4)^2+(4-4)^2+(7-4)^2}{4-1}\\ s^2=\frac{1+4+0+9}{3}\\ s^2=\frac{14}{3}\)

\(s=\sqrt{\frac{14}{3}}\\ s=2.1602469\)

\(s\approx 2.16\)

  !