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Please do not post your AOPS questions here. 

Find the sum of the first 100 terms
of the arithmetic sequence
with the first term 2
and the common difference 5.

$\begin{array}{|rcll|} \hline \text{sum} &=& \dfrac{2+\Big( 2+(100-1)*5 \Big)}{2}*100 \\\\ \text{sum} &=& (2+2+99*5)*50 \\\\ \text{sum} &=& (4+495)*50 \\\\ \text{sum} &=& 499*50 \\\\ \mathbf{\text{sum}} &=& \mathbf{24950} \\ \hline \end{array}$

In the diagram below,
$\angle PQR = \angle PRQ = \angle STR = \angle TSR =\angle A,\\ RQ = 8,~ \text{ and } SQ = 2.$
Find PQ.

$\text{Let $\angle QPR = 180^\circ - 2A$} \\ \text{Let $\angle RQT = 180^\circ - 2A$} \\ \text{Let $\angle TRS = 180^\circ - 2A$} \\ \text{Let $\angle PTQ = 180^\circ - A$}\\ \text{Let $PQ=x$ } \\ \text{Let $QR = QT = 8$ } \\ \text{Let $TR = SR = y$ }$

$\begin{array}{|rcll|} \hline ST&=&QT-SQ \\ ST&=&8-2\\ ST& =& 6 \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \text{In $\triangle [QPT]$:} \\ \hline \dfrac{\sin(180^\circ-2A)}{8} &=& \dfrac{\sin(180^\circ-A)}{x} \\\\ \dfrac{\sin(2A)}{8} &=& \dfrac{\sin(A)}{x} \\\\ \dfrac{\sin(2A)}{\sin(A)} &=& \dfrac{8}{x} \qquad (1)\\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \text{In $\triangle [QTR]$:} \\ \hline \dfrac{\sin(180^\circ-2A)}{y} &=& \dfrac{\sin(A)}{8} \\\\ \dfrac{\sin(2A)}{y} &=& \dfrac{\sin(A)}{8} \\\\ \dfrac{\sin(2A)}{\sin(A)} &=& \dfrac{y}{8} \qquad (2)\\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \text{In $\triangle [QSR]$:} \\ \hline \dfrac{\sin(180^\circ-2A)}{6} &=& \dfrac{\sin(A)}{y} \\\\ \dfrac{\sin(2A)}{6} &=& \dfrac{\sin(A)}{y} \\\\ \dfrac{\sin(2A)}{\sin(A)} &=& \dfrac{6}{y} \qquad (3)\\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline (2)=(3):& \dfrac{y}{8} &=& \dfrac{6}{y} \\\\ & y^2 &=& 48 \\ & y^2 &=& 16*3 \\ & \mathbf{y} &=& \mathbf{4\sqrt{3}} \\ \hline \end{array} \begin{array}{|lrcll|} \hline (2)=(1):& \dfrac{y}{8} &=& \dfrac{8}{x} \\\\ & xy &=& 64 \\\\ & x &=& \dfrac{64}{y} \\\\ & x &=& \dfrac{64}{4\sqrt{3}} \\\\ & x &=& \dfrac{16}{\sqrt{3}} \\\\ & x &=& \dfrac{16}{\sqrt{3}} * \dfrac{\sqrt{3}}{\sqrt{3}} \\\\ & \mathbf{x} &=& \mathbf{\dfrac{16}{3}\sqrt{3}} \\ \hline \end{array}$

$PQ = \mathbf{\dfrac{16}{3}\sqrt{3}}$

"Let f(x) be a polynomial such that f(0) = 4,f(1) = -5, and f(2) = 11. Find the remainder when f(x) is divided by x(x - 1)(x - 2)."

Note:  The equation for r(1) should be:   a + b + c = -5.

Thanks for explaining CV

I wish there was a way we could do this easily without adding the questions to our watch list.

Sometimes I take a pic of the question with Gyazo (which i use for all my pictures)

The metadata is added automatically so I can hyperlink back to the question later on.

Yes   

Thanks Catmg  

[ADE] ≈ 21.108

[BDP] ≈ 22.154

Edit: Nevermind! After rereading it and making some equations, I got it now! Thank you 

In the diagram below, EU = 8, UF = 7,GV = 5, VH = 16, and UV= 8. Find XY.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

UX => m            VY => n

m(8 + n) = 7 * 8         m = 56 / (8 + n)

n(8 + m) = 5 * 16        

n[8 + 56 / (8 + n)] = 80        n ≈ 6.787

m = 56 / (8 + n)                   m ≈ 3.787

XY = √345   ≈ 18.574

Hi, how are you doing? I think that exact question was answered here:

https://web2.0calc.com/questions/algebra-problem_37

O(∩_∩)O

Hi! Perimeter is the distance outside of a shape, so all you have to do is add each side up! (15+27+21) Hope this helps you understand it a bit better 

Attn: Helpplzplzplz:

This is a very standard problem that could have come from anywhere.

Please do not spend your time dobbing on other question askers. 

It really is not a good look.

Answer: $1$

Solution:

You need y² to be as small as possible and x² to be as large as possible. You can set y to 0 to make y² as small as possible. You want x to be as big as it can possibly be, which happens to be 1, in this case. Doing the calulculations result in 1² - 0² = 1 - 0 = $1$.

Answer: $81.25\%$

Solution:

If Pablo gives 20% of his money to Mary, that means he keeps 4/5 of the money. If Mary has 25% more than 4/5 of the original money after she is gives 20% of Pablo's money, then that means that after being gives 20% of Pablo's money, she now has Pablo's original amount. (Read that and the following carefully, I know this answer isn't clear.) Because Mary's original amount of money is Pablo's original amount of money minus 20% of Pablo's money, Mary's original amount of money is 80% of Pablo's money. 20% x 80% = 16%, which means that Mary now has 64% of Pablo's original amount of money, and Pablo has 116% of his original amount. 116 is 81.25% larger than 64, meaning that if Mary gives 20% of her money to Pablo, then he would have more money than Mary by $81.25\%$.

You'll need to graph the equations to see the area in question

Then you can see it is a triangle....you can easily see the base and the height to calculate the area = 1/2 b h

It should be sin(angle A) and sin(angle C).

Sorry :(

First term==2

Common difference ==5 

The sequence is:

2 , 7 , 12 , 17 , 22 , 27 , 32 , 37 , 42 , 47 , 52 , 57 , 62 , 67 , 72 , 77 , 82 , 87 , 92 , 97 , 102 , 107 , 112 , 117 , 122 , 127 , 132 , 137 , 142 , 147 , 152 , 157 , 162 , 167 , 172 , 177 , 182 , 187 , 192 , 197 , 202 , 207 , 212 , 217 , 222 , 227 , 232 , 237 , 242 , 247 , 252 , 257 , 262 , 267 , 272 , 277 , 282 , 287 , 292 , 297 , 302 , 307 , 312 , 317 , 322 , 327 , 332 , 337 , 342 , 347 , 352 , 357 , 362 , 367 , 372 , 377 , 382 , 387 , 392 , 397 , 402 , 407 , 412 , 417 , 422 , 427 , 432 , 437 , 442 , 447 , 452 , 457 , 462 , 467 , 472 , 477 , 482 , 487 , 492 , 497 , >>Number of terms = 100>>Total Sum == 24950

2,000==2 , 2 , 2 , 2 , 5 , 5 , 5 , Total primes factors== 7

2^4 * 3^4 * 5^4 * 7 * 11 * 13 * 17 >> 13,783,770,000 >>== 2000 >>=divisors

Oops!!!

$(3p+1)(t-4)=2tp$

$3tp - 12p + t - 4 = 2tp$

$tp - 12p + t - 4 = 0$

$t(p+1) = 12p + 4$

$t(p+1) = 4(3p + 1)$

$t = 4 \left(\frac{3p+1}{p+1}\right)$

$t = 4 \left(\frac{3p + 3 - 2}{p+1}\right)$

$t = 4 \left(3 - \frac{2}{p+1}\right)$

$t \leq 24$

Continue with https://web2.0calc.com/questions/the-powers-of-coffee-in-mathematicians#r2

You have: $(1, 3, 4, 5, 5, 6), (1, 2, 2, 3, 4, 5, 6)$

For $1,$ you can only have $6,$ and there is $1$ of them, so $1$ way to do this.

For $3,$ you can only have $4,$ and there is $1$ of them, so $1$ way to do this.

For $4,$ you can only have $3,$ and there is $1$ of them, so $1$ way to do this.

For both of the $5,$ you can only have $2,$ and there are $2$ of them, so $2 \times 2 = 4$ way to do this.

For $6,$ you can only have $1,$ and there is $1$ of them, so $1$ way to do this.

In total, there are $1 + 1 + 1 + 4 + 1 = \boxed{8}$

A standard die has 6 faces on it, 1, 2, 3, 4, 5, 6, right? In the case of your 2 dice, they have changed like this:

First die ==1, 3, 4, 5, 5, 6 [It has no 2 on it, but 2 fives]

2nd die ==1, 2, 2, 3, 4, 6 [It has no 5 on it, but 2 twos]. OK, so far?

Form an additions table as follows:

    1,   3,   4,   5,   5,   6

1

2

2

3

4

6

If you make an additions table of 6 X 6 numbers (as is the case with your 2 dice), then you should have 6 * 6 ==36 sums as follows:

(2, 3, 3, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 10, 10, 11, 11, 12)==36 sums

Count the number of "7s" from the above sums and you get ==8 "sevens"

Therefore, the probability of getting a sum of 7, when rolling your 2 "altered" dice is: 8 /36 ==2 / 9

but how?

#2)

When we draw the figure described in the problem, we see that there are vertical angles in the middle of the quadrilaterals. 

The Vertical Angles theorem tells us that opposite angles of two intersecting lines are congruent, so angle COB and angle DOE are congruent. (O is the intersection in the middle of the quadrilateral) Since the sum of the angles in a triangle add up to 180 degrees, we see that triangle COB is 85° +31° +x, and triangle DOE is 47° +x+ y (y is angle BDE). Because both are equal to 180°  in total, 85° +31°+ x = 47°+x+y.

We simplify to get 116° +x=47° +x+y, 116°-47°=y, Angle BDE = 69° 

There are four possible functions f(x).