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n=4624; c=0; a=floor(sqrt(4624));printc," - ",  n," ", a;c=c+1;n++;if(n<4761, goto2, discard=0;

Floor(137 / 2) =68 
m = 68.
69^2 - 68^2 =137
4,761 - 137 =4,624, so:
n=4,624

The y component of the acceleration is   - .350 sin 52 = .=-.2758 m/s^2

When does the y component of the runners motion = 0?

v = v0 + at

0 = 2.88 + (-.2758) t    so  t = 10.44 sec

What is the speed in the x direction at this point in time?

x component of acceleration = .350 cos 52 = .2155 m/s^2

v in x direction = 

v = vo + at

  = 0 + .2155 (10.44) = 2.24 m/s

Hi Melody: Writing a very short line of code, like this one, is a trivial matter, but it finds the answer in milliseconds. I generally await somebody like yourself or one of the other mathematicians to flesh out the details of the problem using Algebra, Calculus, Trig,....etc. I have forgotten most of that stuff since taking them, when dinosaurs roamed the Earth!.

Hi guest,

I find your little lines of code very interesting. 

I have checked your answer just for n=1 and it is certainly true.

But

I'd really like to see somone solve this one mathematically.

I started but got it wrong.  Maybe I just made a silly error or maybe I was on completely the wrong path.  

\(\text{A line has the same slope at all points so }\\ m = \lim \limits_{t\to \infty}\dfrac{y(t)}{x(t)} = -\dfrac 3 2\\ \dfrac{b}{a}=-\dfrac 3 2\\ b=-\dfrac{3a}{2}\\ a+b=3\\ a+\left(\dfrac{-3a}{2}\right)=3\\ -\dfrac{a}{2}=3\\ a = -6\\ b = 9\\ <-6,9>\)

What is unit digit of 7^2018

\(\begin{array}{|rcll|} \hline && \mathbf{{\color{red}7}^{2018} \pmod{10}}\\ &&&\boxed{{\color{red}7^2} \pmod{10} \\ \equiv 49 \pmod{10}\\ \equiv 9 \pmod{10} \\ \equiv 9-10 \pmod{10}\\ \equiv {\color{green}-1} \pmod{10}} \\ &\equiv& \left({\color{green}-1}\right)^{1009} \pmod{10} \\ &\equiv& -1 \pmod{10} \\ &\equiv& -1+10 \pmod{10} \\ &\equiv& \mathbf{9 \pmod{10}} \\ \hline \end{array} \)

The unit digit of \(\mathbf{7^{2018}}\) is 9

There can't be 10 socks helperid

What is unit digit of 7^2018

\(\begin{array}{|rcll|} \hline && \mathbf{{\color{red}7}^{2018} \pmod{10}}\\ &&&\boxed{ {\color{red}7} \pmod{10} \\ \equiv 7-10 \pmod{10} \\ \equiv {\color{green}-3} \pmod{10} } \\ &\equiv& \left({\color{green}-3}\right)^{2018} \pmod{10} \\ &\equiv& (-1)^{2018}3^{2018} \pmod{10} \\ &\equiv& \mathbf{3^{2018} \pmod{10}} \\ &&&\boxed{{\color{blue}3^2} \pmod{10} \\ \equiv 9 \pmod{10} \\ \equiv 9-10 \pmod{10}\\ \equiv {\color{brown}-1} \pmod{10}} \\ &\equiv& \left({\color{blue}3^2}\right)^{1009} \pmod{10} \\ &\equiv& \left({\color{brown}-1}\right)^{1009} \pmod{10} \\ &\equiv& -1 \pmod{10} \\ &\equiv& -1+10 \pmod{10} \\ &\equiv& \mathbf{9 \pmod{10}} \\ \hline \end{array} \)

The unit digit of \(\mathbf{7^{2018}}\) is 9

This question here is giving me a hard time can any help me?

1)

which angle in the quadrilateral with vertices A(2,-4),, B(-5,-2),C(-4,2) and D(4,0) is a right angle?

\(\begin{array}{|rcll|} \hline \overline{CD}^2 = 8^2+2^2 = 68 \\ \overline{CB}^2 = 4^2+1^2 = 17 \\ \overline{BD}^2 = 9^2+2^2 = 85 \\ \hline \end{array}\)

Pythagoras:

\(\begin{array}{|rcll|} \hline \overline{CD}^2+\overline{CB}^2 &\overset{?}{=}& \overline{BD}^2 \quad | \quad \text{If true, then C is a right angle!} \\ 68+17 &\overset{?}{=}& 85 \\ 85 &=& 85 \quad | \quad \text{So C is a right angle!} \\ \hline \end{array}\)

D = R * T    

1424 km  =  (365km / h )   *  T

Thanks, helperid   !!!!

We can start by taking the 2 smaller fractions and turning them into one bigger fraction! (I'm sorry if you don't like fractions, but making fractions makes everything a whole lot easier.) Like this: \(\frac{x(x+2)}{(x+1)(x+2)}+\frac{x(x+1)}{(x+1)(x+2)}\). This doesn't seem like it gets us anywhere but now we can make only one fraction. \(\frac{x([x+1]+[x+2])}{x^2+3x+2}=kx\) Before we divide by x, we have to solve for x=0. (I'm going to skip all of the math, you end up with 0 = 0, so there is nothing to be gained.) Now we can divide by x. \(\frac{x+1+x+2}{x^2+3x+2}\) We can simplify this: \(\frac{2x+3}{(x+1)(x+2)}\).

Honestly, from here, I'm stumped. Hey, CPhill, Pavlov, or Omi, could you finish this one? 

Oops, I think you forgot to put in the equation. Might want to take another look at that 

Let's start by eliminating some choices. A and C can't be correct. For A, 8 socks were chosen 4 times, which means that the proportion was met at least once. Therefore, because the reasoning is invalid, A can't be the right answer. For C, white socks were picked far more often than half the time. Because of this we can eliminate C as a possible answer, leaving only B and D.

Think this one through. We know that 80% of the socks are white socks, so it is fair to say that there are 8 white socks out of 10 total socks. It appears this trial was run 10 times. I think a regular, "add and divide" average should suffice here. So \(\frac{7 \times 3 + 8 \times 4 + 9 \times 3}{10}\) is the long form of our average. The convinient part is that this equals exactly 8, so if we assume that there were 10 socks, the average perfectly matches. Therefore, we can say that D is the right answer. The reason behind C is true, but the 9-sock outcomes were counterbalanced by 3 seven-sock outcomes.

Honestly, they confuse me too, haha!

Very nice, hectictar.......these always confuse me....!!!!!

Last two

(4)  If Bob paid P, then Tim paid (7.80 - P)

So

P < (7.80 - P)       and that's all we can say

(5)   Let the number of men = M     then the women = M + 6

So

M + (M + 6) =30

2m + 6  = 30

2M  = 24

M = 12       and  M + 6  =  18

5

x = men on committee

x+6 = women on committee

x  +  x+6  = 30

2x = 24

x = men = 12      women = 18

3.      Martha and Sally had lunch together on Tuesday. Sally paid $2 more than Martha. The total bill was $5.64

P   = what Martha paid

P + 2  =  what Sally paid

So

P + (P + 2)  = 5.64

Can you take it from here   ????

Here is another way to figure out the answer:

C  =  \(\frac59\)(F - 32)  =  \(\frac59\)F  -  \(\frac{160}9\)

When  F = 0 ,   C  =  \(\frac59\)(0)  -  \(\frac{160}9\)  =  - \(\frac{160}9\)

When  F = 1 ,   C  =  \(\frac59\)(1)  -  \(\frac{160}9\)  =  \(\frac59\) - \(\frac{160}9\)     Notice this value of  C  is  \(\frac59\)  more than it was when  F  was  0 .

When  F = 2 ,   C  =  \(\frac59\)(2)  -  \(\frac{160}9\)  =  \(\frac59\) + \(\frac59\) - \(\frac{160}9\)     Notice this value of  C  is  \(\frac59\)  more than the previous.

So a temperature increase of 1 degree Fahrenheit is equivalent to a temperature increase of 5/9 degree Celsius.

C  =  \(\frac59\)(F - 32)

\(\frac95\)C  =  F - 32

\(\frac95\)C + 32  =  F

When  C = 0 ,   F  =  \(\frac95\)(0) + 32  =  32

When  C = 1 ,   F  =  \(\frac95\)(1) + 32  =  \(\frac95\) + 32       Notice this value of  F  is  \(\frac95\)  more than the previous.

When  C = 2 ,   F  =  \(\frac95\)(2) + 32  =  \(\frac95\) + \(\frac95\) + 32    Notice this value of  F  is  \(\frac95\)  more than the previous.

So a temperature increase of  1  degree Celsius is equivalent to a temperature increase of 9/5 degrees Fahrenheit. (And  9/5 = 1.8). In other words, a temperature increase of 9/5 degrees (not 5/9 degrees) Fahrenheit is equivalent to a temperature increase of 1 degree Celsius.

2.       (Fifteen more than four times a number)  is (6 more than five times the number.) Use "n" for the number 

                         (A)                                              =           (B)

So....

(A)          =       (B)

4n + 15   =   5n + 6   subtract  4N, 6   from both sides

9 = n

1.       The length of a rectangle is 3 more than twice it’s width. The perimeter is 48 feet

I suppose you want to find the dimensions???......if so

2 ( L + W)   =  Perimeter      and  L = 2W + 3

So

2 [ 2W + 3 + W ]  =  48      divide both sides by 2    and simplify

3W + 3  = 24       subtract 3 from both sides

3W  =  21      divide both sides by 3

W = 7      and  L = 2(7) + 3   =  17

P.S.....if you wanted the area......just multiply    L x W

C  =  (5/9)(F - 32)

(1) A temperature increase of 1 degree Fahrenheit is equivalent to a temperature increase of 5/9 degree Celsius.

(2)A temperature increase of 1 degree Celsius is equivalent to a temperature increase of 1.8 degrees Fahrenheit.

(3) A temperature increase of 5/9 degree Fahrenheit is equivalent to a temperature increase of 1 degree Celsius.

The slope of this linear equation  is  5/9

This means that  for every increase in Celsius of 5 degrees, Fahrenheit increases by 9 degrees  ⇒ (A)

Dividihg both quantities in A by 5......for every 1 degree increase in Celsius, Fahrenheit increases by (9/5) = 1.8 degrees

So (2)  is true

Next....divide both quantities in  (A) by 9    and we have that for every 5/9 of a degree change in Celsius, Fahrenheit changes by  1 degree

So ( 1) is also true

So... "D" is correct