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n=1; a=(1457 % n);b=(368%n);c=(1754 %n);if(a==b and b==c, goto5,goto6);printa,b,c,n; n++; if(n<5000, goto1, discard=0;

Rem              n

0 0 0             1
2 2 2             3
8 8 8             9
5 5 5            11
5 5 5            33
71 71 71      99

So, the largest n = 99 and the remainder of 71.

n=1; a=(1457 % n);b=(368%n);if(a==b, goto4,goto5);printa,b,n; n++; if(n<2000, goto1, discard=0;

Rem           n

0 0              1
2 2              3
8 8              9
5 5             11
5 5             33
71 71         99
5 5            121
5 5            363
368 368    1089

So, the largest n =1,089 with the remainder of 368.

Points A(-1, -2)and B(3, 2) are the endpoints of a diameter of a circle graphed in a coordinate plane. How many square units are in the area of the circle? Express your answer in terms of pi.

Think of the horizontal distance on the plane  (from –1 to 3 = 4 units) as the base of a right triangle,

and think of the vertical distance on the plane (from –2 to 2 = 4 units) as the height of that right triangle.

Use Pythagoras' Theorem to obtain the length of the hypotenuse

H² = 4² + 4² = 16 + 16 = 32    therefore the hypotenuse is sqrt(32)

Since that hypotenuse is the diameter of the circle, the radius is half of it, which = (1/2)sqrt(32)

The area of the circle is pi times the radius squared. 

Area = pi times [(1/2)sqrt(32)]²         This would be so much easier if I knew how to draw a radical on this site. 

Anyway, the area ends up being pi times 1/4 of 32 =  8 pi 

.

Solving these equations gives us: x=y*tan(29), and x=(y+100)(tan25).

Use a unit circle to find the appropriate values. y should be approximately 529.9. Then, solve for x by plugging the value of y back in.

An approximation could be $\frac{22}{7}$ or $\frac{355}{113}.$

The slope of the line can be found by: $\frac{y_2-y_1}{x_2-x_1}$ , so we have $\frac{-2-5}{4-3}=\boxed{-7}.$

7/11 = 0.6363636363636363......

So it is   the .63 with aline over it....the line represents the .63 repeats itself indefinitely....

10.75 is a lot of change.....droopy drawers!

Add them all together = 14.42

then divide by all of them (7)    14.42/7 =   $ 2.06 average   <-------- this is the MEAN

to find MEDIAN arrange in ascending order:

0.00   0.00   0.02  0.40   1.25  2.00  10.75     MEDIAN is the MIDDLE number

MODE is the number that occurs most often:   0.00

For the second one....the equation of the line is   y = - x + 1

the shaded area is thus    y <= -x+1     or      y+x -1 <=0

Thank you!! 

1. Find a linear inequality with the following solution set. Each grid line represents one unit.

Thanks

You are right, i overlooked 2^3

$\frac{5e^{\frac{\pi}{2}i}}{2e^{\frac{5\pi}{6}i}}\\~\\ =\frac{5}{2}e^{ (\frac{\pi}{2}-\frac{5\pi}{6}) i }\\~\\ =\frac{5}{2}e^{ (\frac{-2\pi}{6}) i }\\~\\ =\frac{5}{2}e^{ (\frac{-\pi}{3}) i }\\~\\ =\frac{5}{2}(cos(\frac{-\pi}{3})+isin(\frac{-\pi}{3}))$

which answer did you get, Melody?

I believe the answer is either B. or D. 

$\sqrt{37+20\sqrt{3}}=a+b\sqrt{3}$- Original Equation; trying to solve for $a,b$.

Square both sides, to get ${37+20\sqrt{3}}=a^2+2ab\sqrt{3}+3b^2.$  This is just the expansion of $(a+b)^2$ !

Now, we use a bit of matching, on trying to match the variables to the numbers. We see  $20\sqrt{3}$ and $2ab\sqrt{3}$ , so we have $2ab\sqrt{3}=20\sqrt{3}$ , and $ab=10.$

Also, we get $a^2+3b^2=37$ by matching the variables to the numbers, again.  

After trying a few times, we get $a=5, b=2$ , and that sure works !

Check: $(5)(2)=10$ and $5^2+3(2)^2=25+12=37.$

Thus, the answer is $5,2$ .

Do you understand cupcake?  Have you figured out the answer yet?

Melody:

Also 127 mod 8 = 7

What boxes?

You must start the problem by drawing the diagram.

Have you done that?

Take a photo of it using you phone and upload it to the forum if you can. 

We can see if you have drawn it correctly.,

How many positive integers n satisfy 127 = 7 mod n? (n=1 is allowed)

127=7modn

127=kn+7     where k and n are integers and n is bigger than 7

120 = kn

prime factors of 120 are 2^3 *3*5

So what factors are bigger than 7

3    too little

5    too little

3*5     =15

2*5    =10

2*15   =30

4*3    = 12

4*5   =20

4*15  =60

8*3    =24

8*5    =40

8*15  =120

127 = 7 mod n?

10, 12, 15, 20, 24, 30, 40, 60, 120 

So there are 9 of them   

n=1; a=(1457 % n);b=(1754%n);if(a==b, goto4,goto5);printa,b,n; n++; if(n<2000, goto1, discard=0;
Rem       n
0 0           1
2 2           3
8 8           9
5 5          11
26 26      27
5 5           33
71 71      99
269 269  297

So, the largest n = 297 with a remainder of 269

Hi Cupcake,

To answer this question you just need to recognise that 

$n(cos\theta+isin\theta)=ne^{\theta i}$

Convert them both to this form,

do the division

 then convert the answer back to the original form.   

n=1; a=(127 % n);printa,n; n++; if(n<200, goto1, discard=0;

Rem  n
7       8
7       10
7       12
7       15
7       20
7       24
7       30
7       40
7       60

7       120

So, the ten numbers to the RHS.