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10 , 100 , 101 , 110 ,  120 , 130 , 140 , 150 , 160 , 170 , 180 , 190 , 201 , 210 , 301 , 310 , 401 , 410 , 501 , 510 , 601 , 610  , 701 , 710 , 801 , 810 , 901 , 910 >>Total =28 such integers.

Sector area  =   pi *r^2   ( 100 / 360)

50pi  =  pi * r^2  * ( 5/18)      divide out pi

50  (18/5)  = r^2

180  = r^2

r = sqrt 180  =    sqrt (9 * 20)  =   3sqrt (20)

Let P = (0,0)   Q = (13,0)

We can determine the coordiantes of R  by finding the intersection of two circles

x^2  + y^2 =  18^2     →   x^2 + y^2  =  324  (1)

(x -13)^2  + y^2 = 14^2    →  x^2 - 26x + 169  + y^2 = 196  → x^2 - 26x  + y^2 =  27     (2) 

Subtract  (2) from (1)

26x  =  297

x = 297/26

And    (297/26)^2  + y^2  =  324

y^2  = 324  - ( 297/26)^2

y^2  = 130815 / 676   take the positive root for y

y = sqrt (130815) / 26

R  =  ( 297/26 ,  sqrt (130815) /26)

M  =  ( (297 / 26 + 13)   / 2 , sqrt (130815 / 52 )    =   (635/52 , sqrt (130815) / 52 )

PM  =   (1/52)  sqrt  ( 635^2 + 130815)  =  (1/52) sqrt (534050)  ≈  14.05

Let Ian have M  marbles

Then Velibor  has  3M  marbles

When Velibor  gives 63  to Ian  they both have the same number....so....

3M   - 63  =  M + 63

3M - M  =  63 +63

2M =  126

M =  126 /  2    =  63

So Veilbor  started with  189   and  Ian started with  63

So

(3/4) (189  -  x  ) =   63 +  x     

Where x is the number that Ian gives to Veilbor so that Veilbor has (3/4) as many as Ian

Simplify

(3/4) 189  - (3/4)x =  63 + x

(3/4)189 - 63   =  x + (3/4)x

315 / 4 =  (7/4)x

315  =  7x

x =  315 / 7   =     45    

For convenience   let   the side of the tetrahedron = 1....let

A  = ( -1/2 , 0 , 0)

B  = ( 1/2, 0, 0  )

C =  (0, sqrt (3)/2, 0)

D =  (0, sqrt (3)/4 , 6/sqrt (3) )

M =  (0 , (3/8) sqrt (3) , 3/sqrt (3) )  =  (0, (3/8)sqrt (3) , sqrt (3) )

AB  = 1

AM = BM   =  sqrt   [ (-1/2)^2  + (3sqrt (3) / 8)^2  + (sqrt 3)^2   =  sqrt [ 1/4 + 27/64 + 3 ]   = 

sqrt (235) / 8

By the Law of Cosines

AM^2  =  AB^2 + BM^2 - 2 (AB) (BM)cos (ABM)

0  = 1 - 2 ( 1) (sqrt (235) / 8) cos ABM

0 =  1 - [sqrt ( 235) / 8] cos ABM

-1 /  - [  sqrt (235) / 8]  =  cos ABM

8 / sqrt ( 235)  =  cos ABM

I worked on this one for a while yesterday.....I got the same equations that you did

    results in a = 2/3    and b = -4

       so I do not think this parabola can be written as   ax^2 + bx + c = y         where a, b and c are INTEGERS  ( a = 2/3)

~EP

1.     Write a matrix representing the system of equations

                               x+2y+z=5

                              2x-y+2z=15

                             3x+y-z=8

Solve the system by transforming your matrix into reduced row echelon form. State the solution as an ordered triple.

Answer:

(4, -1 , 3)

2.

Multiply the matrices.

                    -1    2    0              2     -3

                     3     -2   4     x      7       -1

                    5    -1     6            1        2

Answer:

12   1 

-4    1

-11   28

3. Solve the system of equations by forming a matrix equation. State your answer as an ordered pair.

    x+2y=-7

   -3x-5y=23

Answer: (-11,2)



explanation : I took the test

Here's another way, without using calculus:

Expand it to \(5x^2 + 462\)

Now, the minimum value occurs when \(x = 0\), because \(x^2\) is always non-negative.

See my answer under your original post

(a)  Let M be the total money that he started with

What he had left after buying the 5 erasers and 11 pencils =  M - (1/3)M =  2/3 M

(b)  This is a little more difficult

Since each eraser cost 3 times as much as a pencil   let the cost of a pencil  = P  and the cost of an eraser = 3P

So  he spends 5 * 3P   =  15P  on erasers  and 11P  on pencils

So  the erasers  represent 15 / ( 15 + 11) =  (15/26)  of  the the (1/3)M  =  (15/26) (1/3)M =  (15/ 78)M =  the cost of 5 erasers

So  each eraser  cost  ( 15/78)M / 5 =  (15/ 390) M  =   (5/130)M =  (1/26)M

He spends (3/4) of his remaining money on some  more erasers  so he spends 

(3/4) (2/3)M = (1/2)M  on erasers

So  the number of additional erasers that he buys =  (1/2)M / (1/26)M  =  26/ 2 =  13 

So, in total, he buys  5 + 13    =   18 erasers

We have the equations

10x + 20y  = 800

x =   2y

The graph here  https://www.desmos.com/calculator/fx0rtrkolm   shows  that   he sells 40 small cups and 20 large cups

Expanding we have

2x^2 + 16x + 30 < 3x^2 -15x - 72       rearrange as

x^2 - 31x  - 102 >  0     factor as

( x - 34) ( x + 3)  >   0

This will be true   when  x < -3   and x > 34

So   the intervals that make this true are

(  -inf, -3)  U  ( 34, inf)

5421

5331

5322

53211

4431

4422

44211

4332

43311

43221

6321

432111

Total—12 partitions that work.

Answer: \(\frac{\sqrt3}{3}\)

Solution:

In the orientation that I* would view this in, A would tbe the point on the 'top'. Side CD as well as point B are on the base of the tetrahedron. Since BCD is an equilateral triangle, the median from CD to B would pass through its center. Call the center O. AOB would be 90 degrees (if it isn't clear, angle ABO is the same as angle ABM). Because there is a right triangle with the desired angle in the right place 'SOH CAH TOA' can be applied, where AB is hypotenuse and BO is adjacent. BO is equal to 2/3 the length of the median MB, which itself is \(\frac{\sqrt3}{2}\) of AB. That makes BO \(\frac{\sqrt3}{3}\) of AB. BO/AB = cos ABM = \(\frac{\sqrt3}{3}\).

*Being bad at geometry, especially 3d geometry, runs in my family. This may not be correct (especially since the question has been asked before on this site and answered with 2/5; no work shown. The answer had -2 points, so it probably wasn't correct, but...)

yup!

Answer: \(x\in(-\infty, -3) \cup (34, \infty)\)

Solution:

Set the equations equal to find critical points:

(2x+10)(x+3)=3(x+3)(x-8)

x=-3 is a critical point, as is x=34.

If this inequality were to be graphed on a number line, it would result in a striped pattern. You can test the middle region (I will test -3

Testing for 0:

(0+10)(0+3)=3(0+3)(0-8)

This region is not shaded.

The regions around it (x<-3 and x>34) are.

Therefore, the answer in interval notation is \(x\in(-\infty, -3) \cup (34, \infty)\).

545 cents - 79*5 cents = 150 cents

150 cents / 25 cents per quarter = 6 quarters

Simplify sqrt(3*324000).

Sqrt[3 x 324,000]==972,000

972000 = 2^5 * 3^5 * 5^3==sqrt[2^5 x 3^5 x5^3] ==sqrt[2^4 x 3^4 x 5^2] x sqrt[2 x 3 x 5]

180 x sqrt[30]

differentiating with respect to x (using chain rule) gets:

\(2(x+12)+2(x+7)+2(x+3)+2(x-14)+2(x-8)\\ =2(5x)\\ =10x\)

There is only 1 critical point \(x = 0\), and since this function is concave up (the double derivative is positive), it is a minimum point. Plug in zero to your equation to get your answer.

Sorry, I thought it was a multiplication sign! Woops!

Thanks for letting everyone know nerdiest :)

https://web2.0calc.com/questions/circles_101#r2

2x^2 + 8x + y^2 - 24y + 62    complete the square on x, y

2(x^2 + 4x + 4) + (y^2 - 24y + 144)  + 62 - 8 - 144

2 (x + 2)^2  + (y - 12)^2   -  90

The min of 90  occurs when x = -2  and  y = 12

maybe