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We need to first find the eqution for the plane

Let vector AB  =  (1, 0 , -1)

Let vector AC  =   ( 1, - 1, 2 )

We can find the normal vector to the plane by taking the cross-product  of  AB x AC

n  =      i    j     k   i     j

           1   0   -1   1    0   =

           1   -1   2   1   -1

[ i * 0 * - 1  +   j * -1 * 1  + k * 1 * - 1 ]  -  [ k * 0 * 1  + i * -1 * - 1 + j * 2 * 1  ]   =

[  0   -  1j - 1k ] -  [ 0  + 1i + 2j ]   =    - 1i  -3j - 1k

The equation  of the plane  using point  A  is  

-1( x - 0)   -3(y -1)  - 1( z  - 1)  = 0

-1x  - 3y  + 3  - 1z + 1  = 0

-1x - 3y - 1z  + 4  = 0

1x + 3y +1z  - 4  = 0

The distance  between  Q  and the plane is given by 

l   1(-2) + 3(3)  + 1( 4)  - 4   l            l  7  l               7

_______________________  =    ______   =      ___  

  √ [ 1^2  + 3^2 + 1^2  ]                    √11               √11

So  d  =  7

THX, Mathleteig  !!!

We wouldn't necessarily have to use Heron's Formula......

Note that  this is an isosceles triangle  with a base of 6  and sides of  5

The altitude  = √ [side ^2  -  (base /2)^2  ]  = √ [ 5^2 - 3^2 ]

So....the area is......

(1/2) base * height

(1/2) 6 *  √[5^2  - 3^2 ]  =

3 * √16  =

3 * 4  =

12 units^2

I completely forgot about Heron's Formula! Thanks CPhill

We have that

( 3a - 1) + (a^2  + 1)  + (a^2 + 2)   = 16         simplify

2a^2  + 3a  + 2  = 16

2a^2  + 3a  -  14  = 0        factor

(2a + 7) ( a - 2)  =  0

Only the second  linear factor produces a positive for a....

a - 2  = 0

a  = 2

So....the sides of the triangle  are

3(2) - 1  =  5

2^2 + 1  =  5

2^2 + 2  =  6

The semi-perimeter, s,  =  [ 5 + 5 + 6 ] /2   =  8

We can find the area by the use of something known as Heron's Formula :

√ [ s ( s - 5) (s - 5) (s - 6)  ]  =

√ [ 8 * 3 * 3 * 2 ]  =

√ [16 * 9  ]  =

√144 =

12 units^2

The answer is 2/5.

Let the vector  v  be   ( 2 - 1, 3 - 1,4 - 2)  =  ( 1 , 2, 2 )

So.....the vector  form of the line  is

(1 + t, 1 +2t, 2 + 2t) =    ( t + 1, 2t + 1, 2t + 2)

So   c  = 2     e  = 2     a   =  1

So    ( c / a , e / a )  =    (   2 / 1 ,   2 / 1)   =    ( 2 , 2) 

Note that  as  x increases by 2, f(x)  increases  by 16

So.....the  absolute  difference  between any two outputs for inputs that are two units apart  is   16

10B + A =1.2[10A + B], solve for A, B

A = 4   and   B = 5

So that: 54 / 45 = 1.2 or 20% increase!.

Your question is not clear enough ! For example, can the digits be repeated such as:

6 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 15
5 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 15
4 + 3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 15
4 + 2 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 15
3 + 3 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 15
3 + 2 + 2 + 2 + 1 + 1 + 1 + 1 + 1 + 1 = 15
2 + 2 + 2 + 2 + 2 + 1 + 1 + 1 + 1 + 1 = 15........etc.

If they can be repeated, then it is a matter of adding partitions of: 5(10 parts) + 10(10 parts) + 15(10 parts).........+90(10 parts), which would be a very involved undertaking.

If they cannot be repeated, that is, you can only use each digit once such as:1234567890, then that might be very easy to calculate. Since the sum of the 10 digits [0 to 9 =45, which is divisible by 5], then the arrangement of the 10 digits wouldn't matter at all, since they will always sum up to 45.

Therefore, if you started with: 1234567890, then you will have:9 * 9! (on the assumption that all 10 digits be different from each other). So, 9 * 9! =3,265,920.

And that is my attempt. Others should take a crack at it!

Here's another way

Call  x  the number  of draws  in which 5 marbles  are added

Call  y   the number  of draws  in which  3  marbles  are added

So

5x +  3y  =  21

x     y

0    7

1    not an tnteger

2    not an integer

3    2

4    not an integer

5    y  is negative

We can stop  here

x + y   is minimized   when x  =  3  and y  = 2

So...as EP found.....5 draws  !!!

Nice work, CU!!!

*****

xxxxxxxxx

Notice that as  x increases by 1,  f(x)  increases by 8

So.....the difference  (in absolute value terms)   =  8

The tranverse axis contains the two foci and two vertices

Thus.....this hyperbola  intersects the x axis

So....the form will be

(x - h)^2            (y  - k)^2

_______  -       ________   =       1

    a^2                    b^2

(h, k)  =  (-3, 2)

2a  = 12            2b  = 8

a  =  6                 b  = 4

So we have

(x  + 3)^2         ( y - 2 )^2

________  -   _________    =      1

    36                   16

No.....the boiling of water (like the melting of ice) occurs at a constant temoerature 100C (lees if the pressure is lowered).....adding more heat will only make it boil faster at 100 C   .....you CAN heat up the steam past the boiling point, but not the water.

If you start with very cold ice, you can add heat to warm it up to 0C where it will begin to melt.....adding heat results in faster ice melting, but no temp increase in the ice.  You CAN heat up the water past the melting point...up to the boiling point as describd above.

First one......

a^2  =  45

a  = √45

a  =  3√5   

2a   =  length of  major axis  =   6√5

b^2  =  27

b  =√27

b =  3√3

2b  =  length of minor axis  = 6√3

Good job, Jenny   !!!

If we had a square  with a diagonal  =  4√2 units.....the side of the square  =  4 units

Then the area  of the square  =  side^2  =  4^2  =  16 units^2

So.....the area of this isosceles triangle will be  1/2  of this

The answer is 5*10^6 = 5,000,000.

15 = 3 * 5

12 = 2 * 2 * 3

16 = 2 * 2 * 2 * 2

^{Take each prime factor the greatest number of times it occurs  (in red)}

^{2 * 2 * 2 * 2 * 3 * 5 = 240 = LCM}

The answer is 2.

The answer is 14.

The age of the turtle is 1.

The probability is 2/3.