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Let's just say that ABCD is a trapezoid. M and N are the midpoints of the non-parallel sides AD and BC, respectively. There is a theorem that states the line MN is the average of the two parallel sides (AB and CD). 

If you try a quadrilateral (cannot be trapezoid or parallelogram) with an extremely long side, you will find that MN < (AB+CD)/2.

This equality works for all convex and real quadrilaterals. 

Hope this helps,

- PM

What is the length of AD or AC? We need the length of some sides to determine what the area is. Many trapezoids that fit your description could have different side lengths, leading to different areas. 

I cannot help you too much because insufficient information was given, but I can tell you that ABCD is an isosceles trapezoid. Angle ADC = Angle BCD, Angle DAB = Angle CBA. Since AC bisects angle DAB, DB also bisects angle ABC.

- PM 

Let's say... 1 red, 1 green, 2 blue.

The new question is... What is the probability of selecting 2 blue balls?

First chance... 1 blue ball has chance 2/4, or 1/2.

Second chance... 1 blue ball has chance 1/3. This is because 1 blue ball is left... and only 3 balls are left.

Multiply these two chances... 1/6 probability.

Thank you, guys!

$450 \cdot (450)^{-1} = 1 \pmod{3599}\\ \text{something to immediately check is does 450 divide 3600?}\\ 3600=450\cdot 8 \\ 8\cdot 450 = 1 \pmod{3599}\\ 8 = (450)^{-1} \pmod{3599}$

not sure where the triangle fits into things

See here :

https://web2.0calc.com/questions/trig-question-please-help-me-12-14-18

See here :

https://web2.0calc.com/questions/okay-i-really-need-some-help-please-trigonometry-12-14-18

If every possible triangle has its own distinct centroid then I think the answer will be closer to

4*3*10C2  +  10*10*10 = 540+1000 = 1540

There might be some double counting here I think but I expect the answer is much bigger than 64

Find the minimum value of

Y=x^2+1/x

Take the derivative  and set to    0

y '  = 2x  - 1/x^2  =  0

So

2x -  1/x^2   = 0

2x = 1/x^2

2x^3 = 1

x^3  =  1/2

x =  ∛(1/2)      ⇒   this is the x value that minimizes the function

The minimum value is

[ (1/2)^(1/3) ] ^2   +   1 / (1/2)^(1/3)  =

(1/2)^(2/3)  + 1 / (1/2)^(1/3)  ≈  1.89

If x,y, and z are positive integers such that x^2+y^2+z^2=174. What is the greatest value of x+y+z?

The greatest value  of x + y + z  is when we have the  triplet  5, 7, 10

And the sum  =  22

What is the sum of the roots of the quadratic 4x^2 - 4x - 4?

Sum of the roots  =     - (-4) / 4   =   1

Find the sum of all solutions to the equation 3x(x+4)= 66

Simplify 

3x^2 + 3x = 66

3x^2 + 3x - 66 = 0

Sum of the solutions (roots)   =   -3/3  = - 1

One root of x^2 + 12x + k = 0 is twice the other root. Find k.

This one is a little tricky.......

The sum of the roots  is    -12/ 1  =  -12

Let one root = a     and the other 2a

So.....the sum of the roots = -12

a + 2a  =  -12

3a = -12

a = -4

2a =  2(-4)  =  -8

Now.... in the form    ax^2 + bx + c

The product of the roots is     c /a

Here    a = 1    and     c = k

So.....the product of the roots   is       k /1    = k

So

(-4) (-8)   = k   =    32

Find all solutions to the equation 10x^2 - 7x - 6 = 0

Again....let's see if we can factor

(5x  - 6) ( 2x + 1) = 0

Set both faxctors to x and solve for x

5x - 6 = 0                         2x + 1 = 0

5x = 6                              2x = - 1

x = 6/5                              x = -1/2

Find all values of x such that 6 = 35/x - 49/x^2

Here.....they are trying to be tricky.......multiply through by x^2   and we get

6x^2 =  35x - 49        rearrange as

6x^2 - 35x + 49= 0      see if we can factor

(3x - 7) (2x - 7)  = 0 

Set both factors to 0   and solve for x

3x - 7  = 0                2x  - 7  = 0

3x  =  7                    2x   = 7

x =  7/3                      x = 7/2

Find the sum of the roots of the quadratic x^2+7x-13=0

In the form ax^2 + bx + c, the sum of the roots =  -b / a

Here       a = 1     and b = 7

So....the sum of the roots  =   -7/1   =  -7

Find all values of x such that 2x^2 +9x- 5 = 0

We could use the quadratic formula to solve, but I like to see if this can factor, first

It will!!!

Factor as

(2x  - 1) ( x + 5)  = 0

Set each factor to 0  and solve for x

2x - 1  = 0                                                    x + 5 = 0

add 1 to both sides                                     subtract 5 from both sides

2x  = 1                                                           x = -5

divide both sides by 2

x = 1/2

The two solutions are in red

1.     e^(2x + 1) = 7        take the Ln of both sides

Ln e^(2x + 1)  = Ln 7       and by a log property, we can write

(2x + 1) Ln e  =  Ln 7      since Ln e  = 1, we can drop this

2x + 1  =  Ln 7             subtract 1 from both sides

2x  = Ln (7)  - 1          divide both sides by 2

x  =   [ Ln (7) - 1 ] / 2      { third answer }

2.  

The graph is shifted 1 unit right

The graph is shifted 2 units up

3.

y=2+5/6x

4x-3y=3

Sub the first equation into the second for y   and we have

4x - 3 [ 2 + 5/6 x ]  = 3        simplify

4x - 6 - 15/6 x  =  3

4x - 6 -  5/2 x  = 3     multiply through by 2 to clear the fraction

8x - 12 - 5x  = 6

3x - 12 = 6         add 12 to both sides

3x  =  18       divide both sides by  3

x = 6

And using  y = 2 + 5/6 x    .....  we can find  y    as

y =  2 + 5/6 (6)   =   2  + 5    = 7

I know how to do the first one.....but.....not the second

1. Suppose g(x) is a polynomial of degree five for which g(1) = 2, g(2) = 3, g(3) = 4, g(4) = 5, g(5) = 6, and g(6) = -113. Find g(0).

The first one sets up a  system of 6 equations with 6 unknowns.....this is very tedious to solve by hand.....so....I'm not!!!....I'll let this website do the "heavy lifting" :

https://matrix.reshish.com/gauss-jordanElimination.php

Here is the system :  

a + b + c + d + e + f = 2

32a + 16b + 8c + 4d + 2e + f = 3

243a + 81b + 27c + 9d + 3e + f = 4

1024a + 256b + 64c + 16d + 4e + f = 5

3125a + 625b + 125c + 25d + 5e + f  = 6

7776a + 1296b + 216c + 36d + 6e + f = -113

We are only interested in the value of "f"   = 121

And this is   =   f(0)

BTW....the polynomial is

-x^5  + 15x^4 - 85x^3 + 225x^2 - 273x + 121

For the first 11 months, she is effectively depositing [ 100 - 10] = $90 per month into her account because of the service charge being applied every month

So...at the end of 11th month, she has  11 * 90   =  $990 in her account

At the start of the 12th month she deposits $100 into her account and now has $1090 in her account.....so....she will not  have to pay the $10 service charge any longer

The account balance at the end of the first year is :   1090 (1.01) = $1100.90 = $1101

At the end of the second year   the account balance is

[ 1101 + 12(100)]*(1.01)   = $2324.01 =  $2324

At the end of the third year, the account balance is

[ 2324 + 12(100) ] (1.01)  = $3559.24  =  $3559

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I'm a little confused, by PM's answer because would you not have 7 different options for the dot in the middle? I'm sorry, I'm new to this subject...

Wow, smart! I never knew you could do it that way... helpful!

Oh yes, you are right! Nice proof there!