All Answers 353101 Answers

There are:
[46,000 x 1] + [46,000 x 3] + [46,000 x 5] + [46,000 x 7] + [46,000 x 9] = 1,150,000 - Total sum of all odd digits between 10,000 and 99,999. 

1) B

2) E 

3)B

Cool.

Solution:

V=W/Q | V = voltage in volts; W = work in Joules; Q = Charge in Coulombs  

6v = W / (4.0E-6)

Solve this for W, and include the unit of (J)oules

GA

First one : 

C) Area(ABCD) + AREA(GHIA) = AREA(DEFG)

Second one :

C) DA^2 = DG^2 + AG^2

First one....this is a 45 - 45 - 90  right triangle....so  CB  = CA

Second one  .....  solve this  for x

3 *x  =  21   

Third one

D) 2π - A

Last one

D) any angle to the center of the side opposite the angle

First one  =   16 ft

Second one

If the tan theta  =  3.....then y  =  3x

So...using the Pythagorean Theorem we haxe that

25  =  sqrt  [x^2 + (3x)^2]     square both sides

625  = x^2 + 9x^2

625  =  10x^2

62.5  = x^2      take the positive root

7.9  =  x

And y = 3x  = 3(7.9)  = 23.7

Last one

C) 15^2 + 16^2 = x^2

THX,  CoolStuffYT !!!

To continue  ......

We know  that   the intersection of the two lines  is  ( -6/17,  -2  10 /17)    = (-6/17,  -44/17)

This is the midpoint between  A  and A'

So......to find A' = (x, y)  we have that

 [-6 + x ] / 2  = -6/17              [ -4 + y ] / 2   =  -44/17 

-6 + x  =   -12/17                     -4 +  y   =   -88/17

x  =  -12/17 + 6                         y  =  -88/17 + 4

x =  -12/17                                y =  -20/17

x  = 90/17

So  A'  =  (90/17, -20/17)

Here's a graph  :  https://www.desmos.com/calculator/trsmplhqlg   

Here are some formulas to help you...

LF = Q1 - 1.5 * IQR
UF = Q3 + 1.5 * IQR

Find the IQR of each, then find the LF and UF of each.

You are very welcome!

:P

Take a look at this link:

https://math.stackexchange.com/questions/1192906/1-2-3-5-7-11-15-22-30-42-56-77-101-135-176-231-partition-numbers-what-is

You are very welcome!

:P

Try re-loading the page or come back later.

Thx Phill

I think this is the answer to your problem :

https://web2.0calc.com/questions/need-answer_18

√[4x^2]  - √x^2   =   6

We can write this in the following manner

l 2x l   -  l x l   =  6

2 l x l   -  ; x l   =   6

l x l   =   6

This is true when  x  = 6  or x  = - 6

What the....

I mean answers

I don't think you can combine them together?

I really don't know that much....

31.31 / 20% = 156.55 - net price before 20% VAT.

156.55  x  1.20 =187.86 - gross price after 20% VAT.

187.86 - 156.55 = 31.31 - The VAT

Simplify

\(\dfrac{6^x\cdot 12^{x+1}\cdot 20^{x+2}\cdot 30^{x+3}\cdot 42^{x+4}\cdot 56^{x+5}\cdot 72^{x+6}\cdot 90^{x+7} }{10!\cdot 8^x\cdot 15^{x+1}\cdot 24^{x+2}\cdot 35^{x+3}\cdot 48^{x+4}\cdot 63^{x+5}\cdot 80^{x+6}} \)

\(\begin{array}{|rcll|} \hline && \mathbf{ \dfrac{6^x\cdot 12^{x+1}\cdot 20^{x+2}\cdot 30^{x+3}\cdot 42^{x+4}\cdot 56^{x+5}\cdot 72^{x+6}\cdot 90^{x+7} }{10!\cdot 8^x\cdot 15^{x+1}\cdot 24^{x+2}\cdot 35^{x+3}\cdot 48^{x+4}\cdot 63^{x+5}\cdot 80^{x+6}} } \\\\ &=& \dfrac{(2\cdot3)^x\cdot (3\cdot4)^{x+1}\cdot (4\cdot5)^{x+2}\cdot (5\cdot6)^{x+3}\cdot (6\cdot7)^{x+4}\cdot (7\cdot8)^{x+5}\cdot (8\cdot9)^{x+6}\cdot (9\cdot10)^{x+7} } {10!\cdot (2\cdot4)^x\cdot (3\cdot5)^{x+1}\cdot (4\cdot6)^{x+2}\cdot (5\cdot7)^{x+3}\cdot (6\cdot8)^{x+4}\cdot (7\cdot9)^{x+5}\cdot (8\cdot10)^{x+6}} \\\\ &=& \dfrac{2^x3^x3^{x+1}4^{x+1}4^{x+2}5^{x+2}5^{x+3}6^{x+3}6^{x+4}7^{x+4}7^{x+5}8^{x+5}8^{x+6}9^{x+6}9^{x+7}10^{x+7} } {10!2^x4^x3^{x+1}5^{x+1}4^{x+2}6^{x+2}5^{x+3}7^{x+3}6^{x+4}8^{x+4}7^{x+5}9^{x+5}8^{x+6}10^{x+6}} \\\\ &=& \dfrac{3^x4^{x+1}5^{x+2}6^{x+3}7^{x+4}8^{x+5}9^{x+6}9^{x+7}10^{x+7} } {10!4^x5^{x+1}6^{x+2}7^{x+3}8^{x+4}9^{x+5}10^{x+6}} \\\\ &=& 3^x9^{x+7} \dfrac{4^{x+1}5^{x+2}6^{x+3}7^{x+4}8^{x+5}9^{x+6}10^{x+7} }{10!4^x5^{x+1}6^{x+2}7^{x+3}8^{x+4}9^{x+5}10^{x+6}} \\\\ &=& 3^x9^{x+7} \dfrac{4^{x+1-x}5^{x+2-x-1}6^{x+3-x-2}7^{x+4-x-3}8^{x+5-x-4}9^{x+6-x-5}10^{x+7-x-6} }{10!} \\\\ &=& 3^x9^{x+7} \dfrac{4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10 }{10!} \\\\ &=& 3^x9^{x+7} \dfrac{4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10 }{3!4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10} \\\\ &=& \dfrac{3^x9^{x+7}} {3!} \\\\ &=& \dfrac{3^x9^{x+7}} {2\cdot 3} \\\\ &=& \dfrac{1}{2}\cdot 3^{x-1}9^{x+7} \\\\ &=& \dfrac{1}{2}\cdot 3^{x-1}3^{2(x+7)} \\\\ &=& \dfrac{1}{2}\cdot 3^{x-1+2(x+7)} \\\\ &=& \mathbf{\dfrac{1}{2}\cdot 3^{3x+13}} \\ \hline \end{array}\)

lets say for example we want to reflect point (-6,-4) which is A 
and lets say we want to reflect it across like line y=-4x-4 

Well you could probably do it graphically.

or

You could say that the gradient of the line is -4 so the gradient of the perpendicular is 1/4

The equation of the perpendicular through (-6,-4) is    

y=0.25x+k

-4=0.25*-6+k

-4=-1.5+k

k=-2.5

so

y=0.25x-2.5

simulaneous solution of the two lines

\(y=-4x-4 \\ y=0.25x-2.5\\ -4x-4=0.25x-2.5\\ -1.5=4.25x\\ x=-6/17\\ y=-2\frac{10}{17}\\ (\frac{-6}{17},-2\frac{10}{17})\)

Now you can find the distance between those two points, then find the other point on the perpendicular that is on other side of the reflection line and the same distance away as the original point was.

These numbers are horrible.  Did you make this question up?

Find the values of k for which \(2x^2 + {\color{red}k}xy + 3y^2 - 5y - 2\) factors into two linear factor.

I have not had direct touble editing but

I am  having trouble with LaTex at the moment.

It is not always rendering properly for me.

When I try to submit a comment It tells me I didn't check the "i'm not a robot" box, and after I check It again and submit it again, It submits the answer but It doesn't let me edit it (shows the comment as if I didnt write it)