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hey I never said I don't trust you I just said you seem pretty shady id still take a bullet for anyone on this site and expect that they do something good with your second chance 

Despite you not trusting me DarkCalc me and my friends support you...anyway you always ask how our day is but never tell us about yours...?how are you friend

[ 6 cos 45  - 3tan 45 ] 

__________________   =

         2 tan 60

6 (√2/2) -  3(1)

___________  =

     2√3

3√2 - 3

______        multiply  top/bottom by √3     

   2√3   

3√6 - 3√3

________   =    divide through by 3

    2 * 3  

√6  - √3

______

    2

f(x)  =  (x^2  - 16)  / ( x - 4)    =  (x + 4) (x - 4) / (x - 4)  =  x + 4

If x  could be 4,  we  would have   ( 4) + 4  =  8

So  k  =  8     will make the function continuous

Mine would be a balding middle aged man with a tshirt with a donut on it.

jk mine is a dog.

Population density  =  population / land area  =

8550405  / 1213.4    ≈ 7047 people / km^2 

Mass  = Density * Volume

Mass / Volume  = Density

The volume of the board is  0.3 * 2.1 * 0.1  = .063 miles^3

So....the density is

0.07 kg  / .063 miles^3   =

1.1 kg / miles^3

x  = 2y + 5    (1)

x^2 + y^2  = 5  (2)

Sub (1) into (2) to find the y intersection of these functions

(2y + 5)^2  + y^2  =  5      simplify

4y^2 + 20 y + 25 + y^2  = 5

5y^2 + 20y +20  = 0      divide through by 5

y^2 + 4y + 4  =  0      factor

(y + 2)^2  = 0        take the square root of both sides

y + 2   = 0

y  = -2

And x  =    2(-2) + 5   =  1

So....(1, -2)  is the tangent point  because it  is the only point that makes both equations true

1 = 2(-2) + 5   is true      and

(1)^2  + (-2)^2  = 5     is also true

Here's a graph :  https://www.desmos.com/calculator/ty1aww3bmp

Given that   \(\frac{8-\sqrt{18}}{\sqrt2}=a+b\sqrt2\)   where  \(a\)  and  \(b\)  are integers, find the values of  \(a\)  and  \(b\) .

----------------

\(\quad\frac{8-\sqrt{18}}{\sqrt2}\\ =\\ \quad\frac{8-\sqrt{18}}{\sqrt2}\cdot\frac{\sqrt2}{\sqrt2}\\ =\\ \quad\frac{8\sqrt2-\sqrt{18}\sqrt2}{\sqrt2\sqrt2}\\ =\\ \quad\frac{8\sqrt2-\sqrt{18\cdot2}}{\sqrt{2\cdot2}}\\ =\\ \quad\frac{8\sqrt2-\sqrt{36}}{\sqrt{4}}\\ =\\ \quad\frac{8\sqrt2-6}{2}\\ =\\ \quad4\sqrt2-3\\ =\\ \quad-3+4\sqrt2\)

So     \(a=-3\)     and     \(b=4\)  .

Awesome job!

He goes through his car door first so that he can get to the other doors.

The area of the cross-section  will be   BC  * CG

Note....BC  = AD    

So...the area is

BC * CG  =

AD * CG =

6 * 12  =

72 cm^2

I can see the messages in my profile, but it says only me or staff members can view these.

Thats because they were removed when someone else cursed.

The width of the rectangle is easy...it  = 8 mm

If we connect the center of each circle, we have an equilateral triangle with sides of 2

And the height of this triangle can be found thusly :

tan 60  = height / 2

√3  =  height/ 2

2√3 mm = height

So....the height of the rectangle  = 2√3 + 2 + 2  =  [2√3 + 4] mm

So...the area  is      width of rectangle * height  of rectangle =

8  ( 2√3 + 4 ) mm^2  =

16√3 + 32 mm^2

Write   \(\frac{12}{\sqrt2}+\sqrt{18}\)   in the form   \(b\sqrt2\)   ,  where   \(b\)   is an integer.

----------

\(\ \quad\frac{12}{\sqrt2}+\sqrt{18}\\ =\\ \quad\frac{12}{\sqrt2}\cdot\frac{\sqrt2}{\sqrt2}+\sqrt{18}\\ =\\ \quad\frac{12\sqrt2}{2}+\sqrt{18}\\ =\\ \quad6\sqrt2+\sqrt{18}\\ =\\ \quad6\sqrt2+\sqrt{2\cdot3\cdot3}\\ =\\ \quad6\sqrt2+3\sqrt2\\ =\\ \quad9\sqrt2\)

volume of box   =   (width)(length)(height)

width   =   15 - x - x

width   =   15 - 2x

length   =   22 - x - x

length   =   22 - 2x

height   =   x

volume of box   =   (width)(length)(height)

volume of box   =   (15 - 2x)(22 - 2x)x

The expression  (15 - 2x)(22 - 2x)x  can be used to determine the volume of the box.

(  And by looking at a graph we can see that the greatest possible volume is about  432 in³ .

The volume of the full tank  is  51 * 39 * 28   =  55692 cm^3 

And 95%  of this is 52907.4 cm^3

So.....the difference between these  is the amount of water required left to fill the tank  = 2784.6 cm^3  (1)

The volume of the log is    (3.14) (5)^2 (44)   = 3454 cm^3   (2)

So....the volume of water that spills out of the tank is just  (2) - (1)  =

3454  - 2784.6  =  

669.4  cm^3

6)The slant height and radius of a cone are 4 and 1, respectively. Unrolling the curved surface gives a circular sector with center angle n degrees. Find n.

If the radius of the cone is 1, the perimeter of the circular sector  is  2pi (radius) =

2pi (1)  =    2pi

So........when we unroll the curved surface, we will have a circular sector with a perimeter of 2pi  and a radius  of 4

So....we can find the central angle [in radians] thusly :

2pi  = radius * theta in radians

2pi  = 4 * theta in radians      divide both sides by 4

pi/2  = theta in radians

And.....pi/2 in radians =  90°

neither did i

5)Find the volume of a cone that has 5 as its base radius and a lateral surface area of 65pi.

The lateral surface area, A,  is given by   

A  = pi * radius * slant height

65 pi  = pi * 5 * slant height       divide both sides by 5 pi

13  = slant height

We can use the Pythagorean Theorem to find the height of the cone

height  =  √ [ slant height^2  - radius^2 ]  = √[13^2 - 5^2]  = √ [144]  = 12

So....the volume of the cone is

(1/3) pi (radius^2) (height )  =  (1/3) pi * 5^2 * 12  =  

(1/3) pi * 300  =

100 pi  units^3

When did you talk about politics? can't find any post about politics in your profile

4)The volume of a cylinder is 100. Another cylinder has twice the height but half the base radius. What is the volume of the second cylinder?

100  = pi (r^2) h

Solving for the height, we have

h  = 100/ (pi *r^2)

So....twice the height  =  200 / (pi * r^2)

So....the volume, V, of the second cylinder  can be expressed as :

V  =  pi (r/2)^2* (200) / (pi * r^2)  =  (pi/pi) (r^2/r^2) (200/ 4)   =  (1)(1)*50  = 50

what?

Like i said up there. I dont realy have a favorite. I just listen to whatever comes on the radio.