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Hey guest! This problem is very non specific in what the value of x can be. As coolstuffYT pointed out, the value of y is minimized at 0(because it's a square term, meaning its either positive or zero). However, the 2x is where things get tricky. Contrary to Coolstuff's answer, the answer is NOT setting x = 2, because the problem clearly defines for us that x > 2, not greater than or equal to. Assuming that the question wants integer values of x and y(it never explicitly says), we set x =3, which gives us:

\(2 * 3 + 0^2 = 6\)

it was B. 1 inch

We are mounds of flesh!

1 kgf/cm2 to ton/square metre = 10 ton/square metre

So multiply by 10.

You are very welcome!

:P

For question one, just measure it.

2 inches is about your whole thumb.

Answer: \(\frac{8}{3}

Explanation:

The largest angle's opposite side is the largest length of the triangle.

So \(2n+12>2n+7>3n-3\).

Take the last two inequalities and get \(n<10\)

Additionally, we can apply the triangle inequality and get \(2n+7+3n-3>2n+12\)

Solve to get \(3n>8\), and n>8/3.

So \(\frac{8}{3} .

You are very welcome!

:P

(I'm not completely sure about my answer, if someone could double check that would be great!)

Click the radian sign.

You are very welcome!
:P

Use the calculator on this site. On the bottom left-hand corner, there are two buttons, one for "degrees" and one for "radians". Click on the one for "radians".

CPhill has answered this one now.

Thanks Chris :)

Answer: 4

Explanation:

Since x and y do not depend on each other... let's let y=0.

So if y=0, then 2x+0^2=2x.

We should minimize x, so make it 2.

2x with x=2 = 4

You are very welcome!

:P

LCM of [12, 13] = 156

156 + 1 = 157 minimum number of eggs.

Thank you Electric Pavlov and CPhill!!! Both your ways were good, I was kinda lost on your way CPhill. I'm not too good at piecing together something that large. I was trying something along the lines of Electric Pavlov's solution, but coudn't figure out how to finish it. Thanks both of you.  Stay safe and healthy!!!

150*0.06 = 9

4x-x=3x

help please! i don't understand these line tests..

\(cos(sin^-1(-0.845))=cos(\frac{1}{sin(-0.845)})=0.77709660144\)

or       \(sin^{-1}\) is the asin key

\(cos(sin^{-1}(-0.845))=cos(asin(-0.845)=0.5347663041\)

  !

sorry! i wrote it wrong it's 

\(cos(sin^-1(-0.845))\)

is the answer different?

The system of equations \(\dfrac{xy}{x + y} = 1, \quad \dfrac{xz}{x + z} = 2, \quad \dfrac{yz}{y + z} = 3\)
has one ordered triple solution \((x,y,z)\).

What is the value of in this solution?

\(\begin{array}{|rcll|} \hline \dfrac{xy}{x + y} &=& 1 \\\\ \dfrac{x + y}{xy} &=& 1 \\\\ \dfrac{x}{xy}+\dfrac{y}{xy} &=& 1 \\\\ \mathbf{\dfrac{1}{y}+\dfrac{1}{x}} &=& \mathbf{1} \qquad (1) \\ \hline \end{array} \begin{array}{|rcll|} \hline \dfrac{xz}{x + z} &=& 2 \\\\ \dfrac{x + z}{xz} &=& \dfrac{1}{2} \\\\ \dfrac{x}{xz}+\dfrac{z}{xz} &=& \dfrac{1}{2} \\\\ \mathbf{\dfrac{1}{z}+\dfrac{1}{x}} &=& \mathbf{\dfrac{1}{2}} \qquad (2) \\ \hline \end{array} \begin{array}{|rcll|} \hline \dfrac{yz}{y + z} &=& 3 \\\\ \dfrac{y + z}{yz} &=& \dfrac{1}{3} \\\\ \dfrac{y}{yz}+\dfrac{z}{yz} &=& \dfrac{1}{3} \\\\ \mathbf{\dfrac{1}{z}+\dfrac{1}{y}} &=& \mathbf{\dfrac{1}{3}} \qquad (3) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline -(1)+(2)+(3) : & -\dfrac{1}{y}-\dfrac{1}{x}+ \dfrac{1}{z}+\dfrac{1}{x}+ \dfrac{1}{z}+\dfrac{1}{y} &=& -1+\dfrac{1}{2}+\dfrac{1}{3} \\\\ & \dfrac{2}{z} &=& \dfrac{-1}{6} \\\\ & \dfrac{z}{2} &=& -6 \\\\ & \mathbf{z} &=& \mathbf{-12} \\ \hline \end{array} \\ \begin{array}{|lrcll|} \hline (1)-(2)+(3) : & \dfrac{1}{y}+\dfrac{1}{x}- \dfrac{1}{z}-\dfrac{1}{x}+ \dfrac{1}{z}+\dfrac{1}{y} &=& 1-\dfrac{1}{2}+\dfrac{1}{3} \\\\ & \dfrac{2}{y} &=& \dfrac{5}{6} \\\\ & \dfrac{y}{2} &=& \dfrac{6}{5} \\\\ & \mathbf{y} &=& \mathbf{\dfrac{12}{5}} \\ \hline \end{array} \\ \begin{array}{|lrcll|} \hline (1)+(2)-(3) : & \dfrac{1}{y}+\dfrac{1}{x}+ \dfrac{1}{z}+\dfrac{1}{x}-\dfrac{1}{z}-\dfrac{1}{y} &=& 1+\dfrac{1}{2}-\dfrac{1}{3} \\\\ & \dfrac{2}{x} &=& \dfrac{7}{6} \\\\ & \dfrac{x}{2} &=& \dfrac{6}{7} \\\\ & \mathbf{x} &=& \mathbf{\dfrac{12}{7}} \\ \hline \end{array} \)

What is the value of \(cos(sin-(0.845)) \)?

Hello Guest!

\(cos(sin(-0.845^o))\approx 1\)

Enter:  cos   sin   -0.845    =         Ergebnis: 0.999999966875

  !

I understand everything except when you raised it all to the sixteenth power, do you mind elaborating further on how you got to 

\(g(x) = a_0 + a_1x + a_2x^2 +..........\)?

Note  that  

{ 1, 2019 , 2, 2018, 3, 2017.....}   will create summed pairs  = 2020  =   101 (20)

The only number  that will not  have a "partner"  is  1010

So....the  max size of this  subset =  2018

cos x = -2/3

Since x  is  in  Q#, then (x/2)  will lie  in Q2  .....and the cosine is negative there....so.....

cos (x/2)  =   - √ [ (1 +cosx) /2 [   =  - √  [ (1 - 2/3)  /2 ]   = - √ [ (3 - 2) / 6 ] =  -√  (1/6)  =

- √6 / 6

Thanks, EP

Here's another method....albeit....maybe not any easier .....LOL!!!!!

The  line intersects  the  y axis  at  15/4  = 3.75

And it  intersects  the  x  axis at x = -5

The distance  between these two points is   sqrt  ( 5^2 + 3.75^2 ) = 6.25

Call the  distance  between  the  y intercept  and the point where the line  intersects  the  upper-most part of the  circle, a

Call the distance  from the  x intercept of the  line  and the point where the line intersects the  "lower" part of the  circle , b

The  distance  between  the  y intercept and the top of the  circle =  6 -3.75 = 2.25

And the distance  from the y intercept to the  bottom part of the  circle  =  3.75 + 6 = 9.75

Likewise......the  distance  from  the  x intercept to the  leftmost point of the  circle  =1

And  the distance  from the  x intercept to  the  rightmost part of the  circle   =11

By  the intercepting  chord  theorem  we have this system

(a) (6.25 + b)  =  (2.25) (9.75)

(a+ 6.25) (b) =  (1) (11)            simplify

6.25a  + ab  =  21.9375     (1)

6.25b  + ab  =  11        (2)          subtract these

6.25 ( a - b)  =  10.9375

(a - b)  =  10.9375/6.25

a - b =  1.75

a = b + 1.75

Subbing  this  into   (2)   we get that

(b + 1.75 + 6.25) ( b)  =11

( b + 8) b  =11

b^2  + 8b - 11 = 0  

b^2  + 8b  = 11

b^2 + 8b + 16 = 11 + 16

(b + 4)^2  =  27        take the positive  root

b + 4 =  √27 = 3√3

b = 3√3  - 4

And 

a = b + 1.75   =   3√3  - 2.25

So.....the  length of  chord  AB  =  

b  + 6.25  + a  =

3√3 - 4 + 6.25  + 3√3  - 2.25   =

6√3  + 6.25 - 6.25  =

6√3

You still have not answered my question:

"It is still not complete. Don't you have the original question?"

No (or is that yes..   ), you are right, YOU shouldn't be disrespectful to people.