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CIRCUMFERENCE = pi x DIAMETER     If the hand is 9cm long (from the pivot point)....that is the RADIUS.  The DIAMETER is 2 x radius = 18cm

Circumference = pi x 18 = 56.55 cm

x/24=arctan (1/(sqrt(2)+sqrt(3)+sqrt(6)+2)    multiply both sides by 24

x = 24 * arctan (1/(sqrt(2)+sqrt(3)+sqrt(6)+2))    =  π

Young person: I want the value of "x", NOT a lecture. Can you come up with it?

Mines good. Im so happy cause this is the last week of school for me and i just got 2 great awards!

Hey! Im good hows your day?

Hey! Same wit me. I haven't been on in a while. Im good btw. Hbu?

arctan is simply an angle.         'arctan x' just means 'the angle whose tangent is x'

arcsin just means 'the angle whose sine is x'       and same for arccos.

So you are just being asked to find an angle,in this case     x    .  And in radians,not degrees.  Why?....

Most trig equations use radians because they are an absolute measure of an angle ,rather than an arbitrary one like degrees.  To convert,there are 2 pi  radians in 360 degrees.   For an explanation about radians,read on.........

Suppose you draw an arc on a circle.(Draw one to see,it  helps) Let the length of that arc be exactly the same length as the radius of that circle. Then the angle made by the sector of that arc is one radian.            To convert to degrees...

..Since a circle's circumference has the length of 2 pi times the radius,there must be 2 pi radians in 360 degrees.

Because the radian was found by using the circle's arc and its radius,the measurement is absolute rather than just arbitrarily splitting the circle up into 360 bits.

1/2 (a)(b) sin(c) where abc are sides of the triangle and angle c is the angle between a and b.

3,420,000 =  3.42 x 10^6

n to the second power ...or  n squared or    \({n}^{2}\)

Or...

N^2-50n-600=0

Subtract 600 from -600 to get 0, 0+600=600.

N^2-50n=600

Plug in values for n, starting with 0.

n=60.

-10 also works.

-10x-10(-50x-10)=-10x-10+500=100+500=600

60x60-(-50x-60)=60x60-3000=3600-3000=600

4 x5 is 20

5 x5 is 25

so it is 

20 : 25

😏😏😏

A coulomb is a huge amount of charge. How many electrons are necessary to produce a -2.0 C charge? (recall the charge of an electron is e = -1.60 × 10-19 C)

 Since the charge of an electron is e = -1.60 × 10-19 C), therefore 1 coulomb has:

1/[-1.60 x 10^-19] =6.25 x 10^18 electron x 2 =1.25 x 10^19 electrons needed to produce -2.0C charge.

10 ?

These are the options possible for this problem:

A.6.0 × 1023 

B.1.25 × 1019 

C.6.3 × 109 

D.6.25 × 1018 

E.1.6 × 1019 

It's easy guest just line up the decimals O.K.  Like I said it's easy.

.....unles he works weekends too....then divide by  7

Apologies to my english!

Hello Guest!

What jobs use linear equations?

Engineers need as Hooke's law :

Hooke's law states so that on a spring that is stretched, the tension force F of the tensioned track is proportional s. The proportionality factor is the so-called spring constant D, which is different from spring to spring. {nl} {nl}

Hooke's law:

{nl} F = f(s) = D * s

This is a linear equation.

Greeting asinus :- )

  !

I developed my rates method when I was teaching maths for nurses.

They have really horrible rates questions.

drops per minute, dose per weight, ml/drop, and a million other rates all have to worked out in combination with each other.  It gets really tricky.

 My method deals with these hard questions really effectively and it  can be done with no understanding of algebra!  

Of course nurses ultimately use stupid 'formulas'.  They are stupid because they don't make sense.  But people learn to use them by rote,  they 'formulas' work and they have been used for a long time, so I guess my method will never seriously catch on :((   What a pity :((  

Not to worry   

Thanks Guest and asinus :)

I commented here 

Hi Chris,

I use graphs a lot but I was not even tempted for this one.

We all develop our own methods for doing things. :)

My answer for this one is really quite comical

e-xcosx

I) find the stationary points

II)determine the nature of each stationary point

Well here is an initial answer anyway ://