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5√32 ?

$$5\sqrt{32} \\
=5\sqrt{16\cdot 2} \\
=5 \underbrace{\sqrt{16}}_{=4} \cdot \sqrt{2} \\
=5\cdot 4 \cdot \sqrt{2} \\
=20 \cdot \sqrt{2} \quad | \quad
\boxed{\ \sqrt{2}= 1.41421356237 \ }\\
= 20 \cdot 1.41421356237\\
=28.2842712475$$

you might be thinking id just say

check in the calculator that is on the home page of this site by just typing 230/4 and hit enter and youll know

but i wont say that but 

no

Since the block isn't accelerating the force will equal the weight of the block, namely 500*9.8 Newtons.

Power is force*speed, so power = 500*9.8*0.15 Joules/second (or Watts).

I'll leave you to crunch the numbers.

.

If the radius of a circle is 4cm, what is that area ?

$$\mathrm{Radius} \quad r = 4\ cm \qquad \mathrm{Area}_{Circle}\quad A =\ ?$$

$$\boxed{ A = \pi \cdot {r^2} }$$

$$A = \pi \cdot 4^2 \ cm^2\\
A = \pi \cdot 16\ cm^2 \\
A = 3.14159265359 \cdot 16\ cm^2 \\
A = 50.2654824574 \ cm^2 \\
A \approx 50.27 \ cm^2$$

This was the question:

Here is one that is just a little different.

Suzie has 10 different coloured beads.  She uses these beads to make a bracelet.

This bracelet does not have a clasp so there is really no beginning or end (but there is clockwise and anticlockwise)

How many different permutations of beads are possible ?

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

Your answer was 10!

If these beadswere placed in a row your answer would have been correct, but when you put the beads in a circle there is no distict start point so you have to put the first bead down to indicate first place.  Now there are 9 beads (not 10 ) that need to be organised around this initial bead.

so the answer is 9!             $${\mathtt{9}}{!} = {\mathtt{362\,880}}$$ so if you go clockwise (or anticlockwise) from any specific point there will be 362 880 possible permutations.    

types of triangles

1. scalene triangles - all sides different, all angles different

2. isosceles triangles - 2 sides equal 2 angles equal

3. equilateral triangles - all sides equal all angles 60 degrees.

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

A. obtuse triangle  - one obtuse angle and 2 acute angles

B. acute triangle  -  all angles acute

C. right angled triangle - one right angle and 2 acute angles.

38 is even so one of the numbers has to be 2

2*19

2 1/3 + 1 1/3

$$2 \frac13 + 1 \frac13 \\\\
= 2 + \frac13 + 1 + \frac13 \\\\
= 2 + 1 + \frac13 + \frac13 \\\\
= 3 + \frac13 + \frac13 \\\\
= 3 + 2\cdot \frac13 \\\\
= 3 + \frac23 \\\\
= 3 \frac23$$

seven * eleven

Draw the triangle first then use the sine rule

I shall take this as a proof

$$e^{i\pi}+1=0$$

$$\\LHS=e^{i\pi}+1\\
LHS=cos\pi+isin\pi+1\\
LHS=-1+i*0+1\\
LHS=0\\
LHS=RHS\qquad Q.E.D.\\$$

I'd pick the point that was symmetrially opposite the y intercept.

For example:

If

Vertex (3,1)

y interecept   7    (0,7)

other point would be  (6,7)

I think it is 300g to mg

300*1000=300,000 mg

I am quite sure that there are unsolved problems and if they are not solved then it cannot be known if they are truly solvable.    Does that make sense?

cos^-1(6.4/9.6)

enter it as      acos(6.4/9.6)

arc cos is the same as inverse cos   

$$\\(y^4)^3=y^4*y^4*y^4=yyyy*yyyy*yyyy=y^{4*3}=y^{12}$$

x-4 cannot be simplified

but

x^-4 =       $$x^{-4}=\frac{1}{x^{+4}}=\frac{1}{x^{4}}=$$

Thanks Heureka,

That clip is great.          

This is the equation of a concave up parabola that is entirely above the x axis.

If you want to know why then ask.    :))

Your answer made me laugh  - thank you.

You are right too - there are no solutions to this question.

2x+7=-8x-9+10x

2x+7=-8x+10x-9

2x+7=2x-9

7=-9

that is rediculous - there are no solutions  :)

$$\\1625=\frac{26}{2 }\times(3+i)\\\\
1625=13(3+i)\\\\
\frac{1625}{13 }=\frac{13(3+i)}{13 }\\\\
125=3+i\\\\
122=i\\\\
i=122$$

i dont know if this is right but i think:

a=80

b=-73

c=83

What is the last digit of pi ?

Not necessairily, you could be working with another type of triangle. Arcsine would be he inverse of Sine, that is it is the value of Sine for a certain angle.

Thank you!