All Answers 358087 Answers

Imagine a right-angled triangle with the opposite = 5 and the adjacent = 1 (so that the angle is tan^-1 5); then the hypotenuse is sqrt(26)  so the cosine of the angle is 1/sqrt(26)  (or sqrt(26)/26 if you prefer).

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Oops!! You're right 4^4 = 256, so probability is 1/256.  

I've interpreted the question to mean what is the probability of getting all four.  

If you interpret it to mean what is the probability of getting at least one (for example) then the probability is 

1 - (1- 1/4)^4 → 1 - 81/256 → 175/256 (≈ 68.3%)

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Hey Alan......check your math   (1/4)^4 = 1/256   This just doesn't seem right.......the MORE tries you make the LESS chance of your desired outcome??

I'm never quite sure about these probabilities, but let's look at it another way      each try has a  3/4 chance of NOT being what you want  (3/4)^4 =81/256 chance you will NOT get what you want......isn't that a 175/256 chance you WILL get what you want?     I do not know.......

Use the Pythagoras theorem:  c = sqrt(12^2 + 35^2)

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28/33

Just multiply the diameter by pi to get the circumference.

25% = 1/4  

Do it four times, each time independent of the others, and the final probability is (1/4)^4 = 1/16 (or 6.25%)

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What is your actual question?

Better solution:

Let x  =  f*y  where f is a finite constant.

Then ln(1+x*y²)/(x²+y²) becomes  ln(1 + f*y³)/((f²+1)y²)

No matter what the size of f, eventually, as y tends to zero, f*y³ will become much smaller than 1, so ln(1 + f*y³) will tend to f*y³, hence ln(1 + f*y³)/((f²+1)y²)  →  f*y³/((f² + 1)y²)  →  f*y/(f²+1)  →   0

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4% x 68 billion=2.72 billion

8/31= 1/?+1?/+1/?

8/31= 1/8 + 1/ 16 + 1/14 6/35

U = energy = 1/2 c v^2   = 1/2 (470 x 10^-6)(15^2) = .052875 j

\(p = \frac{2q^2-1}{3q^2+2} \\ \frac{1}{2} = \frac{2q^2-1}{3q^2+2}\\If \frac{a}{b}=\frac{c}{d},\quad then \,ad=bc. \\ 1(3q^2+2)=2(2q^2-1)\\Simplify\, and\,you\,get:\\ -q^2+4=0 \\ Using\,quadratic\,equation\,method,\, \\ we\,end\,up\,with:\\ q= \pm 2\)     

WHAT DO YOU WANT DONE WITH THESE NUMBERS????

955/12=79.5833......Lazy bones!!!

Solve for y:
y+13/49 = -40

Put each term in y+13/49 over the common denominator 49: y+13/49  =  (49 y)/49+13/49:
(49 y)/49+13/49 = -40

(49 y)/49+13/49 = (49 y+13)/49:
(49 y+13)/49 = -40

Multiply both sides of (49 y+13)/49 = -40 by 49:
(49 (49 y+13))/49 = -40×49

(49 (49 y+13))/49 = 49/49×(49 y+13) = 49 y+13:
49 y+13 = -40×49

49 (-40)  =  -1960:
49 y+13 = -1960

Subtract 13 from both sides:
49 y+(13-13) = -1960-13

13-13 = 0:
49 y = -1960-13

-1960-13 = -1973:
49 y = -1973

Divide both sides of 49 y = -1973 by 49:
(49 y)/49 = (-1973)/49

49/49 = 1:
Answer: |  y = (-1973)/49

Find the derivative of the following via implicit differentiation:
d/dx(-4 x+3 y) = d/dx(-41)
Differentiate the sum term by term and factor out constants:
3 d/dx(y)-4 d/dx(x) = d/dx(-41)
The derivative of x is 1:
3 (d/dx(y))-1 4 = d/dx(-41)
The derivative of y is y'(x):
-4+3 y'(x) = d/dx(-41)
The derivative of -41 is zero:
-4+3 y'(x) = 0
Add 4 to both sides:
3 y'(x) = 4
Divide both sides by 3:
Answer: |  y'(x) = 4/3

Solve for q:
p = (2 q^2-1)/(3 q^2+2)

p = (2 q^2-1)/(3 q^2+2) is equivalent to (2 q^2-1)/(3 q^2+2) = p:
(2 q^2-1)/(3 q^2+2) = p

Multiply both sides by 3 q^2+2:
2 q^2-1 = p (3 q^2+2)

Expand out terms of the right hand side:
2 q^2-1 = 2 p+3 p q^2

Subtract 3 p q^2-1 from both sides:
q^2 (2-3 p) = 2 p+1

Divide both sides by 2-3 p:
q^2 = (2 p+1)/(2-3 p)

Take the square root of both sides:
Answer: |  q = sqrt((2 p+1)/(2-3 p))      or      q = -sqrt((2 p+1)/(2-3 p))
IF p=1/2, then q=+or- 2

0.7 x 4.5 = 3.15 L