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If you gave us an example of what you are trying to solve, we might be in a better position to help you.

30/5 =6

Arcsin(sin(40)*sin(23.22) - cos(40)*cos(23.22)*cos((360*18)/24 - 88))

asin(sin(40)*sin(23.22) - cos(40)*cos(23.22)*cos((360*18)/24 - 88)) = 73.135097037446 degrees

I just put it straight into the web2.0calc      

Thanks Omi :)

Omi thinks that you may have forgotten a ^ symbol.

She has answered a possible alternate version of your question.

25^((log3 log3 25)/log3 25)=2log2 x

I am confused....   Is this the question that you are asking?

\(25^{\frac{(log_{10}3 *log_3 25)}{log_3 25}}=2log_{10}(2 x)\\~\\ 25^{log_{10}3} =2log_{10}(2 x)\\~\\ \)

if 3000 is 60% of a total, what is the total

60% of X = 3000

X = 3000 /  60%

X = 3000 / 0.60

3000/0.6 = 5000

Or you could have said

60% is 3000 so 

10% is 3000/6 so

100% is 3000/6 * 10

=5000

How do you solve Cos^2x + 3*Sin^2x + sqrt3*Sin^2x = 1  ?

How do you solve Cos^2x + 3*Sin^2x + sqrt3*Sin2x = 1

\(Cos^2x + 3Sin^2x + \sqrt3*Sin2x = 1\\ Cos^2x +Sin^2x+ 2Sin^2x + \sqrt3*Sin2x = 1\\ 1+ 2Sin^2x + \sqrt3*Sin2x = 1\\ 2Sin^2x + \sqrt3*Sin2x = 0\\ 2Sin^2x + \sqrt3*2sinxcosx = 0\\ 2Sinx(sinx + \sqrt3*cosx )= 0\\ sinx=0 \qquad or \qquad sinx + \sqrt3*cosx=0\\ sinx=0 \qquad or \qquad sinx =- \sqrt3*cosx\\ sinx=0 \qquad or \qquad tanx =- \sqrt3\\ x=0,\;\; 180^0,\;\; 120^0,\;\;300^0\\ x=180n \qquad or\quad x=180n-60\qquad n\in Z\\ \mbox{All angles measurements are in degrees} \)

Solve for x:
cos^2(x)+3 sin^2(x)+sqrt(3) sin(2 x) = 1

Simplify trigonometric functions:
2-2 sin(pi/6-2 x) = 1

Subtract 2 from both sides:
-2 sin(pi/6-2 x) = -1

Divide both sides by -2:
sin(pi/6-2 x) = 1/2

Take the inverse sine of both sides:
pi/6-2 x = (5 pi)/6+2 pi n_1  for  n_1  element Z
   or  pi/6-2 x = pi/6+2 pi n_2  for  n_2  element Z

Subtract pi/6 from both sides:
-2 x = (2 pi)/3+2 pi n_1  for  n_1  element Z
   or  pi/6-2 x = pi/6+2 pi n_2  for  n_2  element Z

Divide both sides by -2:
x = -pi/3-pi n_1  for  n_1  element Z
   or  pi/6-2 x = pi/6+2 pi n_2  for  n_2  element Z

Subtract pi/6 from both sides:
x = -pi/3-pi n_1  for  n_1  element Z
   or  -2 x = 2 pi n_2  for  n_2  element Z

Divide both sides by -2:
Answer: |  x = -pi/3-pi n_1  for  n_1  element Z
            or  x = -pi n_2  for  n_2  element Z

(a) Superior Construction Pte Ltd is a successful company dealing with many major projects in Singapore. Recently, it has submitted its biddings for two major Government projects. Project A worth about $120 million and the company believes it has 40% chance of securing the project. Project B worth $1.8 billion and there is 30% chance the company can win the project.

P(A)=0.4 worth $120 million

P(b)=0.3 worth $ 1.8 million

Both projects are independent of each other. What is the probability that the company:

i. will secure Project A or B but not both  

0.4*0.7+0.3*0.6 = 0.46

ii. will not secure Project A or will not secure Project B

1-P(will secure both) 

1-0.4*0.3 = 0.88

I'll give b and c a go .....

(b) Do you agree that “if two events are mutually exclusive then these two events will be independent”? Why?

NO if they are mutually exclusive it means that if one happens the other cannot happen so they would not be independent.  They are totally dependent on each other.

(c) Provide one business-related example each, with explanation, for mutually exclusive and independent events.

Mutually exclusive:

Is Fred going to become the CEO or is Suzie going to become the CEO (given that there will only be one CEO)

Independant events:

Will it rain today  and will Suzie become the CEO.

what is sqrt(3x) times sqrt(6x)

\(\sqrt{3x} \times \sqrt{6x}\\ =\sqrt{3x} \times \sqrt{3x}\times \sqrt{2}\\ =3x\sqrt2\)

OR

\(=\sqrt{3x} \times \sqrt{6x}\\ =\sqrt{18x^2}\\ =\sqrt{9x^2}\times \sqrt2\\ =3x\sqrt2\\\)

You have not said how long the track is, or whether there is a pole at the end or not. 

1/0 is undefined so you cannot have any function of it !

I have assumed that the average speed of the truck is always 15miles/hour - even on the return journey.

Hi again 315.   :)

Say the average round distance for Haul A is X miles

Then the average round distance for Haul B will be      

60%*(x+5) + 20%*(X-2) +20%*X

=0.6X + 0.6*5 -0.2X - 0.2*2 +0.2X

= X     + 0.6*5           - 0.2*2

This question is only concerned with the difference in the average distance so which is

=0.6*5           - 0.2*2

=    3             -   0.4

=   2.6 miles       The average truck will travel 2.6miles further per load for plan B

Now I think that there are too many things that are poorly worded or not explained in this question BUT

this is my take.

What is the extra cost for part A.  

I have to assume that the $2.35 / m^3   includes transport and material costs.

2.35*2*10^6 = 4.7*10^5 = 4.7 million dollars

Extra costs for area A is 4.7 million dollars

Now what are  the extra costs for part B

Q  how many truck loads will it take to fill the 14*10^6 m^3 tip

A  14*10^6 divided by 20 = 7*10^5 loads     

Q and how many extra distance will this be for B?        

A  2.6*7*10^5 = 1.82*10^6     miles

Q  At 15mph how much extra time will this take?

A   1.82*10^6 divided by 15  = 121333 and 1/3 hours

Q  At $35 pr hour how much extra cost will this be?

A  121333 and 1/3 hours *35 = $ 4246666.67      

Extra Cost for area B is    4.245 million dollars

Extra costs for area A is   4.7 million dollars

So, based on the information given,  area B is cheaper.

what is 2 and 1/4 times 1 and 1/3

2 1/4 x 1 1/3 =

Simplify the following:
(2+1/4) (1+1/3)

Put 1+1/3 over the common denominator 3. 1+1/3  =  3/3+1/3:
(2+1/4) 3/3+1/3

3/3+1/3  =  (3+1)/3:
(2+1/4) (3+1)/3

3+1 = 4:
(2+1/4)×4/3

Put 2+1/4 over the common denominator 4. 2+1/4  =  (4×2)/4+1/4:
4/3 (4×2)/4+1/4

4×2  =  8:
((8/4+1/4)×4)/(3)

8/4+1/4  =  (8+1)/4:
4/3 (8+1)/4

8+1 = 9:
9/4×4/3

9/4×4/3 = (9×4)/(4×3):
(9×4)/(4×3)

(9×4)/(4×3) = 4/4×9/3 = 9/3:
9/3

The gcd of 9 and 3 is 3, so 9/3 = (3×3)/(3×1) = 3/3×3 = 3:
Answer: |=3

You can also use "Summation Formula" to get the same result, which is basically summing up the PV of 1st., 2nd, 3rd,...........48th payment:∑[386/1.007^(n+1), n=0 to 47]=$15,690.34.

Judy had a number of raisins.  After eating one, she gave half the remainder to her sister.  After eating another raisin, she gave a third of what was left to her brother.  Judy now had only six raisins left.  How many raisins did she start with?

Since Judy has 6 raisins Left.

6 + 3 =9 what she had left before giving 1/3 to her brother.

9+1 =10 Number befor eating one and before giving 1/3 to her brother

10 x 2 =20 Number of raisins before she gave half to her sister.

20+1 =21 Number of raisins she started with.

There are many ways of solving this problems. This is the most childish and simplest!!!!!.

An arc subtends a central angle measuring 2 π/3 radians. What fraction of the circumference is this arc?

2Pi=360, so

2Pi/3=360/3 or 1/3 of the circumference.

9.8)(0)+1/2Vi^2=(9.8)(100)+1/2(72)^2 solve for vi please show steps

Changed your "Vi" variable to just "V"

Solve for V:
V^2/2+0. = 3572.

V^2/2+0. = V^2/2 and 3572. = 3572:
V^2/2 = 3572

Multiply both sides by 2:
V^2 = 7144

Take the square root of both sides:
Answer: |  V = 2 sqrt(1786)     or     V = -2 sqrt(1786)

A car dealership offers you no money down on a new car. You may pay for the car for 4 years by equal monthly end-of-the-month payments of $386 each, with the first payment to be made one month from today. If the discount annual rate is 8.40 percent compounded monthly, what is the present value of the car payments?

PV=P{[1 + R]^N - 1.[1 + R]^-N} R^-1. This is the formula you use to calculate this:

The PV of all the 48 payments=$15,690.34. This is effectively the price of the car.

Assuming "Φ" is a property of a distribution:

CDF[normal distribution | mean | mu = 0
standard deviation | sigma = 1, 2.2]

=0.986097 according to W/A.

arctan (15.4 / 11.8) = about  52.54°