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(i) Price elasticity iseems to be  %change of quantity divided by %change in price.

Q = 200 - 4P - 0.2P^2

When P = 15  Q = 95

If P increases by 1% (to 15.15) then Q becomes  93.495  a change (decrease) of 1.6%

So presumably the price elasticity of supply is 1.6%/1%  or 1.6  (or -1.6, since Q decreases?)

(This is largely guesswork based on a quick search for the meaning of price elasticity!.)

It turns into b***s if you type b***s. That actually is B a L L s

If Bowser drew a green one, which means he is lying......

There is 75% that Mario's initial draw is red

And 66 2/3 % that Mario's second draw is red.

Multiply them together = 50%.

There are 80% chance that he has a 50% probability to draw 2 red ones in a row.

Therefore 80% x 50% = 40%

He has a 40% probability now.

If Bowser drew a red one, which means he isn't lying.....

There is 50% that Mario's initial draw is red

And 33 1/3 % that Mario's second draw is red.

Multiply them together = 16 1/6 %

There are 20% chance that he has a 16 1/6% probability to draw 2 red ones in a row.

Therefore 20% x 16 1/6% = 3 1/3%

He has a 40% + 3 1/3% probability now.

Therefore Mario has a 43 1/3% probability to draw 2 red b***s in a row.

On calculator: a/b 

In LaTeX: \dfrac{a}{b} and \frac{a}{b} where the \dfrac command makes bigger fractions.

e.g. : \(\dfrac2{3}\) Using \dfrac

\(\frac{2}{3}\) Using \frac

\(\dfrac{3}{8}(16x-48)+3x\\ =3(2x-6)+3x\\ = 6x - 18 + 3x\\ = 9x - 18\)

If {a,b} and {b,a} is counted the same.......

Starting with 1:

{1,3}, {1,4}, ...... {1,9}, {1,3,5}, {1,3,6}...... ,{1,3,9}......

Subtotal 7+5+3+1 subsets = 16 subsets.

Starting with 2:

{2,4}, .......{2,9}, {2,4,6}, {2,4,7}... {2,4,9}...... Too lazy to type them all XD

Subtotal 6 + 4 + 2 = 12 subsets

Starting with 3: 

{3,5}, {3,6} .... {3,9}, {3,5,7}, ..., {3 , 5, 9}.........

Subtotal 5 + 3 + 1 = 9 subsets

There we have found a rule for this.

Total subsets = 4 x 4 + 4 x 3 + 3 x 3 + 3 x 2 + 2 x 2 + 2 x 1 + 1 x 1 = 16 + 12 + 9 + 6 + 4 + 2 + 1 = 50 subsets.

\(\underbrace{something something}_{No\; solution}\)? I don't know what you mean but I am trying to solve this.

$90000 per year(per annum) = $(90000/12) per month = $ 7500 per month!!

Which is same as Guest number 1's answer.

(Method 1)There are 10 multiples of 3 in 1- 30, So there are 10 multiples of 3 in 50 - 80. Then we count from 81.

81, 84, 87, 90, 93, 96, 99.

There are 17 multiples of 3 between 50 and 100.

Then we have 99 is not an even number.

51, 54, 57, 60..... , 96 <---- 16 numbers there.

Then we divide 16 by 2 which is 8

There are 8 even multiples of 3 between 50 and 100.

(Method 2) Even multiples of 3 must be divisible by 6 because those multiples of 3 are divisible by 2 and 3.

Starting from 54, we have

54, 60, 66, 72, 78, 84, 90, 96 Which is 8 numbers.

72 is not divisible by 7.

Therefore answer is C.

\((40,000-3000-(-18,000+10000))\times1-0.4\\ = 40000 - 3000 + 18000 - 10000-0.4\\ =45000 - 0.4\\ = 44999.6\\ \text{It doesn't equal 25400}\)

\(\dfrac{4\pi r^3}{3}=3000\\ 4\pi r^3 = 9000\\ \pi r^3 = 2250 \\r=\sqrt[3]{\dfrac{2250}{\pi}}\)

r is the radius of the sphere.

\(\text{Surface area}\\ =4 \pi r^2\\ = 4(\dfrac{2250}{\pi})^{\frac{2}{3}}\\ \approx 320.195399866\;\;\text{cm}^2\)

This is a approximate value of pi. But starting from ..... 897932 ....... it's not so accurate. The previous digits are correct though......

Kmax= (6.63*10^-34J*s) (7.09*10¹⁴s) - 2.17*10^-19J

Im not sure if the 10¹⁴s is suppose to be 10¹⁴s^-1 or not.

Someone told me it should but it doesnt say on the problem. 

\(\begin{array}{|rcll|} \hline K_{max} &=& ( 6.63*10^{-34}J*s) ( 7.09*10^{14}s^{-1} ) - 2.17*10^{-19}J \\\\ K_{max} &=& 6.63*7.09*10^{-34}*10^{14} J*s*s^{-1} - 2.17*10^{-19}J \\\\ K_{max} &=& 6.63*7.09*10^{-34+14} J*s*s^{-1} - 2.17*10^{-19}J \\\\ K_{max} &=& 6.63*7.09*10^{-20} J*s^1*s^{-1} - 2.17*10^{-19}J \\\\ K_{max} &=& 6.63*7.09*10^{-20} J*s^{1-1} - 2.17*10^{-19}J \\\\ K_{max} &=& 6.63*7.09*10^{-20} J*s^{0} - 2.17*10^{-19}J \quad | \quad s^{0} = 1 \\\\ K_{max} &=& 6.63*7.09*10^{-20} J*1 - 2.17*10^{-19}J \\\\ K_{max} &=& 47.0067*10^{-20} J - 2.17*10^{-19}J \\\\ K_{max} &=& 4.70067*10*10^{-20} J - 2.17*10^{-19}J \\\\ K_{max} &=& 4.70067*10^{1-20} J - 2.17*10^{-19}J \\\\ K_{max} &=& 4.70067*10^{-19} \ J - 2.17*10^{-19} \ J \\\\ K_{max} &=& (4.70067- 2.17)*10^{-19} \ J \\\\ K_{max} &=& 2.53067*10^{-19}\ J \\ \hline \end{array}\)

e= (1.7*10^3J)-(3.3*10^2J)                  I got 1.4*10^-3J but i dont think it is right. 

            (1.7*10^3J)

\(\begin{array}{|rcll|} \hline e &=& \dfrac{(1.7*10^3J)-(3.3*10^2J)}{(1.7*10^3J)} \\\\ e &=& \dfrac{1.7*10^3J}{1.7*10^3J} -\dfrac{3.3*10^2J}{1.7*10^3J} \\\\ e &=& 1 -\dfrac{3.3*10^2{\color{red}\not} J }{1.7*10^3{\color{red}\not} J } \\\\ e &=& 1 -\dfrac{3.3*10^2 }{1.7*10^3 } \\\\ e &=& 1 -\dfrac{3.3*10^{2-3} }{1.7} \\\\ e &=& 1 -\dfrac{3.3*10^{-1} }{1.7} \\\\ e &=& 1 -\dfrac{0.33}{1.7} \\\\ e &=& 0.80588235294 \\ e &=& 8.0588235294 *10^{-1}\\ \hline \end{array}\)

Write the slope intercept form of (-3,-1) parallel to y=5/3x-1

Yes, it is.

3x - 4 = -19    Add 4 to both sides of equation

3x= -15              Divide both sides by 3

x = -5

$ 10.60

It is called " Exponential Decay Formula", or sometimes "Radioactive Decay Formula". Here is another version of it but is little more involved, because you have to use Euler's constant and find "constant of decay" "k", in this formula:A=A0e^kt. But, the result of this is exactly the same as the one I gave you.

P.S. I must admit that I'm very, very surprised that you are tasked to deal with radioactive materials and yet they haven't taught you how to calculate the decay of these elements?. That is very odd.

By the way, why do you find such a simple thing "difficult"? Obviously, you don't have a "knack" for Math. I feel sorry for you. You should consider seeing a tutor in Math, or in Physics. Good luck to you.

What exactly is this formula called? When I look it up, not a thing comes up about it.

Here the first number after 50 divisible by 3 is  51. But that is odd.

The next number divisible by 3 is 54.

Hence the first term is 54.

Then the numbers increase by a common difference 6.

The last even number divisible by 3 less than 100 is 96.

So

96 = 54 + (n-1) 6

96-54 = (n-1)6

42 = (n-1)6

42/6= n-1

7 = n-1

7+1 = n

n = 8    ......Answer