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Factors of 120  :

1 | 2 | 3 | 4 | 5 | 6 | 8 | 10 | 12 | 15 | 20 | 24 | 30 | 40 | 60 | 120

7 is the smallest

good catch Melody.

terrible use of variables in the OP.

Thank you!

\(P[6] = \dfrac{9}{10}\cdot \dfrac 1 6 + \dfrac{1}{10} \dfrac{18}{100}= \dfrac{21}{125} \)

\(P[\text{biased|6}] = \dfrac{P[\text{6|biased}]P[\text{biased}]}{P[6]} = \\ \large \dfrac{\frac{18}{100}\frac{1}{10}}{\frac{21}{125}} = \dfrac{3}{28}\)

\(P[\text{biased|!6}] = \dfrac{P[\text{!6|biased}]P[\text{biased}]}{P[\text{!6}]}=\\ \dfrac{\frac{82}{100}\frac{1}{10}}{1-\frac{21}{125}} = \dfrac{41}{416}\)

You can do the last one... see if you learned anything

Thank you very much!! :)

Thank you so much! I apologize for the late reply.

Thank you so much for your help! I apologize for the late reply.

Can u make it more clear?

Second one

To find out what the original expression was under the radical, we can evaluate this :

( 2^3)*12  * (a^3)^3  *  (b^2)^3 * b   =

8*12  * a^9  * b^6*b =

96a^9b^7

So   "A" is correct

First one

Notice that the last term  has to be  "+25"

So....we can eliminate  A and D  because their final terms  will yield  " -25"

Since all therr terms of the polynomial are positive, the factors will only have a "+" involved

Look  at C

(2x + 5) (x + 5)   = 2x^2 + 5x + 10x + 25  = 2x^2 + 15x + 25

So....this is the correct choice......

Note that the radius of the largest circle is  OC + CB = 32 + 36 =  68

The radius of the smallest circle is (1/2)CB = (1/2)(36) = 18

The radius of the intermediate circle is the tricky one to figure !!!

Note that AO is the radius of the largest circle = 68

And OC = 32

So....AO + OC = the diameter of the intermediate circle = 68 + 32 = 100

So....the radius of this circle = 50

So....the shded area is

Area of largest circle - area of the other two circles =

pi  * [ 68^2 - (50^2 + 18^2) ]  =

pi  * [ 4624 - ( 2500 + 324) ]   =

pi * [ 4624 - 2824]  =

pi * [ 1800] units^2   ≈  5654.87 units^2

Here's a graphical solution for the second one :

https://www.desmos.com/calculator/fnueaahoif

If we let  x be the number of pies and y be the number of batches of brownies.....we can graph the inequality 

8x + 12y ≤ 112

Lookng at the graph....we can forget the points inside the shaded area....these combinations will not yield enough desserts

Note that the red dot - representing 3 pies and 8 batches of brownies - is closer to the boundary of the inequality than the points (2, 9)   and (5,7)

So   

(3, 8)  yields = 120 desserts

(2, 9) yields = 124 desserts

(5, 7) yields = 124 desserts

So.....we want to choose   3 pies and 8 batches of brownies

Thanks to EP  for spotting my earlier error  !!!

Checkout this site:

https://www.symbolab.com/

It shows problems step by step and you can find practice.

I am hoping for my family to be well and America too. Also. when you get older, you'll realize that present, themselves, aren't really that important. There will at least be some presents and that's fine.

I think it is because I didn't consider the space between the barrels....

Notice that......if we had 1 barrel....the actual area occupied by the barrel is a  2.5 x 2.5 square  =

6.25 ft^2

So    52.5 / 6.25  = 8.4 barrels....( round to 8 barrels)

To get a feel for this....look at the following image :

A quick google search seems to say that Jesus was born 6-4 BC   (There are historical records which seem to document his birth and life....whether you believe he was the Son of God is another matter....but that does not change the legitimacy of the question)        Let's say 5 BC   ...if he WERE alive today he would be

5 + 2018 = 2023 years old.

poof

He was born in the year 3 B.C.E.

That's an interesting way of solving this problem, heureka......how was that method derived???

Second one

Which of the combos provides for 112 desserts?

3 pies * 8 pieces/pie   +  8 batches * 12 pieces/batch  = 112?

        24 pieces   + 96 pieces = 120    does NOT equal 112        This will work with some desserts left over.........try the others to see if ONE of them equals 112 pieces......pick the one that provids at LEAST 112 pieces with the least amount left over.

We look at the  digit after the 5.   Since it is a "9," we round the "5" to a "6"

So we have : 46,000

T= mv^2/r      Solve for 'v'    1st : Multiply both sides of the equation by 'r'

Tr = mv^2      Then divide both sides by 'm'

Tr/m = v^2     Take the sqrt of both sides

sqrt( Tr/m) = v