All Answers 352688 Answers

It matters to me Chris.

Thank you so much! I just wish I had been able to write it out! :) 

Don't worry about it, LF......!!!

90.    We have 40 marbles....and we are drawing with replacement.......

Blue then Red  =

(4/40) * (12/40)  =

(1/10) * (3/ 10)

3 / 100

91.  Two yellows  =  (8/40) * (8/40)  =  (1/5) * (1/5)  =  1/25

92.  Not a green on the first pick...= 24/40  =  3 / 5

3. 

\( \sqrt{5 + x} + \sqrt{20 - x} = 7\)

We probably can  just solve this by inspection....note.....x = 4 seems to work

So

(5 + 4) (20 - 4)   =  

9 * 16 =

144

Thanks for explaining it! 

Noice guys! Thanks to CPhill and guest!

Don't worry about it, LF....as long as I can see it, the form doesn't matter

We can write the numerator  as  x^(5/2)  y ^(3/2)

And the denominator is x^(2) y ^(3)

So....using   aⁿ / a^m  =  a^n-m   we have

x^(5/2 - 2)  *  y^ (3/2 - 3)   =

x^(5/2- 4/2) * y^(3/2 - 6/2)  =

x^(1/2) * y ( -3/2) ⇒   "B"

Yes, very good, CPhill! 

No problem! 

Thank you for presenting your question nicely :)

I actually already have this answer!! Haha!! I figured it out!! Thanks though! 

Thanks, tertre.....may I add.....the events are dependent......if we were drawing "with replacement," they would be independent....but here, the probability of drawing a diamond on the second draw changes depending upon what card was drawn first

Thanks so much! 

2.

\(y=\frac{x-2}{x^2-7x+10}\)

Factor the denominator

y =        (x - 2)

     ___________    

       (x - 2) (x - 5)

This will not be defined  for any values that make the denominator  0

Thus....it will not be defined when x  = 2   or  x = 5

Thus.....two values

Please post your question one at a time.

and please type your worded question in so that they can be copied into the answer easily. 

Using the digits 1, 2, 3, 4, 5, how many even three-digit numbers less than 500 can be formed if each digit can be used more than once?

The hundreds digit can be 1,2,3 or 4   that is 4 choice       (less than 500)

the tens has 5 choices.               

The units digit has to be even so that is 2 choices

so there are  4*5*2=  40 possible numbers

I’ve been much happier doing my own thing. I found what I can do and such yadayada, that doesn’t really matter. Thanks Alan.

It's been awhile since I did this...so...here goes nothing

v = (3, 4)     w  =  (-2, 7)

We want to project  v onto w

proj_w v  =   [( v dot w) /( length of w)^2 ] * w  =  [ 22/ 53) * (-2, 7)  =  (-44/53, 154/53)

This  is   v1

To  find v2   we have    [ v - v1] =  (3, 4) - (-44/53, 154/53)  =  (203/53, 58/33)

Here's a pic :

The "why" of this always gave me some trouble, too. If I can remember what my  Calc teacher said, if we shine a beam onto v from the "top" of the diagram perpendicular to v2, it will "project" a perfect "shadow" of v onto v2. Likewise....if we shine a beam from the "right" of the diagram onto v such that the beam is perpedicular to v1, it will "project" a perfect "shadow" of v onto v1.

Sorry, hectictar.....that's about as good of an explanation as I can supply......!!!!

I think this is used in Physics to break up components of work  (as well as some other applications, too...)

How in the world did you get that!!

There are: 4 numbers to choose from for the 1st digit from the left. Any number beginning with 5 is not allowed, because that number would be over 500!!

There are: 5 numbers to choose from for the 2nd digit from the left.

There are: 2 numbers to choose from for the 3rd and last digit.

So, that is: 4 x 5 x 2 =40 numbers in total.

Here are ALL the permutations possible:

{1, 1, 1} | {1, 1, 2} | {1, 1, 3} | {1, 1, 4} | {1, 1, 5} | {1, 2, 1} | {1, 2, 2} | {1, 2, 3} | {1, 2, 4} | {1, 2, 5} | {1, 3, 1} | {1, 3, 2} | {1, 3, 3} | {1, 3, 4} | {1, 3, 5} | {1, 4, 1} | {1, 4, 2} | {1, 4, 3} | {1, 4, 4} | {1, 4, 5} | {1, 5, 1} | {1, 5, 2} | {1, 5, 3} | {1, 5, 4} | {1, 5, 5} | {2, 1, 1} | {2, 1, 2} | {2, 1, 3} | {2, 1, 4} | {2, 1, 5} | {2, 2, 1} | {2, 2, 2} | {2, 2, 3} | {2, 2, 4} | {2, 2, 5} | {2, 3, 1} | {2, 3, 2} | {2, 3, 3} | {2, 3, 4} | {2, 3, 5} | {2, 4, 1} | {2, 4, 2} | {2, 4, 3} | {2, 4, 4} | {2, 4, 5} | {2, 5, 1} | {2, 5, 2} | {2, 5, 3} | {2, 5, 4} | {2, 5, 5} | {3, 1, 1} | {3, 1, 2} | {3, 1, 3} | {3, 1, 4} | {3, 1, 5} | {3, 2, 1} | {3, 2, 2} | {3, 2, 3} | {3, 2, 4} | {3, 2, 5} | {3, 3, 1} | {3, 3, 2} | {3, 3, 3} | {3, 3, 4} | {3, 3, 5} | {3, 4, 1} | {3, 4, 2} | {3, 4, 3} | {3, 4, 4} | {3, 4, 5} | {3, 5, 1} | {3, 5, 2} | {3, 5, 3} | {3, 5, 4} | {3, 5, 5} | {4, 1, 1} | {4, 1, 2} | {4, 1, 3} | {4, 1, 4} | {4, 1, 5} | {4, 2, 1} | {4, 2, 2} | {4, 2, 3} | {4, 2, 4} | {4, 2, 5} | {4, 3, 1} | {4, 3, 2} | {4, 3, 3} | {4, 3, 4} | {4, 3, 5} | {4, 4, 1} | {4, 4, 2} | {4, 4, 3} | {4, 4, 4} | {4, 4, 5} | {4, 5, 1} | {4, 5, 2} | {4, 5, 3} | {4, 5, 4} | {4, 5, 5} |. These are ALL the permutations with repeats under 500. If you count the bold ones you should get 40.

As  x  gets closer to  π/2  (from the left)  tan x  gets larger.

A line that is  π/2  radians with the positive  x  axis is a straight vertical line.

So to make  tan x  as large as possible, we want to make  x  as close to  π/2  as possible...

The closest that  x  can be within the domain is  π/4 .

When  x  = π/4 ,      f(x)  =  f(π/4)  =  5 tan( π/4 )  =  5(1)  =  5

If we look at a graph, here...

90.

The total number of marbles: \(12+8+4+16=40\)

Blue Marble: \(\frac{4}{40}=\frac{1}{10}\)

Red marble: \(\frac{12}{40}=\frac{3}{10}\)

So, the answer is just \(\frac{1}{10}*\frac{3}{10}=\boxed{\frac{3}{100}}\)

89. The probability of choosing a dime first is \(\frac{2}{10}=\frac{1}{5}\) .

Next, picking a quarter without replacement is just \(\frac{3}{10}\).

Multiplying, we get \(\frac{1}{5}*\frac{3}{10}=\frac{3}{50}\)

81. (5x)^1/2=square root of 5x, which is \(\sqrt5x, B\)

No problem, 

CPhill, can I use your glasses? 

Solution:

This triangle is isosceles, and so the altitude to the side with length 6 must hit that side at its midpoint. Thus our triangle is divided into two right triangles with hypotenuse \(5\) and one side of length \(3\). Thus each of these is a \(3-4-5\)  triangle, and each one has area \(\frac{3 \times 4}{2} = 6\), for a total area of \(\boxed{12}\).