All Answers 355920 Answers

You need to proof read your equation .......

5.57%   =  .0575

quarterly interest =  .0575/4 = .014375

due at the end of 'n' years would be (put the number of years for 'n' and compute):

         5000 (1 + .014375)⁴ⁿ  

nvm!

Yeppers !   Math on......

15 , 22 , 29 , 36 , 43 , 50 , 57 , 64 , 71 , 78 , 85 , 92 , 99 , Total =  13 such numbers

uugghh!!....so my main error really was coming to the correct expression for the area, from there I could manage.....Thanx once again ElectricPavlov!

ElectricPavlov,

yes, I understand all that, I was just having trouble with understanding the difference between multiplying two functions to yield a specific outcome, and deviding those functions for a similar outcome....

Let the shorter leg = S
The longer leg        = S + 8
The Hypotenuse     = S + 16

Pythagoras’s Theorem:
[S + 16]^2 = [S + 8]^2 + S^2

Solve for S:
(S + 16)^2 = S^2 + (S + 8)^2

Expand out terms of the right hand side:
(S + 16)^2 = 2 S^2 + 16 S + 64

Subtract 2 S^2 + 16 S + 64 from both sides:
-64 - 16 S - 2 S^2 + (S + 16)^2 = 0

Expand out terms of the left hand side:
-S^2 + 16 S + 192 = 0

The left hand side factors into a product with three terms:
-(S - 24) (S + 8) = 0

Multiply both sides by -1:
(S - 24) (S + 8) = 0

Split into two equations:
S - 24 = 0 or S + 8 = 0

Add 24 to both sides:
S = 24 or S + 8 = 0

Subtract 8 from both sides:

S = 24 - The short side; 24+8=32 - The longer side;
32 + 8 =40 - The Hypotenuse

Check: 40^2 = 32^2 + 24^2
             1600 = 1600

Area =2x  * ( 4-x)    over the domain  x = 0 to x = 4

Area =  8x - 2x^2

you want to find the maximum value of this function (which is a dome shaped parabola) from x = 0 to x = 4

  you can graph it ...or calculate the max value of x    that produces the max area (the vertex)

     - b /2a = -8/(2* -2) = 2         so x value of vertex = 2      plug this in to the area equation to find y = area = 8 units^2

Or if you want to do the derivative and set = 0

   -4x + 8 = 0

    x = 2     this is where the slope of the parabola is 0 ...this   is the x value which gives the greatest area....the vertex of the parabola....

                     you will have to plug this value of x into the area equation to calculate what the max area is when x = 2

                       8x-2x^2   = y  = area = 8 units²

Here is a graph of the area function:

https://www.desmos.com/calculator/mzd3s0ecs2

No matter what the value of "a" and its multiples is, the GCD of: (a + 2) and (a + 8) will always be = 2

The 'domain expressions' are the areas ..or values...of   'x'   that satisfy your equation..

   like from x = 3 to infinity     (  3 < x < + inf)

       or   from   x = 1.5 to 3     which would be  ( 1.5 < x < 3)

Thank you ElectricPavlov...I will study that a bit closer....do appreciate..

Thank you kindly...just was not sure about the domain expressions..do appreciate.

In the second graph you can see where the individual functions are + or -   ....

   or where they are both -    or   both +

If you look closely wou will see   +  and -   up until approx x = '1' then a little bit of both + until x ~ 1.5  then   + - again up until 3

       then both +   above 3

   this is graphed in the upper picture  you can see where the division of the two is negative   then a small bit of positive...then negative up to x=3   then positivr above x = 3 

Numerically, there will be different answers.

Example:

-12 * 3 = -36

-12/3 = -4

But I think the domain expressions will be the same.....

Alan,

I'm sorry, no..I do not see what I should see there..

ElectricPavlov,

so then...there will be no difference between the answers of 

\(f(x).f'(x)<0 \)

and 

\({f(x) \over{f'(x)}}<0\)

because in both instances the criteria is the same...one equation must be negative or positive, while the other is positive or negative, respectively?

Look at the graph and see if you can decide from that:

Here are graphs of the individual functions:

Agree!   Sorry....didn't see 'isosceles' in the Q !     ~ EP

Same story, one of the two has to be negative and the other has to be positive .....and neither can be zero.

For GREATER than zero both have to be positive  ....or both have to be negative.

DO you need more than that? 

oh yes...off course..You know Melody....I am working through some grade 12 papers and a lot of things are starting to really open up and sink in..I simply cannot believe that things are not as difficult as they seem, it really is common sense in sooo many cases, but having said this, I really want to thank ALL of you guys who have remained patient throughout my silly questions, and in many cases having explained things more than once. I finally understand all of this now..

From your 3 sets, you can have this many combinations:

[(1, 2, 3), (1, 2, 5), (1, 2, 6), (1, 2, 7), (1, 2, 9), (1, 2, 10), (1, 2, 11), (1, 3, 5), (1, 3, 6), (1, 3, 7), (1, 3, 9), (1, 3, 10), (1, 3, 11), (1, 5, 6), (1, 5, 7), (1, 5, 9), (1, 5, 10), (1, 5, 11), (1, 6, 7), (1, 6, 9), (1, 6, 10), (1, 6, 11), (1, 7, 9), (1, 7, 10), (1, 7, 11), (1, 9, 10), (1, 9, 11), (1, 10, 11), (2, 3, 5), (2, 3, 6), (2, 3, 7), (2, 3, 9), (2, 3, 10), (2, 3, 11), (2, 5, 6), (2, 5, 7), (2, 5, 9), (2, 5, 10), (2, 5, 11), (2, 6, 7), (2, 6, 9), (2, 6, 10), (2, 6, 11), (2, 7, 9), (2, 7, 10), (2, 7, 11), (2, 9, 10), (2, 9, 11), (2, 10, 11), (3, 5, 6), (3, 5, 7), (3, 5, 9), (3, 5, 10), (3, 5, 11), (3, 6, 7), (3, 6, 9), (3, 6, 10), (3, 6, 11), (3, 7, 9), (3, 7, 10), (3, 7, 11), (3, 9, 10), (3, 9, 11), (3, 10, 11), (5, 6, 7), (5, 6, 9), (5, 6, 10), (5, 6, 11), (5, 7, 9), (5, 7, 10), (5, 7, 11), (5, 9, 10), (5, 9, 11), (5, 10, 11), (6, 7, 9), (6, 7, 10), (6, 7, 11), (6, 9, 10), (6, 9, 11), (6, 10, 11), (7, 9, 10), (7, 9, 11), (7, 10, 11), (9, 10, 11)], >Total distinct combinations = 84

You will have to go through each combination, add up the 2 smaller numbers and see if their total is greater than the 3rd number. Example: (1, 2, 5). 1 + 2 = 3 which smaller than 5. Therefore, it cannot make a triangle. On the other hand, (7, 9, 10) will make a triangle since 7 + 9=16 which is > 10. You will have to do this for each and every one of the above 84 combinations. Then you will have to count all the ones that will make a triangle and divide that number by 84. And that will be your Probability.

Mr.James' class:

2 / [2+3] x 30 = 12 boys

3 / [2+3] x 30 = 18 girls

Mr. Singh's class:

1 /[1+2] x 30 = 10 girls

2/[1+2] x 30 = 20 boys.

For 2 dice, the probability of rolling an EVEN sum of: 2 or 4 or 6 or 8 or 10 or 12 are as follows:

1 + 3 + 5 + 5 + 3 + 1 =18 / 6^2 =1 / 2

The probability of rolling a sum of 5 or 10 is as follows:

4 + 3 = 7 / 6^2 = 7 / 36

Finally, the overall probability of rolling an EVEN sum, or a sum of 5 or 10 is:

[18 + 7] / 6^2 = 25 / 36

S Q UA R E. Ordinarily, 6 letters would give you 6! = 720 permutations. But since you want the letters "UA" always together, then as Melody often says, "rope them together", or consider them as one letter. So, instead of having 6 letters or 6! permutations, you will simply have 5! = 120 permutations.