All Answers 355974 Answers

Let say Donna has x candies

Now, she kept half the number

of candies plus 3 candies

i.e. (x/2 + 3)

She gave remaining candies to Jane

i.e. x - (x/2 + 3) = x/2 - 3

Jane ate 1/3 of the candies 

plus 4 candies

i.e. 1/3 (x/2 - 3) + 4 = x/6 + 3

Jane gave remaining candies to

Kaye

i.e. (x/2 x 3) - (x/6 + 3)

= x/3 - 6

Kate ate 1/4 of the candies

ie 1/4 (x/3 - 6) = x/12 - 3/2

remaining candies are

(x/3 - 6) - (x/12 - 3/2)

= x/4 - 9/2 = 42

=> x/4 = 42 + 9/2

=> x = 4 (42 + 9/3)

     x = 186

Now Donna kept half the

number of candies plus 3 candies

=> x/2 + 3

=> 186/2 + 3

=> 96

Donna ate 96 candies

There are 52 possible draws ...there are 4 {Jack, Queen, King} and 2{red Ace} - so there are 14 out of the 52 will be "winners" ...the odds are 14 in 52 (aka 14:52) ...which (divide both numbers by 2) can also be expressed as 7:26 ...or (multiply by 100/52) is a little under 27 per 100 (ie. 27%)

(136, 163, 233, 316, 323, 332, 361, 613, 631)>>Total = 9 ways

Probability is: 9 / 6^3 ==9 / 216==1 / 24

ABC,DEF ==692,307

692,307 x 4 ==2,769,228

307,692 x 9 ==2,769,228

Find the sum of all values of x such that f(x)=0

Hello Guest!

\(x\in \{-0.5,8\}\\ \color{blue}The\ sum\ is\ 7.5.\)

  !

Thanks Tiggsy,

I never would have thought of that by myself.

It is always great to see you on the forum.   I hope you enjoyed Christmas :)

----------------

Attn:   jsaddem

I get two possible answers,  do you know what they are?   

Could you interact with us please.

Find the side length s of the square PQRS.

Hello Guest!

\(A=30cm^2,\ a=6cm\)

\(A_{\triangle}=\frac{1}{2}ah=\frac{1}{2}\cdot 6cm\cdot h=30cm^2\\ \color{blue}h=10cm\)

\(h:(h-s)=a:s\\ hs=ah-as\\ s(a+h)=ah\\ s=\dfrac{ah}{a+h}=\dfrac{6cm\cdot 10cm}{6cm+10cm}\)

\(s=3.75cm\)

The side length of the square PQRS is s = 3.75cm.

  !
 

T==Tortoise,  E==Elephants

T ==E  +  52..............................(1)

T + 10 ==2[E + 10 + 10]...........(2), solve for E, T

{Note: the 2 elephants will have aged 10 EACH !}

Use substitution to get:

T==74 years old

E==22 years - combined age of the 2 elephants.

Check: 74 + 10 ==84 years - 10 years from now [it will be twice as old as 2 elephants combined]

              22 + 10 + 10 ==42 years - combined age of the 2 elephants.

t = e + 52     or     e = t-52

t+10 =  2(e+10)

t+ 10 = 2 ( t-52 + 10)

t = 94 y/o

Looks more like 9 to me.  you can't count that first one because 3 x 2 x 6 does not equal 18  

Problem: Let J and K be given points along a certain line.  There are exactly two other distint points on U and V on this line such that UJ/UK = 7 and VJ/VK = 7.  Let W be the midpoint of UV.  Calculate the ratio WJ/WK.

Answer: 28

The sum of all possible values of g(8) is 33.

This simplifies to (3*sqrt(2) + 5)/2.

$3/ 5 bar  * 100 bar = $ 60

= -2x^2 -6x - 36

 -2 ( x^2+3x )    - 36

= -2 ( x + 1.5)²  - 36  + 4.5

= -2(x +1.5)² - 31.5      <===== finish from here

Can't have a log of a negative number

9 1/2 *  8  = 76 in ²

Jason started with   3/5      Aaron had 2/5

3/5 * 3/4     :   3/5 * 1/4 + 2/ 5

9/20 :    3/20 + 2/5

9/20 : 11/20

9 : 11    is   J: A       A:J =   11:9

what i mean is that you type something like \(x=1\) to define a variable called "x". then typing in \(x+x\) would return 2. there should also be a command to delete all the variables (for example, \(\text{deleteVariables}\).)

MOMMY WAKE UP! The presents are here :D.

 
Merry Christmas

The area of overlap is 72 in^2.

The set of all a is [-3,2].

The answer is sqrt(5).

ABC,DEF = 165,414.

Answer #1 is not correct.

Best is to differentiate wrt x, turning the integral equation into a differential equation.

\(\displaystyle \frac{d}{dx}f(x)= -\{1.[f(x)]^{2} - 0.[f(0)]^{2} +\int^{x}_{0}\frac{\partial}{\partial x}[f(t)]^{2}dt , \)

\(\displaystyle \frac{d}{dx}f(x)= - [f(x)]^{2}.\)

Now integrate that.

There will be a constant of integration, so substitute back into the original equation to determine its value.