All Answers 358093 Answers

Let the No. of Cards with Raphael = x

      the No. of Cards with James = y,

According to Question, x + y = 480 ----- (1)

Now No. of Cards given by Raphael to James = x/4

Cards with Raphael = 3x/4 ------- (2)

and No. of Present cards with James = [y + 3x/4] --- (3)

Now James given 1/3 of his total cards,

               1/3[y + x/4] = y/3 + x/12 = (4y + x)/12 ----- (4)

New position of Cards left with James

       = [y + x/4] - [(4y + x)/12]
       = y + x/4 - 4y/12 - x/12

       = (12y + 3x - 4y - x)/12 = (2x + 8y)/12 ------ (5)

and Raphael will left with

3x/4 + [(4y + x)/12] = 3x/4 + (4y+x)/12 = (9x + 4y + x)/12

= (10x + 4y)/12 ----- (6)

equating (5) and (6)

2x + 8y = 10x + 4y 

or 8x = 4y => y = 2x ------ (7)

put this value y in (1)

x + 2x = 480

3x = 480

x = 160 cards

y = 320 cards 

James had 320 cards in the beginning 

P(6) = 122.

The area is 8*pi - 4, so a + b = 8 + 4 = 12.

The minimum possible value of x^2 + y^2 is 14.

An example quadratic is x^2 - 6x + 8/25.

The smallest possible value of a is 7269, as in 7269/14538 = 1/2.

The probability is 5/24.

The area of PQRS is 44.

There are 7! arrangements where the two O's are togehter, so by complementary counting, the answer is (8! - 7!)/2 = 17640.

The probability is 1/66.

The area is 6*sqrt(3).

I

wwkk...

There are 10 such non-congruent triangles.

There are 33 possible values of the third side.

They are all 3-digit numbers, so I'm going to put three lines to represent the three digits in the number.

___ ___ ___

The first digit has five possible options: 5, 6, 7, 8, 9

The second digit has two possible options: 3, 4

The third digit uses the remaining number from the second space: either 3 or 4;;; 1 option

_5_  _2_  _1_

We multiply these together to get 5 * 2 * 1 which is equal to 10 numbers

(if you want, you can list them out as well. they are 534  543  634  643  734  743  834  843  934  943)

- h

First, we can prove that in order to make the nonzero difference between two perfect squares as small as possible, their square roots must differ by 1. We can prove this as follows:

Let the two squares be \(a^2\) and \(b^2\). Without loss of generality, let us assume \(b \geq a, \) so we can write \(b \) as \(a+k.\) Now, to compute the difference, we get \(b^2-a^2=(a+k)^2-a^2=a^2+2ak+k^2-a^2=2ak+k^2.\) Obviously, in order to minimize the nonzero difference, \(k \neq 0, \) and \(k \) must be as small as possible.

Because we want \(2ak+k^2 \) to be 1, we can factor this as \(k(2a+k)\) to see that either \(k=2a+k=-1,\) or \(k=2a+k=1.\) 

If  \(k=2a+k=-1,\)  \(2a-1=-1,\) so \(a=0,\) which is a contradiction.

If \(k=2a+k=1,\) \(2a+1=1,\) so \(a=0,\) which is a contradiction.

Thus, two nonzero squares \(a^2\) and \(b^2\) cannot differ by 1.

There are four prime numbers (2, 3, 5, 7) in the first 10 whole numbers.

The probability of rolling a prime is  4/10  and the probability of not rolling a prime is  6/10.

The probability that exactly three of the dice show a prime number is:

                      ₄C₃ · (4/10)³ · (6/10)¹

If the legs have values  x  and  y     its area is  ½·x·y     and the sum of its legs is  x + y

The area is equal to three times the sum of its legs:     ½·x·y  =  3(x + y)

Solving:                                                                                xy  =  6(x + y)

                                                                                            xy  =  6x + 6y

                                                                                      xy - 6y  =  6x

                                                                                     y(x - 6)  =  6x

                                                                                              y  =  6x / (x - 6)

Substituting values for x to get integral values for y gives:  

     (7, 42)     (8, 24)     (9, 18)     (10, 15)     (12, 12)     (15, 10)     (18, 9)     (24, 8)     (42, 7)

Removing the duplicates:     (7, 42)     (8, 24)     (9, 18)     (10, 15)     (12, 12) 

Finding the areas of these triangles and summing their answers will give the final answer.

m/w ==19/7...................................(1)
[m+a]/w ==3/1..............................(2)
[m+a]/[w - b]==7/2........................(3)
[m+a +7.5] / [w - b -2.5]==45/11...(4), solve for a, b, m, w

Use eliminations and substitutions to get:
a==10, b==5, m==95, w=35

a^3 - b^3 ==10^3 - 5^3==1,000 - 125 =875

40.5 / 2.25 =  18 pens

What real values of x are not in the domain of f(x) = 1/(abs(x^2 + 3x - 4) + abs(x^2 + 9x - 10))?

\(f(x) =\frac{ 1}{|x^2+3x-4|+ |x^2 + 9x - 10|}\\ f(x) =\frac{ 1}{|(x+4)(x-1)|+ |(x+10)(x-1)|}\\ x\ne 1 \)

What is f(f(f(17)))?

Hello Guest!

\(f(17)=17+13=30\\ f(f(17))=f(30)=30+13=43\\ {\color{blue}f(f(f(17)))=}f(43)=43+13=\color{blue}56\)

  !

What is the value of 1 - 2x + 3x^2 - 4x + 5x - 6x + 7x^2 - 8x + 9 in terms of x? Express your answer in the form ax^2 + bx + c where a, b, and c are numbers.

Hello Guest!

\(f(x)=ax^2+bx+c\)

\(f(x)=10x^2-15x+10\)

\(a=10,\ b=-15,\ c=10\)

If   f (x) = 0,  then:

\(x^2-1.5x+1=0\\ x=0.75\pm \sqrt{0.5625-1}\)

\(x\in \{0.75-0.661438i,\ 0.75+0.661438i\}\)

  !