All Answers 356020 Answers

Price of Workbook = $6

Price of Textbook = (3/5)(6) + 6 = $9.6

Let T = number of Textbooks

     W = number of Workbooks

      S = total number of textbooks and workbooks

               S = T + W --> equation (1)
    * 0.35S = T --> equation (2)

    * 6W = 108 + 9.6T --> equation (3)

 equation (2) in equation (1)

      S = T + W

        S = 0.35S + W

         0.65S = W --> equation (4)

Total sales:

        6W + 9.6T = (108 + 9.6T) + 9.6T

         6W = 108 + 9.6T

                    * T = 0.35 (equation (2))

          6W = 108 + 9.6 (0.35S)

          6W/6 = (108 + 3.36S)/6

            W = 18 + 0.56S --> equation (5)

   equation (4) = equation (5)

      0.65S = 18 + 0.56S

      0.65S - 0.56 S = 18

             0.09S/0.09 = 18/0.09

                  S = 200 books

                200 books was sold altogether

Adam = $x money

Firdaus = $y

(x - 9)/y = 2/7 -(1)

2y = 7(x - 9) = -(1)

    2y = 7x - 63 -(1)

 x/(y - 9) = 7/11

   11x = 7(y - 9)

   11x = 7y - 63 

2 (11x - 7y = -63)

7 (2y - 7x = -63)

  14y - 49x = -63 * 7

  22x - 14y = -63 * 2 

| - - - - - - - - - - - - - - - - - -
|       -27x = -567
|            x = 21
|            y = 42
|   x + y = $63
|
|
|

I think so. 

Here is my reasoning: \(x^2+y^2= (x+y)^2-2xy\), since we know x+y is an integer, and x^2+y^2 is also an integer, then -2xy must be an integer. 

maybe someone else can do this as well?

If we were to randomly arrange the cards such that the first 3 cards would be for the first person and so on, there would be 9! ways to do so.

Now, if we were to distribute the 3 red cards such that each person gets one, there are 6 ways that could happen. Then, we can distribute the remaining 6 cards using the same above method to get 6! ways.

That means the final probability is 6*6!/9!=1/84

:D

I plugged it into a calculator, because I personally can't solve it. Here is what I found:

\(a = {\sqrt{30}\over4} \)

\(b = {2\sqrt{30}\over15}\)

\(c = {\sqrt{15}\over\sqrt2}\)

\(d ={ \sqrt{30}\over5}\)

\(e ={ 2\sqrt{30}\over3}\)

It can't be. If b were 0, then ab would equal 0, not 1. 

Also, don't say that... Constructive Feedback. 

As Melody said, "Learning is a journey..."

2573 = 31*83

This is a composite number.  It has 4 factors.   1, 31, 83 and 2573

I discovered this by entering    

factor(2573) =       into the web2.0calc.   

that means that

a = 1

b = 0

c = 2

d = 1

e = 3

i'm pretty sure i did something wrong... i'm not that good at math anyway

How many points with integer coordinates are in the interior of the triangle?

Hello Guest!

\(m=-1\\ f(x)=-1(x-6)+0\\ f(x)=-x+6\)

\(n_p=1+2+3+4=10\)

10 points with integer coordinates are in the interior of the triangle.

  !

\( y= 2x^2 - 7x + 1\ ||\ \ y = 8x^2 - 25x + 13\). Give all points where they intersect.

Hello Guest!

\(f(x) = 2x^2 - 7x + 1\\ g(x)=8x^2 - 25x + 13\\f(x)=g(x)\\ f(x)-g(x)=-6x^2+18x-12=0\\ x^2-3x+2=0\\ x=\frac{3}{2}\pm \sqrt{\frac{9}{4}-2}\\ \color{blue}x\in\{1,2\}\)

\(\mathbb L=\{P_1(1,-4);P_2(2,-5)\}\)

  !

It is good that you have shared your answer but why not tell us what you think  f(-2)  and g(-2) are .....

Let A be the point (0,0)

Let B be the point (22,0)

the line equidistance from thes points is x=11,  the centre will line on this line

now AC=12 and BC=14   Find C using locus

\(x^2+y^2=144\qquad (x-22)^2+y^2=196\\ subtract\\ (x-22)^2-x^2=196-144\\ -44x+484=52\\ x=9.\bar{81}\\ x^2=96\frac{48}{121}\\ x^2+y^2=144\\ y^2=144-96\frac{48}{121}\\ y^2=47\frac{73}{121}\\ y^2=\frac{5760}{121}\\ y^2=\frac{2^7*3^2*5}{121}\\ y^2=\frac{2^6*3^2*5*2}{121}\\ y=\frac{24\sqrt{10}}{11}\\ C(\frac{108}{11}, \frac{24\sqrt{10}}{11})\)

The centre will be the point on the line x=11 that is equidistant to  A(0,0), B(22,0)  and  C  let the point be Q(11,y)

I think I will let you take over from there.

Here is the pic

LaTex:

x^2+y^2=144\qquad (x-22)^2+y^2=196\\
subtract\\
(x-22)^2-x^2=196-144\\
-44x+484=52\\
x=9.\bar{81}\\
x^2=96\frac{48}{121}\\
x^2+y^2=144\\
y^2=144-96\frac{48}{121}\\
y^2=47\frac{73}{121}\\
y^2=\frac{5760}{121}\\
y^2=\frac{2^7*3^2*5}{121}\\
y^2=\frac{2^6*3^2*5*2}{121}\\
y=\frac{24\sqrt{10}}{11}\\
C(\frac{108}{11}, \frac{24\sqrt{10}}{11})

Prime Factors of 2573 = 2573 is a prime number and cannot be factorize

The only factors of 2573 are 1 and 2573

2573 divided by 1, 2573, gives no remainder. They are integers and prime numbers of 2573, they are also called composite number.

ahh, now i see/understand what you did (i was the first answer guest) 

nice job

for the second question shouldn't it be <= 2?? 

after i typed it in wolfram alpha, it gave t<=2 for all real numbers 

hmm

\(a^2+2ab+b^2=6ab\\ (a+b)^2=6ab\\ a+b=\sqrt{6ab}\\ a^2-2ab+b^2=2ab\\ (a-b)^2=2ab\\ a-b=\sqrt{2ab}\)

So

 \(\frac{a^2-b^2}{ab}\\ =\frac{\sqrt{6ab}\cdot\sqrt{2ab}}{ab}\\ =\sqrt{12}\cdot\frac{ab}{ab}\\ =2\sqrt{3}\)

(notice that a and b are both positive so the answer must be positive)

Also, this question was asked quite a while ago:

simplifying a/b-b/a gives a^2/ab - b^2/ab = \({a^2-b^2}\over ab\) which is the first equation divided by ab...except the subtraction sign... anyone else want to solve this? right now i'm trying to find a way to solve this..

also according to wolfram alpha a/b-b/a = \(\pm 2\sqrt{3}\)

Domain:  set of all values for the independent variable (x).

Range:  set of all values for the dependent variable (y).

f(x)  =  | 1 - x |     Domain = all real numbers

                           Range = all real number >= 0.

f(t)  =  sqrt( 2 - t )     Domain = all real numbers >= 2

                                Range = all real numbers >= 0

You can do this on the web2.0calc!

just enter

factor(2023) and hit enter.

factor(2023)= = 7*17^2

How about presenting your question properly, without the unrendered LaTex.

Sorry, for the typo. It shouls read:

2023 ==7  x  17  x  17 [Not 19 !]

Why did you say 2023 is a prime number?

2023 / 7 ==289

289 / 19 ==19 

19 / 19 == 1

So, the prime factors of 2023 are: 7  x  19  x  19

This will happen when 

x^2-5x+2 = 0      ( so long as x^2-4x+2 isn't 0 as well. )

use the quadratic formula

I don't know either ......

You have made 2 mian logic errors here.

1) You have said that all the numbers can be used more than once, which they can't.

2) You also said that order chosen counts, you already understand that you made that mistake.  

Learning is a journey :)