Math Is Fun Forum

sure it is!  just look where australia is on a map: in a corner at the end of the world!!

like a boy that has distanced himself from society cuz nobody wants him... sad really. 

hi yaz white sheep!!!!

1a) yes ur right!!!  hooray for you!!!!!

1b) awwww..... u missed it by one mistake.  use the properties of multiplication by negatives:
           -1(-1)=+1          -1(1)=-1           -(a-b)=-a+b
      try and finish the problem knowing the above.

2a) (a+b)(a-b)
      = a(a) + a(-b) + b(a) + b(-b)
      = a² -ab + ab -b²
      = a² - b²

2b) (a+b)(b+c)(c+d)
      = (ab + ac + b² + bc)(c+d)
      = abc + abd + ac² + acd + b²c + b²d + bc² + bcd

3a) i will let u solve this one.  use the following properties:
             a/a = 1      the a's cancel out   

            (a-b)/(a-b) = 1    same thing here

3b) 1 - z = 4z - 9
      step 1) tranfer all of the z's onto one side:
                 1 -z + z = 4z + z - 9
                 1 = 4z + z - 9
                 1 = 5z - 9
           (remember: if you want to move something to the other side you must do the opposite of it's         
            sign to both sides)

        step 2) move all of the constants to the other side and solve for z:
                 1 + 9 = 5z - 9 + 9
                 10 = 5z
                 (10)/5 = (5z)/5
                   2 = z

2d)    (z-1)/(z+3) = (2z+3)/(2z-3)
                 
        use the following property:
           a/b = c/d
           a(d) = c(b)
           
          (z-1)(2z-3) = (2z+3)(z+3)

try and work out the last one.  if you run into trouble you know where to get help!!

c yaz!

hahahah that's funny!!!

if u think that is too easy try this one:

∫(x²-29x+5)/[(x-4)²(x²+3)]dx

this tuff question was on my calculus2 final, and no, i did not solve it

hi yaz maths student!!!!

erm... typing and explaining how to find the angles will take me a good amount of time so i'm just gonna give u the answers.

1) E=90° X=60° E=30°
2a) A=150° E=15° F=15°
2b) A=90° E=45° F=45°
2c) A=30° F=75° E=75°

*sings*  It's fun to stay at the y-m-c-a.  It's fun to stay at the y-m-c-a.  They have everything...

hi there Nuu.... uu... ria!!!

just for reference reasons, here is the graph of logx :

as we can see, logx will go to ∞ when x gets larger and larger.  but ∞ is not a number and consequently we cannot evaluate log(∞).  however, if we know what happens to x as it approaches ∞, we can easily compute the limit of logt.  here is how:

           lim(x->∞ ) logx
                    as x -> ∞, we know t = x -> (∞) = ∞
           lim(t->∞ )logt = ∞

hope this helps!!!

hi yaz josh!!!

i'm going to take a wild guess and say this is problem #32... am i right??? 

f + g = (1+x)^(1/2) + (1-x)^(1/2)
      that's it!!!!  why??  we cannot add radicals.

f - g = (1+x)^(1/2) - (1-x)^(1/2)
       that's it!!!  why??  we cannot subtract radicals.

fg = (1+x)^(1/2) * (1-x)^(1/2)
       use the property (ab)^(1/2) = (a)^(1/2) * (b)^(1/2)
       [(1+x)(1-x)]^(1/2)
        (1-x^2)^(1/2)

f/g = (1+x)^(1/2) / (1-x)^(1/2)
        use the property: (a/b)^(1/2) = (a)^(1/2) / (b)^(1/2)
        [(1+x) / (1-x)]^(1/2)

that's it buddy!!

kylekatarn!!!!!!!!!!  it looks like u made a simple mistake in question c integration:

4∫sin2θdθ = -2cos2θ + c

what is the name of the formula u used for question d first part???  i've never seen it!!!!

it would be lonely w/ out me!!!!!!!!!!!

i would not wanna try it ur way kylekatarn

hi yaz!!!!!

4)
a)
         i. y'= 3x^2 - 12
         ii. here we wanna find out where the tangents are horizontal soooo... we set it equal to zero!
                    3x^2 - 12 = 0
                    3x^2 = 12
                      x^2 = 4
                      x = +2, -2

             so le'ts plug these guys back in2 the original equation:
                 (2)^3 - 12(2) + 7 = -9 --------> (2, -9)
                 (-2)^3 - 12(-2) + 7 = 23 ------> (-2, 23)

        iii. here we wanna find out when the graph is decreasing and increasing soo we choose points!
                let's try -1000000000000
                  3(-1000000000000)^2 - 12 = some positive number!!!  so increasing!!!

                let's try 0
                   3(0)^2 - 12 = some negative number!!!  so decreasing!!!

                let's try 1000000000000
                   3(1000000000000)^2 - 12 = some positive number!!!  so increasing!!!

               sooooo.... the graph is increasing from (-infinity, -2) U (2, infinity)
                                             and decreasing from (-2, 2)

b) i. since u know this one i'll go hed and skip it!

    ii. we know the total surface area = 2040
         so let's come up w/ an equation for it!!!!!

          we know the base and lit of the box are squares, so the surface area of a square is x^2 but we have two of these so let's go hed and multiply that by 2 so it looks like this 2x^2

           if we cut open the box and spread out the pieces we will end up w/ 4 rectangles and 2 squares.  so the surface area of a rectangle is x*(height) and we have four of these so let's multiply that by 4 so it looks like this 4xh.

now we have an equation!!!!  surface area = 2x^2 + 4xh = 2040

oh noes!!!!  but we wanna maximize the volume and only got the surface figured out... bummer.

we know the volume of the box = x^2 * h. 

now lets solve for h in the surface euqation:
510/x - 1/2 * x = h

and plug that in in2 our volume equation:
volume = x^2 (510/x - 1/2 * x)
            = 510x - 1/2 * x^3

iii) let's find the derivative of the volume!!!!
            =510x - x^3 * 1/2
            = 510 - 3/2 * x^2

now we wanna find out the maximum volume of the box, but how do we do that???  well, let's find out the largest point possible!!!!  set the derivate of the volume equal to zero

             510 - 3/2 x^2 = 0
                         x = +(340)^(1/2) or -(340)^(1/2)
                                  but we know the volume can't be negative!!!  silly!!!  so let's only keep positive answers!!
               
finally, we must prove this is the largest value, so in order to do this, we wanna c how the graph behaves.  let's find out where it deacreses and increases by choosing points

    lets try 5
                   510 - 3/2 * (5)^2 = some positive number!!!  increasing baby!!!!
    lets try 2000000000000000
                    510 - 3/2 * (2000000000000000)^2 = some negative number!!!  oh dear!!!

   finally, we know the graph is increasing from zero to (340)^(1/2) and decreasing from (340)^(1/2) all the way to infinity land!!!  so x = (340)^(1/2) must be the maximum point!!!

good day!