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Wednesday, May 18, 2011

hw #8-3 Practice Packet

If you do EVERY PROBLEM, GREAT!!

REQUIRED:
Do all problems 8-1 thru 8-4
Do 8-5 #1,3,5 only (no surface area of cones on test)
Do 8-6 #1-4, 7, 11
Do 8-7 #1-7 Odds
Do 8-8 #1-3

Please spread the work to your math-buddies that the assignment was reduced... thanks!

42 comments:

  1. This question was asked earlier...

    Anonymous said...

    im confused on how to draw a net and I already watched the video tutors...

    Answer:
    In the case of a rectangular prism or square pyramid, a net is simply a drawing of the unfolded panels of the box.

    The case of a cylinder is similar... instead of panels you can actually envision the "unfurling" of the cylinder wall. We have demonstrated this in class several times and in our lab the other day.

    If you can ask a more specific question perhaps I can help you more.

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  2. is the surface area of cones going to be on the test?? rj

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  3. rj doesnt understand 8-7 prob 7. i tried everything but i keep getting different answers. help now!!!!

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  4. Hi mr, chamberlain I seem to be a little confused on lateral area?? I understand S.A and volume

    -sam

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  5. For RJ

    You just reverse the order of operations so dividing 88 by 8 and 4.6 to get the answer

    PS: for the other problems do not forget pie.

    Marc

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  6. to rj i got a very strange number on 8-7 #7 but it is just algebra I think 4.6(8)(x)=88
    I multiplied and got 36.8 then divided that from 88 and got 2.391304348??

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  7. For problem 8-7 #7

    I got 7.2m is that correct??

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  8. From the last comment how did you get that answer?

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  9. How do you do the set-up the problem for 8-5 #7? I had the surface area of a cone is= pie (r2)= pie (r) (slant)

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  10. are you going to post an answer key...?!

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  11. for problem 8-7 #7 i got 2.4 (2.39)

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  12. is the sphere stuff going to be on the test

    luke

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  13. Sphere stuff on the test... you will have the formulas... so, e-z-schmeezy, right?

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  14. Lateral area is a very simple concept... From a geometric perspective, "lateral" is synonymous with "sidewall"... So the lateral area of a prism is the area of the sidewalls - you don't include area of the base or the top. For a pyramid or cone, it is the area of the sidewall and you don't include the area of the base (there is no top). Ca-peesh??

    The book gives you a description and formula on pg 375, but who needs a stinkin' formula when you understand and can construct it?

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  15. 8-7 #7 was described pretty well above, but I'll give it another try as well.

    You are given the volume of a rectangular pyramid, with l=8m, w=4.6m, V=88m^3

    We know the formula for Volume is V=(1/3)lwh, yes?

    I'll demonstrate two options:

    Option 1:

    V=(1/3)lwh so, let's substitute what we know
    88=(1/3)(8)(4.6)h let's multiply (simplify)
    88=12.27h
    h=88/12.27
    h=7.17 (rounded)

    Option 2:
    An industrious algebra student might choose to create a formula for height from the original V=(1/3)lwh formula.

    If you multiply both sides by 3 and divide by lw, you get h=3V/lw... what fun!

    So, just substitute and solve...

    h=3(88)/[(8)(4.6)]
    h=264/36.8
    h=7.17 (rounded)

    I prefer option 2, but both methods work JUST FINE!

    Ca-peesh??

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  16. are you posting the answer key??

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  17. NO SURFACE AREA OF CONES ON THE TEST, SO YOU CAN SKIP 8-5 #7,9, OK???

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  18. I WILL POST A PARTIAL ANSWER KEY... LET ME KNOW IF ANY PARTICULAR PROBLEM IS GIVING YOU TROUBLE.

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  19. Practice 8-4
    #1) 2(5*8)=80 (front&back)
    2(12*8)=192(right&left)
    2(5*12)=120(top&base)
    Surface Area (S.A.)= 80+192+120=392cm^2
    #2) 2[1/2(6*8)]=48 (area of two triangles)
    you'll need to use the Pythagorean theorem to find the length of the hypotenuse (which is 10m)
    10*15=150 (side right panel)
    8*15=120 (back left panel)
    6*15=90 (base panel)
    S.A.= 48+150+120+90=408m^2

    #3) A cube!!!
    6 faces, all 14*14
    so, 6(14^2)=6(196)=1176in^2

    #5) Area of base=Pi*r^2=3.14*(3.5)^2
    =3.14*12.25=38.47in^2
    Length of sidewall = Circumference of Circle
    Circumference=Pi*d=3.14*7=21.98
    Area of sidewall (L.A.)=lw=21.98*28=615.44in^2
    S.A.=Twice the Base plus the L.A.
    S.A.=2(38.47)+615.44=76.94+615.44=692.38in^2

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  20. Practice #8-5

    #1
    Area of Base = 6^2 = 36cm^2
    Area of Triangular Panel=(1/2)6*11=33cm^2
    4 triangles= 4*33=132cm^2
    S.A.=36+132=168cm^2


    #3
    Area of Base = 7.5^2 = 56.25in^2
    Area of Triangular Panel=(1/2)7.5*10=37.5in^2
    4 triangles= 4*37.5=150in^2
    S.A.=56.25+150=206.25in^2

    #5
    Area of Base = 1.5^2 = 2.25cm^2
    Area of Triangular Panel=(1/2)1.5*3.4=2.55cm^2
    4 triangles= 4*2.55=10.2cm^2
    S.A.=2.25+2.55=4.8cm^2

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  21. ahhhh mr. c for number 5... its the lateral area not the surface area i got 10.2 is it correct? evan

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  22. #8-6

    #1
    V=lwh
    V=24*7*8=1344cm^3

    #2
    27*27*27=27^3=19,683in^3

    #3
    Two ways to look at this:
    It is a triangular prism with a base area of (1/2)12*16=96yd^2 and 24 yards "high"
    so, 96*24=2,304yd^3

    OR

    The triangular prism is HALF THE VOLUME of the rectangular prism that measures 12*16*24. It's volume would be 12*16*24=4608yd^3, and half of 4608=2,304yd^3.

    A very nice cross-check!

    #4 is a little tricky (not too bad). You need the Pythagorean theorem (again) to find the length of the missing side of the triangle. In this case a^2=c^2-b^2 or a^2=17^2-15^2

    a^2=289-225
    a^2=64
    a=8

    Now,
    Area of one triangle is (1/2)8*15=60in^2
    Let's call that the base.
    The volume is the area of the base * the height
    60*11=660in^3

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  23. 8-6 #7

    Area of the base=pi*r^2=3.14*7^2=3.14*49=153.86ft^2

    Volume is Area of the base * height

    153.86*42=6,462.12ft^3

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  24. i know 8-6 number 4 is really easy but i cant get it i just cant evan

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  25. For 8-5 #5... Evan is correct... as you can see, since I showed my work(!!!!), I did specify that the area of the side panels was 10.2cm^2.

    Is anyone else paying attention? Any problems?

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  26. I found a typo in 8-4 #5 you said that the diameter is 7 in even though on the problem that is the radius. There fore you put the radius as 3.5 not 7 in your work. :)

    - Sam Rollenhagen *cough* *cough* extra credit

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  27. For #8-7, you should do both a pyramid and a cone.

    #1
    V=(1/3)lwh
    V=(1/3)6*6*5
    V=(1/3)180
    V=60cm^3

    #2
    V=(1/3)(pi)(r^2)(h)
    V=(1/3)(3.14)(7.4^2)(15.6)
    V=(1/3)(3.14)(54.76)(15.6)
    What I like to do here, instead of multiplying by .33, is to multiply the other numbers and divide by three. As you know, dividing by 3 is equivalent to multiplying by 1/3, and it is slightly more accurate than multiplying by .33 (you lose precision because of the repeating decimal).
    V=[(3.14)(54.76)(15.6)]/3
    V=2,682.36/3
    V=894.12 m^3 (rounded)

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  28. anyone on 8-6 number 4 its a triangular prism why cant i do it evan

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  29. Practice 8-4

    #1)
    2(5*8)=80 (front&back)
    2(12*8)=192(right&left)
    2(5*12)=120(top&base)
    Surface Area (S.A.)= 80+192+120=392cm^2

    #2)
    2[1/2(6*8)]=48 (area of two triangles)
    you'll need to use the Pythagorean theorem to find the length of the hypotenuse (which is 10m)
    10*15=150 (side right panel)
    8*15=120 (back left panel)
    6*15=90 (base panel)
    S.A.= 48+150+120+90=408m^2

    #3) A cube!!!
    6 faces, all 14*14
    so, 6(14^2)=6(196)=1176in^2


    #5)
    Area of base=Pi*r^2=3.14*(7)^2=3.14*49=153.86in^2
    Length of sidewall = Circumference of Circle
    Circumference=Pi*d=3.14*14=43.96
    Area of sidewall (L.A.)=lw=43.96*28=1,230.88in^2
    S.A.=Twice the Base plus the L.A.
    S.A.=2(153.86)+1,230.88=307.72+1,230.88
    S.A.=1,538.6in^2

    Did I get it right, Sam?

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  30. Evan, did you see my explanation of 8-6 #4?

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  31. just did thanks

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  32. wait nooooooo wrong one no wonder i was soooooooo off donde esta where is it
    ^
    ll
    ll
    ll where is it up there?

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  33. nevermind i am a genius and a big boy and i figured it out ALL BY MYSELF! small victory

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  34. 8-8
    #1
    Circle with radius 10cm

    S.A.=4(pi)r^2
    =4(3.14)(10^2)
    =1,256cm^2

    V=(4/3)(pi)r^3
    Hint: Multiplying by 4/3 is equivalent to multiply by 4 and then dividing by 3

    So you could do this (more accurate):
    V=[(4)(3.14)(10^3)]/3
    V=[12,560/3]
    V=4,186.66cm^3

    OR this:
    V=(1.33)(3.14)(10^3)
    V=4,176.2cm^3

    See how much difference 4/3 vs. 1.33 can make??

    I will accept either method/solution on the test.

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  35. it's a good thing this blog is anonymous... i don't wanna know who "donde esta" is!!

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  36. i aced that soooooo easy cuz im a big boy now and id did it ALL BY MYSELF small victory evan

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  37. donde esta means WHERE IS IT in SPANISH ERROR on comment number 34

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  38. mr c 60x11 isnt 660 (in 8-6 #4) -amanda !

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  39. catch my mistake sam haha it is 660 whoops.

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  40. minus 10 pts for Amanda! is that ok with everyone?

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  41. 8-5 #5 confuses me

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