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Stormswan

  • one year ago

can someone check my answers please?? C: I will give a medal!

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  1. Stormswan
    • one year ago
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    A segment with endpoints A (2, 1) and C (4, 7) is partitioned by a point B such that AB and BC form a 3:2 ratio. Find B. (3.2, 1.6) (3.2, 4.6) (3.8, 3.4) (4.4, 5.8)

  2. Stormswan
    • one year ago
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    I think the answer is B

  3. Stormswan
    • one year ago
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    Find the perimeter of a quadrilateral with vertices at C (−2, 1), D (2, 4), E (5, 0), and F (1, −3). Round your answer to the nearest hundredth when necessary. 12 units 16 units 20 units 24 units

  4. Stormswan
    • one year ago
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    and im not sure on this one... but i think its C?

  5. Stormswan
    • one year ago
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    that's for #2??

  6. anonymous
    • one year ago
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    for the first one the change in x is 2 and the change in y is 7 we can find the distance by r^2= x^2 + 7^2

  7. Stormswan
    • one year ago
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    okay, could you explain how to do this please? I really need to learn this because I have a bad grade in the class and I need to catch up.

  8. Stormswan
    • one year ago
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    you know what? i dont need help with those anymore... but i need help with a few others, if anyone is willing to help??

  9. Stormswan
    • one year ago
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    A cylinder has a radius of 6 feet and a height of 9 feet. Which of the following can be used to calculate the volume of a cone that fits exactly inside the cylinder? 1 over 3(3.14)(92)(6) (3.14)(92)(6) 1 over 3(3.14)(62)(9) (3.14)(62)(9)

  10. Stormswan
    • one year ago
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    this is the question.

  11. freckles
    • one year ago
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    |dw:1437893489289:dw|

  12. freckles
    • one year ago
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    my drawing got cut off

  13. Stormswan
    • one year ago
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    aw, okay. what did it say??

  14. freckles
    • one year ago
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    basically I was saying you can use the slope of that line along with AB/BC=3/2 to solve for B \[\frac{\sqrt{(x-2)^2+(y-1)^2}}{\sqrt{(x-4)^2+(y-7)^2}}=\frac{3}{2} \\ \text{ and use } \frac{y-1}{x-2}=\frac{y-7}{x-4}=\frac{7-1}{4-2}\] \[\sqrt{\frac{(x-2)^2+(y-1)^2}{(x-4)^2+(y-7)^2}}=\frac{3}{2} \\ \text{ \square both sides } \\ \frac{(x-2)^2+(y-1)^2}{(x-4)^2+(y-7)^2}=\frac{9}{4} \\ \text{ use the slope thing } y-1=\frac{6}{2}(x-2) \text{ which simplifying a bit gives }\\ y-1=3(x-2) \\ \text{ also we have from the slope thing that } y-7=3(x-4) \\ \\ \text{ so now we have the equation } \\ \frac{(x-2)^2+9(x-2)^2}{(x-4)^2+9(x-4)^2} =\frac{9}{4} \\ \text{ combine like terms on \top and bottom } \\ \frac{10(x-2)^2}{10(x-4)^2}=\frac{9}{4} \\ \text{ remember } \frac{10}{10}=1 \\ \frac{(x-2)^2}{(x-4)^2}=\frac{9}{4} \\ (\frac{x-2}{x-4})^2=\frac{9}{4} \\ \text{ this should be pretty easy \to solve from here }\]

  15. freckles
    • one year ago
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    once you solve that equation for x you can use y=3(x-2)+1 to find y by pluggin in that x

  16. freckles
    • one year ago
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    for the perimeter question I would use the distance formula 4 times once for each side and then add up all those side measurements

  17. Stormswan
    • one year ago
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    is C right?

  18. freckles
    • one year ago
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    for which one?

  19. Stormswan
    • one year ago
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    ummm im talking about the question i just recently asked... not the ones before.

  20. freckles
    • one year ago
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    oh the volume of a cone is pi*r^2*h/3 you are given h is 9 and r is 6 just plug them in

  21. Stormswan
    • one year ago
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    so it's D.

  22. freckles
    • one year ago
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    it is none of them have no idea where they get 62 or 92

  23. freckles
    • one year ago
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    perhaps they meant 6^2 or 9^2 or something but as the choices stand none of them are correct

  24. Stormswan
    • one year ago
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    what. this is confusing. this is a practice worksheet.. so it doesnt actually count as a part of my grade.. but i'm trying to find out how to do these so that I can get a good grade on the real assignments..

  25. Stormswan
    • one year ago
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    okay.... then how about another question....

  26. freckles
    • one year ago
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    so it really is written as 62 or 92?

  27. Stormswan
    • one year ago
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    yeah.

  28. freckles
    • one year ago
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    not \[6^2 \text{ or } 9^2?\]

  29. Stormswan
    • one year ago
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    Pyramid A is a square pyramid with a base side length of 7 inches and a height of 6 inches. Pyramid B has a volume of 3,136 in3. How many times bigger is the volume of pyramid B than pyramid A? 15 32 54 75 So on this i think the answer is.... B

  30. ganeshie8
    • one year ago
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    @Stormswan it would be less confusing if you limit to one question per post

  31. Stormswan
    • one year ago
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    oh.

  32. freckles
    • one year ago
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    we can do the question before it just won't exactly match any of your choices

  33. Stormswan
    • one year ago
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    after this question i will make a new post deal? :D

  34. Stormswan
    • one year ago
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    ummm @freckles no, it's okay. let's try the next one. i think i got that one right :)

  35. freckles
    • one year ago
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    but choice D didn't have a division by 3 in it

  36. Stormswan
    • one year ago
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    idk xD i'm just as confused as you are. but i'll just stick with that answer. Can you check if my next question is right? :)

  37. freckles
    • one year ago
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    I didn't do it but if you found it by calculating Volume of B/Volume of A where volume A.. equals.. \[\text{ volume of } A=\frac{ \text{ base length } \cdot \text{ base width } \cdot \text{height }}{3}\] then you did it correct

  38. freckles
    • one year ago
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    like you know base length=base width since the base is a square

  39. freckles
    • one year ago
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    so V of A=(7*7*6)/3 compute this then do volume of B divided by volume of A

  40. Stormswan
    • one year ago
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    okay, good! thanks :)

  41. freckles
    • one year ago
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    I assume good means what I just asked you to do confirmed your answer of 32

  42. Stormswan
    • one year ago
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    i'll open a new post with another question, if you dont mind helping @freckles but yeah, I figured it out :)

  43. freckles
    • one year ago
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    32 is correct because 3136/98 is 32 where Volume of B =3136 and Volume of A=98

  44. freckles
    • one year ago
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    alright I might come look at it

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