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Math Help - almost done with angles and circles

  1. #16
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Outragexl10 View Post
    How to find x and y from those two...for some reason its not coming to mind...
    ok, so you have problems solving the simultaneous equations? am i to assume that you got the process then? as in, you understand how i formed the two equations

    There are two mainstream methods to solve systems of equations (there are more than two ways, but these two get used more often), Elimination, and Substitution.

    I will do Elimination, it is your responsibility to look up the rest when you have time.

    The objective of Elimination, as its name suggests, is to eliminate on variable by adding (or subtracting) one equation from the other. Of course for this to work, the variable you want to eliminate has to have the same coefficient in both equations. To get that to happen, you may have to multiply one (or both equations) by something. Let's see how this works.

    2x + 3y = 185 ................(1)
    3x + y = 120 ..................(2)

    Let's eliminate y. To do this i must multiply equation two by 3 (or -3 if you prefer). we get:

    2x + 3y = 185 ....................(1)
    9x + 3y = 360 .....................(3) = (2)*3

    Now we have 3y in both equations, we can therefore, subtract one from the other, and the y's will go away. we obtain:

    7x = 175 .................(3) - (1)

    Now we solve for x, we get

    \boxed { x = 25 }

    Now we plug in that value for x into either of the original equations to find y

    3x + y = 120

    \Rightarrow y = 120 - 3x

    \Rightarrow y = 120 - 3(25)

    \Rightarrow \boxed { y = 45 }
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  2. #17
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    Quote Originally Posted by Jhevon View Post
    ok, so you have problems solving the simultaneous equations? am i to assume that you got the process then? as in, you understand how i formed the two equations

    There are two mainstream methods to solve systems of equations (there are more than two ways, but these two get used more often), Elimination, and Substitution.

    I will do Elimination, it is your responsibility to look up the rest when you have time.

    The objective of Elimination, as its name suggests, is to eliminate on variable by adding (or subtracting) one equation from the other. Of course for this to work, the variable you want to eliminate has to have the same coefficient in both equations. To get that to happen, you may have to multiply one (or both equations) by something. Let's see how this works.

    2x + 3y = 185 ................(1)
    3x + y = 120 ..................(2)

    Let's eliminate y. To do this i must multiply equation two by 3 (or -3 if you prefer). we get:

    2x + 3y = 185 ....................(1)
    9x + 3y = 360 .....................(3) = (2)*3

    Now we have 3y in both equations, we can therefore, subtract one from the other, and the y's will go away. we obtain:

    7x = 175 .................(3) - (1)

    Now we solve for x, we get

    \boxed { x = 25 }

    Now we plug in that value for x into either of the original equations to find y

    3x + y = 120

    \Rightarrow y = 120 - 3x

    \Rightarrow y = 120 - 3(25)

    \Rightarrow \boxed { y = 45 }
    I see now, i haven't used this for 2 years and is why i couldn't recall this, but i see how you do it now... thanks so much

    This is why i also prefer not to get straight answers, because it helps me learn... thanks again for explaining it to me
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  3. #18
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    Quote Originally Posted by Outragexl10 View Post
    ...Write the equation of the circle with center (-2, -3) and tangent to the line 2x + 5y = 10.
    (x – 2)2+(y – 3)2 = 29
    (x – 2)2+(y – 3)2 = 5
    (x + 2)2+(y + 3)2 = 29
    (x + 2)2+(y + 3)2 = 5
    none of these
    ...
    Hello,

    according to the text the equation of the circle is

    (x+2)^2+(y+3)^2=r^2

    If the line is a tangent to the circle the distance of the centre to the line must be as large as the radius.

    The distance of a point P(x1, y1) to a line Ax + By + c = 0 is calculated by:

    d=\frac{Ax_1 + By_1 + C}{\sqrt{A^2+B^2}}

    d=\left| \frac{2 \cdot (-2) + 5 \cdot (-3)-10}{\sqrt{2^2+5^2}} \right| = \frac{29}{\sqrt{29}}=\sqrt{29}

    Thus circle #3 is the right one
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  4. #19
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    Quote Originally Posted by Outragexl10 View Post
    ...Describe the locus of points that are 9 cm from point B.
    a segment 18 cm long
    a circle with a radius of 9 cm
    a sphere with a radius of 9 cm
    a circle with an area of 81 sq cm
    ...
    Hello,

    Definiton: The circle (line) is the set of points (in a plane) which have the same distance to a fixed point, the centre of the circle.

    Definition: The sphere (surface) is the set of points (in a space) which have the same distance to a fixed point, the centre of the sphere.

    As you can see there are two answers possible.
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  5. #20
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    Quote Originally Posted by Outragexl10 View Post
    ...Find the surface area of the frustum. Show all work.
    ...
    Hello,

    the surface of a frustum consists of the top circle, the base circle and the curved surface. It is calculated by:

    A_{frustum}=\pi r_1^2 + \pi r_2^2 + \pi \cdot s \cdot (r_1+r_2) . Plug in the values you know:

    A_{frustum}=\pi \cdot 3^2 + \pi \cdot 6^2 + \pi \cdot 7 \cdot (3+6) = 108 \pi
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  6. #21
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    Quote Originally Posted by earboth View Post
    Hello,

    Definiton: The circle (line) is the set of points (in a plane) which have the same distance to a fixed point, the centre of the circle.

    Definition: The sphere (surface) is the set of points (in a space) which have the same distance to a fixed point, the centre of the sphere.

    As you can see there are two answers possible.
    Whereas I agree that, thus far, the problems have dealt merely with planar geometry the most general answer to the question must be considered to be the sphere with a radius of 9 cm. Since this is the "most correct" answer listed it should be the one chosen.

    -Dan
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  7. #22
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Outragexl10 View Post
    Given a cylinder with a surface area of 60 units squared and a height of 7 units. Find the volume of a sphere that has the same size radius as the given cylinder. Show all work.
    Note that, for a cylinder, the surface area, SA, is given by:

    SA = 2 ( \mbox { Area of the circular base } ) + ( \mbox { Height} \times \mbox {Circumference })

    \Rightarrow 60 \pi = 2 \pi r^2 + 2h \pi r

    \Rightarrow 2 \pi r^2 + 2h \pi r - 60 \pi = 0

    since h = 7 we have:

    2 \pi r^2 + 14 \pi r - 60 \pi = 0

    \Rightarrow r^2 + 7r - 30 = 0 ...........I divided through by 2 \pi

    This is a quadratic equation, you should be able to solve for r

    Now, the volume, V, of a sphere is given by:

    V = \frac {4}{3} \pi r^3

    use the value you got for r in the above formula and you will have your answer
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  8. #23
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    Hello again, Outragexl10!

    Now I see that you've solved many of them already.
    Oh well . . . maybe I can offer a different approach.


    5) .[6 points] . In circle V,\:RT is a diameter.



    a) Find m\angle T and m\angle R.

    b) If ST = 22, find the length of the radius to the nearest tenth.
    (a) Since \Delta RST is inscribed in a semicircle, \angle S = 90^o.

    Hence, \angle T and \angle R are complementary: . (x^2 + 30) + (5x + 24) \:=\:90

    . . which simplifies to:. . x^2 + 5x - 36 \:=\:0

    . . which factors: . (x - 4)(x + 9) \:=\:0

    . . and has the positive root: . x\,=\,4

    Therefore: . \begin{array}{ccccccc}\angle T & = & 5x + 24 & = & 5(4)+24 & = & 44^o \\ \angle R & = & x^2 + 30 & = & 4^2 + 30 & = & 46^o\end{array}


    (b) Let r = VR = VT. .Then: . RT = 2r

    In right triangle RST, we have: . \sin46^o \:=\:\frac{22}{2r}\quad\Rightarrow\quad r \:=\:\frac{11}{\sin46^o}

    Therefore: . r \;=\;15.2917995 \;\approx\;15.3




    7) .[4 points] . PQ is a common external tangent of circle R and circle S.
    The radius of circle S is 12 feet and the radius of circle R is 5 feet. . TU = 8 ft.
    Find the area of quadrilateral PQRS.


    Note that radii SP and RQ are perpendicular to tangent PQ.

    Note also that: ST = SP = 12,\;RU = RQ = 5
    . . Hence: . SR \:=\:12 + 8 + 5 \:=\:25

    Invert the quadrilateral.
    Code:
        S *
          |     *      25
        7 |           *
          |                 *
        V + - - - - - - - - - - - * R
          |           x           |
        5 |                       | 5
          |                       |
          * - - - - - - - - - - - *
          P                       Q

    Through R draw a line parallel to PQ, meeting SP at V.
    . . Let x = VR

    In right triangle SVR, we have: . x^2 + 7^2 \:=\:25^2

    . . Then: . x^2 + 49 \:=\:625\quad\Rightarrow\quad x^2 \:=\:576\quad\Rightarrow\quad x \:=\:24

    Quadrilateral PQRS is a trapezoid with parallel bases 5 and 12
    . . and height 24.

    Therefore, its area is: . a \;=\;\frac{24}{2}(5 + 12) \;=\;204 ft².

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  9. #24
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    In circle L,is a tangent andis a diameter. Find the measure of arc MN. Show all work.


    WORTH 4 points (Not yet scored by your instructor)



    i still have no clue about this one???
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  10. #25
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Outragexl10 View Post
    In circle L,is a tangent andis a diameter. Find the measure of arc MN. Show all work.


    WORTH 4 points (Not yet scored by your instructor)



    i still have no clue about this one???
    Soroban gave you this theorem earlier in this post:

    If two tangents, or a tangent and a secant, or two secants are drawn to a circle,
    . . the included angle is one-half the difference of the two intercepted arcs.
    i think that's what you need to use
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