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Math Help - perimeter, flowcharts, and equations

  1. #1
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    Exclamation perimeter, flowcharts, and equations

    Help please!

    Solve for x:
    (x+2)(x-5)=6x+x^2-5

    x^2+2x-15=0

    2x^2-11x=-3



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  2. #2
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    Quote Originally Posted by peachgal View Post
    Help please!

    Solve for x:
    (x+2)(x-5)=6x+x^2-5

    x^2+2x-15=0

    2x^2-11x=-3
    Hello,

    if a quadratic equation has the form:

    ax^2+bx+c=0 then the solutions are: x_{1,2}=-\frac b{2a} \pm \frac1{2a} \cdot \sqrt{b^2-4ac}

    Transcribe all the equations to the form given above and use the formula. As an example:

    2x^2-11x=-3~\iff~2x^2-11x+3=0~\implies~x_{1,2}=-\frac{-11}{2 \cdot 2} \pm \frac1{2 \cdot 2} \cdot \sqrt{121-4 \cdot 2 \cdot 3} ~\implies~ x_{1,2}=\frac{11}4 \pm \frac14 \cdot \sqrt{97}

    The first equation isn't a quadratic equation!

    = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =



    I can't draw a flow chart here so I'll list the steps which has to be done:
    1. Calculate \angle(f)=62^\circ. Sum of intern angles of a triangle is 180.
    2. Calculate \angle(b)=86^\circ. Sum of intern angles of a triangle is 180.
    3. Corresponding intern angles are equal thus the triangles are similar.
    4. Proportion of corresponding sides is a constant:
    \frac{x}{2.4}=\frac64. Solve for x.

    Use proportions:
    \frac x3=\frac{5+2}{5} . Solve for x.

    You are dealing with an isoscele right triangle thus \theta = 45^\circ

    = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

    [/QUOTE]

    to a) Since the angle at the base of the right triangle is 45 you are dealing with an isoscele triangle. Therefore the other leg of the right angle is 7 mm too and the length of the hypotenuse is 7 \cdot \sqrt{2}\ mm

    Therefore the perimeter is p=7 + 7 + 7 \cdot \sqrt{2} = 7(2+\sqrt{2})

    to b) The length of top and base line together is 30 m. That is only possible if the base line is translated by 6 m to the right. The parallelogramm consists of 2 right triangles with the legs 12 m and 9 m and a rectangle with l = 9m and w = 6 m. The hypotenuse of the right triangles is: 12^2 + 9^2 = 15^2. Thus the perimeter of the parallelogramm is p = 2 \cdot 18 + 2 \cdot 15 = 66

    to c) Split the figur into a rectangle with l = 7'' and w = 6'' and a right triangle. One leg of the right angle is 11'' - 7'' = 4''.
    Since one angle is 60 the angle at the top of the right triangle is 30. Therefore the hypotenuse is twice as long as the opposite side of the 30-angle. h = 8''
    The vertical leg v of the right triangle is:
    8^2 - 4^2 = 48~\implies~v = 4 \cdot \sqrt{3}
    The perimeter of the figur is:
    p = 7'' + 6'' + 11'' + 8'' + 4 \cdot \sqrt{3}'' - 6''=26'' + 4 \cdot \sqrt{3}''
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