Help please!

Solve for x:

(x+2)(x-5)=6x+x^2-5

x^2+2x-15=0

2x^2-11x=-3

http://i4.photobucket.com/albums/y10.../flowchart.jpg

http://i4.photobucket.com/albums/y10.../perimeter.jpg

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- Dec 4th 2007, 04:22 PMpeachgalperimeter, 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

http://i4.photobucket.com/albums/y10.../flowchart.jpg

http://i4.photobucket.com/albums/y10.../perimeter.jpg - Dec 5th 2007, 04:38 AMearboth
Hello,

if a quadratic equation has the form:

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

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

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

The first equation isn't a quadratic equation!

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

http://i4.photobucket.com/albums/y10.../flowchart.jpg

I can't draw a flow chart here so I'll list the steps which has to be done:

1. Calculate $\displaystyle \angle(f)=62^\circ$. Sum of intern angles of a triangle is 180°.

2. Calculate $\displaystyle \angle(b)=86^\circ$. 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:

$\displaystyle \frac{x}{2.4}=\frac64$. Solve for x.

Use proportions:

$\displaystyle \frac x3=\frac{5+2}{5}$ . Solve for x.

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

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

http://i4.photobucket.com/albums/y10.../perimeter.jpg[/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 $\displaystyle 7 \cdot \sqrt{2}\ mm$

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

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: $\displaystyle 12^2 + 9^2 = 15^2$. Thus the perimeter of the parallelogramm is $\displaystyle 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:

$\displaystyle 8^2 - 4^2 = 48~\implies~v = 4 \cdot \sqrt{3}$

The perimeter of the figur is:

$\displaystyle p = 7'' + 6'' + 11'' + 8'' + 4 \cdot \sqrt{3}'' - 6''=26'' + 4 \cdot \sqrt{3}''$