1. perimeter, flowcharts, and equations

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

x^2+2x-15=0

2x^2-11x=-3

2. Originally Posted by peachgal

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