By the Segment Addition Postulate,

$\displaystyle AB + BC + CD + DE = AE$

$\displaystyle x +( x + 3) +( 2x) +( x-6) = 40$

$\displaystyle 5x-3 = 40$

$\displaystyle 5x-3 +3 = 40 + 3$

$\displaystyle 5x = 43$

$\displaystyle 5x \div 5 = 43 \div 5$

$\displaystyle x = 8.6$

$\displaystyle AB = x = 8.6$

The length of a line segment can be determined using the distance formula:

$\displaystyle D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$\displaystyle D=\sqrt{(3-6)^2+(5-1)^2}$

$\displaystyle D=\sqrt{-3^2+4^2}$

$\displaystyle D=\sqrt{9+16}=\sqrt{25}=5$

To find the length of this line, you can subtract $\displaystyle 10-1$ to get $\displaystyle 9$. Since the y-coordinates are the same, you don't have to take any vertical direction into account. Therefore, you only look at the x-coordinates!

We can use the distance formula:

$\displaystyle D=\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2}$

$\displaystyle =\sqrt{(-3-5)^2+(5-4)^2}$

$\displaystyle =\sqrt{64+1}$

$\displaystyle =\sqrt{65}\approx 8.06$

The radius is the distance from the center of the circle to anypoint on the circle. So we can use the distance formula in order to find the radius of the circle:

$\displaystyle D=\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2}$

$\displaystyle =\sqrt{(4-3)^2+(4-8)^2}$

$\displaystyle =\sqrt{1+16}$

$\displaystyle =\sqrt{17}\approx 4.12$

A rectangle has two diagonals with the same length. So we should find the length of $\displaystyle AD$. We can use the distance formula:

$\displaystyle D=\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2}$

$\displaystyle =\sqrt{(4-7)^2+(1+1)^2}$

$\displaystyle =\sqrt{9+4}$

$\displaystyle =\sqrt{13}\approx 3.6$

The largest line that can be drawn within a circle is the diamater. The diameter is equal to twice the radius. Given that the radius is equal to 12 inches, the largest line that could be drawn (the diameter) would be equal to 24 inches. 

The distance on a number line is the number of whole number places between two numbers.  When you are going from a negative value to a positive one or vice versa, you must find the distance from $\displaystyle 0$ that both points are then add them together.  $\displaystyle -3$ is $\displaystyle 3$ spaces away from $\displaystyle 0$ and $\displaystyle 12$ is $\displaystyle 12$ spaces away.  To find the distance you then add your two values of $\displaystyle 3$ and $\displaystyle 12$ to get $\displaystyle 15$.

The Pythagorean Theorem states:

$\displaystyle a^{2 } +b^{2} =c ^{2}$  where $\displaystyle c$ represents the hypotenuse and $\displaystyle a$ and $\displaystyle b$ represent the measurements of the legs of the right triangle.

$\displaystyle 30^{2} = 30 \times30 = 900$

$\displaystyle 40^{2} = 40 \times 40 = 1600$

$\displaystyle 900 + 1600 = 2500$

$\displaystyle \sqrt{c^{2}} = \sqrt{2500}$

$\displaystyle c = 500 cm$

The Pythagorean Theorem states that

$\displaystyle a^{2} +b^{2} = c^{2}$

Where $\displaystyle a$ and $\displaystyle b$ are the measurements of each leg and $\displaystyle c$ is the hypotenuse.

$\displaystyle 3^{2} = 3 \times3 =9$

$\displaystyle 8^{2} = 8\times 8 = 64.$

$\displaystyle 9 + b^{2} = 64$

$\displaystyle (9-9) + b^{2} = 64-9$

$\displaystyle b^{2} = 55$

$\displaystyle \sqrt{b^{2}} = \sqrt{55}$

$\displaystyle b\approx 7.4cm$