Unit 2 Quiz Review (Pythagorean Theorem, Distance, Area)

How is the distance formula correctly written:

• $d\ =\ \sqrt{\left(y_{1\ }-\ y_2\right)^2\ -\left(x_{2\ }-\ x_1\right)^2}\$

• $d\ =\ \sqrt{\left(x_{2\ }-\ x_1\right)^2\ +\ \left(y_{2_{\ }}-\ y_1\ \right)2}$

• $d\ =\ \sqrt{\left(y_{1\ }-\ y\ _2\right)^2\ +\ \left(x_{2\ }-\ x_1\right)}^2$

• d = $\sqrt{\left(x\right)^2-\ \left(y\right)^2}$

Tags

CCSS.HSG.GPE.B.7

A(3,1) B(-2,-1) written correctly is:

• $\sqrt{\left(-2-3\right)^2\ +\ \left(-1-1\right)^2}$

• $\sqrt{\left(1-3\right)^2\ +\ \left(-1-2\right)^2}$

• $\sqrt{\left(-2-3\right)^2\ -\ \left(-1-1\right)^2}$

• $\sqrt{\left(-2-3\right)^2\ -\ \left(-1-1\right)^2}$

Tags

CCSS.HSG.GPE.B.7

A(2,0) B(-2,4) is written as

• $d\ =\ \sqrt{\left(4-2\right)^2\ -\ \left(0-2\right)^2}$

• $d\ =\ \sqrt{\left(4-0\right)^2\ -\ \left(-2-2\right)^2}$

• $d\ =\ \sqrt{\left(-2\ +0\right)^2\ +\ \left(4+2\right)^2}$

• $d\ =\ \sqrt{\left(-2-2\right)^2\ +\ \left(4-0\right)^2}$

Tags

CCSS.HSG.GPE.B.7

What is the distance between AB?

• 0

• undefined

• 5

• 4

Tags

CCSS.6.G.A.3

Find the distance to the nearest tenth.

• 9.7

• 10.7

• 11.7

• 12.7

Tags

CCSS.HSG.GPE.B.7

What is the distance between the two points to the nearest tenth?

• 6.3

• 6.5

• 6.7

• 16

Tags

CCSS.HSG.GPE.B.7

What is the distance between AB

• 0

• 8

• 9

• 7

Tags

CCSS.6.G.A.3

Find the distance.

• $\sqrt{60}$

• $\sqrt{61}$

• $\sqrt{11}$

• $\sqrt{5}$

Tags

CCSS.HSG.GPE.B.7

Find the distance between (-5, -8) and (-1, -16).

• $\sqrt{592}$

• $\sqrt{100}$

• 10

• $\sqrt{80}$

Tags

CCSS.HSG.GPE.B.7

a² + b² = c^{2
a = 9 b = 40}

• c = 41

• c = 65

• c = 25

• c = 5

Tags

CCSS.8.G.B.7