# Working with square roots warm up 1.√3 + √3 = 2.√4 +√4 = 3.√5 + √5 = 4.√1 + √1 = 5.(√3) (√3) = 6.(√5) (√6) = Simplify 7. √24 = 8.√18 = 9.√81 = 10.√150.

## Presentation on theme: "Working with square roots warm up 1.√3 + √3 = 2.√4 +√4 = 3.√5 + √5 = 4.√1 + √1 = 5.(√3) (√3) = 6.(√5) (√6) = Simplify 7. √24 = 8.√18 = 9.√81 = 10.√150."— Presentation transcript:

Working with square roots warm up 1.√3 + √3 = 2.√4 +√4 = 3.√5 + √5 = 4.√1 + √1 = 5.(√3) (√3) = 6.(√5) (√6) = Simplify 7. √24 = 8.√18 = 9.√81 = 10.√150 = 1. 2 √3 2.4 3.2 √5 4.2 5.√9 = 3 6.√30 7.2 √6 8.3 √2 9.+ or – 9 10. 5√6

Let’s try some more 1.12 2 + 3 2 2.√36 + √4 3.(√5)(√10) 4.(√18)(√3) 5.(√6)(√8)(√12) 1.153 2.8 3.5√2 4.3√6 5.24

Geometry 1 st Semester BOE Review

Pythagorean Theorem a 2 + b 2 = c 2 a and b are LEGS of a right triangle. LEGS are those segments that create a right angle. c is the HYPOTENUSE. The HYPOTENUSE is the segment across from the right angle.

LEG HYPOTENUSE RIGHT ANGLE a b c If a = 5 and b = 8, find c If a = 2 and c = 8, find b If b = 1 and c = 5, find a c = √89 b = √60 = 2 √15 a = √24 = 2√6 LET’S PRACTICE

Cool things you should remember about triangles ♥if c 2 = a 2 + b 2 you have a right triangle ♥if c 2 > a 2 + b 2 you have an obtuse triangle ♥if c 2 < a 2 + b 2 you have an acute triangle ♥in a right triangle, the hypotenuse is always the longest side.

What if… What if you had a square with a side measure of 6. What would the length of your diagonal be? 6√2 What if you had an equilateral triangle with side measure of 10. What would the altitude of the triangle be? 5 √3 What would the area of the triangle be? 25 √3

Systems of Equations 2 ways to solve: ♥Elimination ♥Substitution

Elimination The goal is to see if you can eliminate a variable. 2x + y = -1 7x -5y = 5 Multiply the first equation by 5: 10x + 5y = -5 7x - 5y = 5 Add them together: 17x = 0 solve: x = 0 Substitute x into an original equation to find y: 2(0) + y = -1 y = -1 Substitute them both into the next equation to make sure your answers are correct: 7(0) – 5(-1) = 5 5 = 5

Substitution See if you can isolate a variable and solve for the remaining one. 2x + y = -1 7x -5y = 5 Solve the first one for y: y = -2x -1 Substitute into the second one. 7x – 5(-2x – 1) = 5 Solve 7x + 10x + 5 = 5 17x = 0 x = 0 Substitute into the first equation and solve y = -1 Substitute x & y into the 2 nd equation to be sure it works 2(0) + (-1) = -1 -1 = -1

Download ppt "Working with square roots warm up 1.√3 + √3 = 2.√4 +√4 = 3.√5 + √5 = 4.√1 + √1 = 5.(√3) (√3) = 6.(√5) (√6) = Simplify 7. √24 = 8.√18 = 9.√81 = 10.√150."

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