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Write the equation of a line that is perpendicular to the given line and that passes through the given point. –3x – 6y = 17; (6, 3) y = x – 9 y = 2x – 9 y = –2x – 9 y = x + 0 3. is the relationship shown by the data linear? if so, model the data with an equation. x y 1 5 5 10 9 15 13 20 the relationship is linear; y – 5 = (x – 1). the relationship is not linear. the relationship is linear; y – 5 = (x – 1). the relationship is linear; y – 1 = (x – 5). write an equation in point-slope form for the line through the given point with the given slope. (–10, –1); m = y + 10 = (x + 1) y – 1 = (x – 10) y – 1 = (x + 10) y + 1 = (x + 10) 5. write an equation for each translation of . 6.5 units up y + 6.5 = | x | y = | 6.5 x | y = | x | + 6.5 y = | x | – 6.5 6. write an equation for each translation of . 5.5 units right y = | x | + 5.5 y = | x – 5.5 | y = | x | – 5.5 y = | x + 5.5 | 7. which equation translates y = | x | by 8 units to the left? y = | x | – 8 y = | x | + 8 y = | x – 8| y = | x + 8|

Prove the converse of the pythagorean theorem using similar triangles. the converse of the pythagorean theorem states that when the sum of the squares of the lengths of the legs of the triangle equals the squares length of the hypotenuse, the triangle is a right triangle. be sure to create and name the appropriate geometric figures.

For every corresponding pair of cross sections, the area of the cross section of a sphere with radius r is equal to the area of the cross section of a cylinder with radius and height 2r minus the volume of two cones, each with a radius and height of r. a cross section of the sphere is and a cross section of the cylinder minus the cones, taken parallel to the base of cylinder, is the volume of the cylinder with radius r and height 2r is and the volume of each cone with radius r and height r is 1/3 pie r^3. so the volume of the cylinder minus the two cones is therefore, the volume of the cylinder is 4/3pie r^3 by cavalieri's principle. (fill in options are: r/2- r- 2r- an annulus- a circle -1/3pier^3- 2/3pier^3- 4/3pier^3- 5/3pier^3- 2pier^3- 4pier^3)