Algebraic Reasoning Spring Final Exam Review Day 3

Simplify the product.

8p(–3p² + 6p – 2)

• –5p³ + 14p² – 6p

• 48p² – 16p – 24p³

• 14p² – 6p – 5p³

• –24p³ + 48p² – 16p

Simplify the product.

-n(6n + 3)

• -9n²

• -6n² + 3n

• -6n - 3

• -6n² - 3n

Simplify the product using the distributive property.

(3h - 7)(3h - 6)

• 9h² + 3h - 42

• 9h² - 3h - 42

• 9h² - 39h + 42

• 9h² - 39h + 42

• 16x- 64

• -16x - 64

• x² + 16x + 64

• -8x + 16

Divide.

(14m⁹ - 4m⁸ + 8m⁶) / 2m³

• 7m⁶ - 4m⁸ + 8m⁶

• 7m⁹ - 2m⁸ + 4m⁶

• 7m⁶ - 2m⁵ + 4m³

• 7m⁶ - 2m⁵ + 8m⁶

Divide.

(16x³ - 8x² + 20x) / 4x²

• 4x² - 2 + 5/x

• 4x + 5/x

• 4x - 2 + 5/x

• 4x - 2

You live near a bridge that goes over a river. The underneath side of the bridge is an arch that can be modeled with the function y = -0.000495x² + 0.619x where x and y are in feet. How high above the river is the bridge (the top of the arch)? How long is the section of bridge above the arch?

• The bridge is about 193.52 ft above the river and the length of the bridge above the arch is about 625.25 ft

• The bridge is about 193.52 ft above the river and the length of the bridge above the arch is about 1250.51 ft

• The bridge is about 1250.51 ft above the river and the length of the bridge above the arch is about 193.52 ft

• The bridge is about 1250.51 ft above the river and the length of the bridge above the arch is about 625.25 ft

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The function y = -0.024x² + 0.55x models the height y, in feet, of your pet frog's jump, where x is the horizontal distance, in feet, from the start of the jump. How far did the frog jump? How high did it go? Round your answer to the nearest hundredth.

• The frog jumped about 3.15 ft far and about 22.92 ft high.

• The frog jumped about 22.92 ft far and about 11.46 ft high.

• The frog jumped about 22.92 ft far and about 3.15 ft high.

• The frog jumped about 11.46 ft far and about 3.15 ft high.