College Algebra Final

Factor:
 2x² - 3x - 2

• (2x - 2)(x + 1)
• (2x + 1)(x - 2)
• (2x - 1)(x - 2)
• (2x - 1) (x - 1)

Tags

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

Factor:
9x² - 6x - 8

• (3x+2)(3x-4)
• (3x+2)(3x+4)
• (3x-2)(3x-4)
• (x-1)(9x+8)

Tags

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

CCSS.HSA.APR.B.3

What is the factored form of:
 25x² - 9

• (5x+3)(5x-3)
• (x+5)(x-3)
• (x-3)(x+5)
• (5x+3)(3x+5)

Tags

CCSS.HSA.APR.C.4

Factor by grouping
x³ - 4x² -4x + 16

• (x+4)(x+2)
• (x-4)(x² -4) 
• (x+4)(x² -4)
• (x-4)(x-2)

Tags

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

Use the zero-factor property to solve.
x² + 2x - 8 = 0

• x = -4 or x = 2
• x = - 4
• x = 4 or x = -2
• x = -2

Find the slope between the two given points: (-2,1) and (5,7)

• 7/6
• 2
• 6/7
• -2

Tags

CCSS.8.EE.B.5

Write the equation of the line with slope 4 through the point (-2, 5).

• y = 4x - 13
• y = 4x + 13
• y = 4x - 3
• y = 4x + 3

Tags

CCSS.HSA.CED.A.2

CCSS.HSG.GPE.B.5

Given the line 2x-3y=9, write the equation of a line perpendicular to it that passes through the point (4, -1)

• y = -3/2x - 7
• y = 2/3x + 4
• y = 2/3x - 1
• y = -3/2x + 5

Tags

CCSS.HSG.GPE.B.5

5(n-2)=10(3n+9)

• -4
• 4
• 2.5
• 25

Tags

CCSS.HSA.REI.B.3

CCSS.HSA.CED.A.1

CCSS.HSA.REI.A.1

pV = nRT
solve for R

• R= pV/nT
• R= p/n + V/T
• R= pV - nT
• R= (pV-n)/T

Tags

CCSS.HSA.CED.A.4