Part 2 Algebra II Fall Semester Exam Review

Write the system of equations that would determine when the ball and balloon are at the same height.

• $s\left(t\right)=40t+64$ $s\left(t\right)=-16t^2+8t\$

• $s\left(t\right)=-16t^2+40t\$ $s\left(t\right)=64t+8$

• $s\left(t\right)=-16t^2+64t\$ $s\left(t\right)=8t+40$

• $s\left(t\right)=-16t^2+40t\$ $s\left(t\right)=-16t+64$

• $10x+50y+100z=260$ $150x+500y+800z=2250\ \$ $\ x=y+1$

• $10x+50y+100z=260\$ $150x+500y+800z=2250\$ $\ y=x+1$

• $10x+50y+100z=260\$ $150x+500y+800z=2250$ $y=z+1$

• $150x+500y+800z=2250\$ $10x+50y+100z=260\$ $x=z+1$

• $x+y\le100$ $3x+5y\ge400\$

• $3x+5y\le400\$ $x+y\ge100$

• $x+y\ge400$ $3x+5y\le100$

• $x+y\le400$ $\ 3x+5y\ge100$

Solve the system of equations

• (-4, -2, 0)

• (2,0,4)

• (0,2,4)

• (4,2,0)

• x=4 and z=2

• x=4 and z=0

• x=2 and z=4

• x=2 and z=0

• (2,4) and (5,1)

• (-6,0) and (-4,0)

• (-2,4) and (-5,1)

• (6,0) and (4,0)

Tags

CCSS.HSA.REI.C.7

• At .264 seconds, the elevator and ball are both at a height of 22.639 ft and at 4.736 seconds, the ball and elevator are both at a height of 67.361 feet.

• At .264 seconds the elevator and ball are both at a height of 22.639 ft.

• At .264 seconds, the ball and elevator are both at 12.6 feet.

• At 4.736 seconds, the ball and elevator are both at a height of 67.361 feet.

• The graph will not change.

• The graph will have a new y-intercept of (0, 4).

• The graph will have a new x-intercept of (6,0)

• The graph will have a new y intercept of (0,4) and a new x intercept of (6,0).

• $f\left(x\right)=\frac{8}{5}\left|x+1\right|$

• $f\left(x\right)=\frac{5}{8}\left|x\right|-1$

• $f\left(x\right)=\frac{5}{8}\left|x+1\right|$

• $f\left(x\right)=\frac{8}{5}\left|x\right|-1$

Match the system to the graph (solution set is on the left)

• y ≤ 3x   and  y ≤ -2x + 4
• y > 3x   and  y > -2x + 4
• y < 3x   and  y > -2x + 3
• y > 3x   and  y < -2x + 4

Tags

CCSS.HSA.REI.D.12