Solve: 3x+7=22
x = 5
Rewrite in standard form:
2x2 -5x = 12
2x2 -5x -12 = 0
In which quadrant is the point (−3,2)?
Quadrant II
Find the slope through (1,2) and (5,10).
m = 2
A rectangle is 12m long and 7m wide. How much fencing is needed?
38m
Solve |2x -1| = 7
x = -3 or x = 4
Solve by factoring:
x2 + x = 12
(x+4)(x-3) = 0 so x = -4,3
Find the midpoint between (−4, 5) and (2,−3).
(-1,1)
Find the equation and slope of the line through (3,−2) and (3,6).
x = 3, slope = undefined
A circle has diameter 18 ft. Find its area in terms of π.
r = 9ft, so 81π ft2 or 254.5 ft2
Solve and write in interval notation: −4(2x−3)≥20
x ≥ -1 or (-inf, -1]
Use the discriminant to classify:
2x2 -4x + 7 = 0
Δ = (−4)2−4(2)(7)=−40, so no real solutions
Find the midpoint between (−4,5) and (2,−3).
√29
Find the slope-intercept equation through (−2,6) and (4,−3).
y = -3/2x + 3
A closed cylinder has radius 4 cm and height 10 cm. Find its total surface area.
351.85 cm2
Solve and state all restrictions:
3/x + 2/(x-1) = 1
restrictions: x cannot equal 0,1
solution: x = 3 ± √ 6
Solve using the quadratic formula:
x2 - 6x + 2 = 0
x = 3 ± √7
Point P=(−4,5) and midpoint M=(2,−1) are given. Find the other endpoint Q.
Q = (8, -7)
Find the equation of the line perpendicular to y=−2x+5 through (3,−4).
y = 1/2x - 11/2
A cone has radius 6 cm, height 8 cm, and slant height 10 cm. Find its total surface area.
96π cm2 or 301.6 cm2
Solve and check for extraneous solutions:
√(x+5) = x-1
x = 4
Solve by completing the square: x2 − 8x + 5 = 0
(x−4)2 = 11, so x = 4 ± √11
A point lies on x2+y2=25, has x=3, and is in Quadrant IV. Find the point.
y2 = 16 and y<0 so (3,-4)
Rewrite in standard circle form and identify the center and radius: x2+y2+8x−10y+25=0
(x + 4)2 + (y - 5)2 = 16
center: (-4,5)
radius: 4
A closed cylindrical container has radius 3 in and height 12 in. Find both its total surface area and volume.
Surface Area: 90π in2
Volume: 108π in3