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Published byうのすけ つねざき Modified over 5 years ago
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Write the equation of a line in point-slope from that passes through (2, 5) and (5, 11).
y− 5 = 2 (x− 2) m= = =−2 y− 11 = − 2 (x− 2) m= = =0 y− 5 = −2 (x− 2) Two wrongs and a right 5−11 5−2 11−5 5−2 6 3 −6 3 5−11 5−2 −6 3
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y ≥ x2 + 3 Two wrongs and a right
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Solve the multi-step equation:
4(–x – 2) – 5 = 13 –4x + 8 – 5 = 13 –4x + 3 = 13 –4x = 10 x = – = – 5 2 –4x – 8 – 5 = 13 –4x – 13 = 13 –4x = 0 x = 0 –4x = 26 x = – = – 13 2 Two wrongs and a right 4 2 4 2
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Solve the absolute value inequality:
x + 5 < 15 –15 < x + 5 < 15 –20 > x > 10 –20 < x < 10 –20 < x < 20 Two wrongs and a right
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Find the slope of the line that passes through (–2, 5) and (–2, 1).
m= = 4 −4 =−1 Find the slope of the line that passes through (–2, 5) and (–2, 1). m= = 4 0 Slope is undefined m= = 0 4 =0 Two wrongs and a right 5−1 −2−2 4 −4 5−1 −2− −2 4 0 −2 − −2 5−2 0 4
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Given f(x) = 5x – 2, find the value of x when f(x) = 8.
f(8) = 5(8) – 2 f(8) = 38 f = 4.75 Given f(x) = 5x – 2, find the value of x when f(x) = 8. 8 = 5x – 2 10 = 5x 2 = x Two wrongs and a right
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(x – 3)2 – 5 = 11 (x – 3)2 = 16 x – 3 = 4 x = 7 x – 3 = ±4 x = 3 ± 4 x = 7 or x = –1 (x – 3)2 = 6 x – 3 = ±6 x = 3 ± 6 x = 9 or x = –3 Two wrongs and a right
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Simplify the expression:
Two wrongs and a right
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y3x −2 y4x3 y3y4 x2x3 y12 x6 y7 x5 y7 x1 1 x1y7 Two wrongs and a right
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−3−4 + −6+1 −7 + −5 7 + 5 12 7 + 7 14 −1 + −5 1 + 5 6 Two wrongs and a right
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x2 – 5x + 6 = 2 (x – 3)(x – 2) = 2 x – 3 = 2 or x – 2 = 2 x = 5 or x = 4 x2 – 5x + 4 = 0 (x + 4)(x + 1) = 0 x + 4 = 0 or x + 1 = 0 x = –4 or x = –1 (x – 4)(x – 1) = 0 x – 4 = 0 or x – 1 = 0 x = 4 or x = 1 Two wrongs and a right
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x 2 + x 3 =6 3x + 2x = 36 5x = 36 x = 36 5 3x + 2x = 6 5x = 6 x = 6 5
Two wrongs and a right x 2 x 3 x 2 x 3 36 5 36 6 6 5 x 2 x 3 3x 6 2x 6 5x 6 5 36 36
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Parabola with vertex (2, –3)
Two wrongs and a right
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Line with slope – 2 3 Two wrongs and a right 2 3 2 3 2 3
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Line graphed is to y = 3x + 1
Two wrongs and a right
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Graph of y = 2 (x – 3)2 – 5 Two wrongs and a right
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Graph of y = x Two wrongs and a right 5 5 5
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Points B and D are tangent to the circle.
2x2 + 2x = 7x + 12 2x2 – 5x – 12 = 0 (2x + 3)(x – 4) = 0 2x = –3 or x = 4 x = −3 2 or x = 4 x = 4 (2x – 3)(x + 4) = 0 2x = 3 or x = –4 x = 3 2 Two wrongs and a right
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SU is the diameter of the circle. Find the m ST .
m TU = 2(61) = 122 m ST = 122 m ST = 180 – 122 = 58 Two wrongs and a right
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Two wrongs and a right Solve for x. 2x + 30 = 5x – 36 66 = 3x x = 22
7 7
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Find the area of the trapezoid.
A = 330 m2 A = ( ) 10 A = 250 m2 A = ( ) ( ) A = 550 m2 Two wrongs and a right 2 2 2
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Find the volume of the cylinder.
Two wrongs and a right
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Angles A and B are supplementary. m∠A=36; find m∠B.
36 + m∠B = 90 m∠B=54 36 + m∠B = 180 m∠B=144 36 + m∠B = 360 m∠B=324 Two wrongs and a right
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Two wrongs and a right Find the Cos K Cos K = 5 12 Cos K = 12 13
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Two wrongs and a right Find the mMBD. 180 – (56 + 78) 180 – 134 46
Triangle MBD is isosceles, and thus, mMBD = mMDB = 62. mAMC = 180 − ( ) mAMC = 46 mAMC = mDMB = 46 mMDB = 180 − ( ) mMDB = 180 − (108) mMDB = 72 Two wrongs and a right
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Two wrongs and a right The rectangle has area = 432 and has 12
congruent squares. What is the perimeter? Let the side of one of the small squares = x. Thus (4x) (3x) = 432 12x2 = 432 x2 = 36 x = 6 Perimeter = 6 × 4 = 24 Perimeter = 6 × 14 = 84 The rectangle has area = 432 and has 12 congruent squares. What is the perimeter? It is NOT possible to find the perimeter. Two wrongs and a right
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Two wrongs and a right ABC is a right triangle. AM BC. mABC = 55.
Find mMAC. mABC + mACM + 90° = 180° mACM = 180 – 90 – 55 = 35° mMAC + mACM + 90 = 180° mMAC = 180 − 90 − mACM = 180 − 90 − 55 = 35° mACM = 180 – 90 – 55 = 45° = 180 − 90 − 45 = 45° = 180 − 90 − 35 = 55° Two wrongs and a right
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Parallelogram WXYZ is mapped to its image WXYZ by a reflection
across the x-axis. Two wrongs and a right
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The following triangles are
congruent by ASA. Two wrongs and a right
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In ABC, AC AC. Find the length ( y ) of BC.
22 14 = y+15 y 22y = 14(y + 15) 22y = 14y + 210 8y = 210 y = 26.25 14 30 = y y+15 30y = 14(y + 15) 30y = 14y + 210 16y = 210 y = In ABC, AC AC. Find the length ( y ) of BC. 22 14 = 15 y 22y = 210 y = 9.54 Two wrongs and a right
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