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The number of marbles in 13 bags
# 1 The number of marbles in 13 bags The number of marbles you had before purchasing more bags. The number of marbles in each of the 13 bags.
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# 2 The number of children Jamie has. The number of months. The monthly tuition for the science club The one-time materials fee. The total cost for one child to join the science club.
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3π₯ 2π₯ β2 +5 2π¦ β3 5π¦ +5 6 π₯ 2 β4π₯ 10 π¦ 2 +10π¦ +15π₯ β10 β15π¦ β15
# 3 Use the area model. 3π₯ 2π₯ β2 +5 2π¦ β3 5π¦ +5 6 π₯ 2 β4π₯ 10 π¦ 2 +10π¦ +15π₯ β10 β15π¦ β15 6 π₯ 2 +11π₯β10 10 π¦ 2 β5π¦β15
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# 4 Multiply Exponents Multiply Exponents π¦ 6 β π¦ 4 = π¦ 10 π§ 6 β π§ 3 = π§ 9 Add Exponents Add Exponents To raise to a power; multiply exponents. To multiply; add exponents.
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Simplify the numerator and denominator.
# 5 3 π β β 4 =1 Simplify the radicals. 1 3 π β2β3 27β2 =1 Simplify the numerator and denominator. 9 3 π 9 =1 Re-write the equation with all terms in the same base. 3 π = 3 0 Use the properties of exponents. πβ2=0 π=2
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Use the properties of exponents.
# 6 3 π β β 45 =3 Simplify the radicals. 3 π β 9 =3 3 π β3 =3 Use the properties of exponents. 3 π =3 Use the properties of exponents again. πβ3=1 π=4
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# 7 = π 3 5 π π₯ π π 3 π. π π₯ π
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# 8 π. (7 π¦ 3 ) 1 2 π. π 5 3 π. π π π
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Simplify each of the following.
# 9 Simplify each of the following. a. π₯ π₯ 3 4 b π π π β π 3 2 c. 3π π π π 5 When dividing the same base, subtract the exponents. = π₯ 1 4 β 3 4 = π 2 5 β β π 1 2 β 3 2 = π 5 6 β π 3β5 = π₯ β 2 4 = 1 9 π π β2 =2 π π β 2 2 = π₯ β 1 2 =2 π 1 π β1 = π π 2 = 1 π₯ 1 2 = 2π π Do not leave a negative exponent in the answer.
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π. real; the 3 is irrational
# 10 π. real; the 3 is irrational π. real; 4β 3 5 = rational π. real; irrational π. real; = 9 9 =1 rational, integer, whole
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2π‘+8=7+5π‘β20 20π€β10=8+2π€β12+12 2π‘+8=5π‘β13 20π€β10=8+2π€ β2π€ β2π€ β2π‘ β2π‘
# 11 Distribute Distribute 2π‘+8=7+5π‘β20 20π€β10=8+2π€β12+12 Combine Like Terms Combine Like Terms 2π‘+8=5π‘β13 20π€β10=8+2π€ β2π€ β2π€ β2π‘ β2π‘ Subtract 2t from each side of the equation. 8=3π‘β13 18π€β10=8 Subtract 2w from each side of the equation. +13 +13 +10 +10 3 3 Divide each side of the equation by 3. 21=3π‘ 18π€=18 18 18 Add 13 to each side of the equation. Divide each side of the equation by 18. Add 10 to each side of the equation. 7=π‘ π€=1
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# 12
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# 13 23+3π₯=5π₯+7 β3π₯ β3π₯ 23=2π₯+7 β β7 16=2π₯ 8=π₯ 8 hours
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# 14 0.08π= π β0.05π β0.05π 0.03π=7.50 250=π
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36 total cardβ14=22 baseball cards
# 15 π₯+ 2π₯β6 =36 +6 +6 Add 6 to each side of the equation. Combine like terms. π₯+2π₯=42 3 3π₯=42 3 Divide each side of the equation by 3. x represents the number of football cards! π₯=14 36 total cardβ14=22 baseball cards
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Divide each side of the equation by 4.
# 16 21+7 π₯+7 28 π₯ 21 84=π₯+ 2π₯β7 +(π₯+7) Combine Like Terms 2π₯β7 35 4 4 84=4π₯ Divide each side of the equation by 4. 2(21)β7 21=π₯ Draw and label the triangle.
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Change the inequality direction when dividing by a negative.
# 17 β1 4 1 6 26 4 9 Change the inequality direction when dividing by a negative. 3 56 π. π+ 1 3 > 1 2 e. d. f. π₯+12β2 π₯β22 >0 5πβ3<β8 7<3π₯β5β€22 2π+9>17 π. β3π₯β€β9 π. 17β₯π₯β9 βπ βπ 3 6 β 2 6 = 1 6 β9 β9 +π +π π₯+12β2π₯+44>0 12<3π₯β€27 5π<β5 2π>8 β π π β π π π₯β₯3 26β₯π₯ βπ₯+56>0 β56 β56 π<β1 π>4 βπ₯>β56 β β1 4<π₯β€9 π> 1 6 π₯<56
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The plant height must be 79 inches or taller.
# 18 (4) Clear the denominator by multiplying each side of the equation by 4. π₯ 4 β₯73 (4) π₯β₯292 Combine Like Terms. 213+π₯β₯292 Subtract 213 from each side of the equation. β213 β213 π₯β₯79 The plant height must be 79 inches or taller.
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Mrs. Hawk can have no more than 28 questions.
# 19 Clear the denominator by multiplying each side of the equation by 6. (6) π₯ 6 β€15 (6) π₯β€90 Combine Like Terms. 62+π₯β€90 Subtract 62 from each side of the equation. β62 β62 π₯β€28 Mrs. Hawk can have no more than 28 questions.
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(π) π. ππ₯βππ= ππ¦ π (π) π. πβπ=π+π π₯ βπ βπ π(ππ₯βππ)=ππ¦ π π βπ=π π₯ π π
# 20 (π) π. ππ₯βππ= ππ¦ π (π) π. πβπ=π+π π₯ βπ βπ π(ππ₯βππ)=ππ¦ π π βπ=π π₯ π π β π π =π₯ π(ππ₯βππ) π =π¦ (π) π. ππ₯βππ= ππ¦ π (π) π. πβπ=π+π π₯ βπ βπ βπ=π π₯ β1 β1 π(ππ₯βππ)=ππ¦ ππ₯βππ ππ₯βππ π=βπ π₯ π= ππ¦ ππ₯βππ
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Substitute the values for x and y into the equation.
# 21 Substitute the values for x and y into the equation. π΄π₯+3π¦=48 π΄(β3)+3(14)=48 Now solve for A. β3π΄+42=48 β42 β42 β3π΄=6 β3 β3 π΄=β2
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Substitute the values for x and y into the equation.
# 22 Substitute the values for x and y into the equation. 2π₯+π·π¦=13 2(β18)+π·(7)=13 Now solve for D. β36+7π·=13 7π·=49 π·=7
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b. Solve the equation for d. c. Solve the equation for b.
# 23 Marcy has $150 to buy packages of hot dogs and hamburgers for her booth at the carnival. At her local grocery store she found packages of hot dogs cost $6 and packages of hamburgers cost $20. a. Write an equation that can be used to find the possible combination of hot dog and hamburger packages Marcy can buy using her budget of exactly $150. (Hint: Use d to represent the number of hot dog packages and b to represent the number of hamburger packages.) b. Solve the equation for d. c. Solve the equation for b. c. π= 150β6π 20 a. 6π + 20π = 150 (or equivalent) π. π= 150β20π 6 (or equivalent)
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# 24 Andrew needs exactly 120 pencils and markers for his class to play the review game he created. His supply closet has boxes of pencils with 12 in each and boxes of markers with 6 in each. Write an equation that can be used to find the number of boxes of pencils and boxes of markers Andrew will need to take to reach his total of 120. Re-write the equation so that is could be used to find the number of pencil boxes. Re-write the equation so that is could be used to find the number of marker boxes. a. If p = the number of pencil boxes and m = the number of marker boxes, 12π+6π=120 b. π=β 1 2 π+10 (or equivalent) c. π=β2π+20 (or equivalent)
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# 25 This is not a function because β4 (domain) is assigned to more than one value in the range. This is a function because every element of the domain is assigned to exactly one element in the range. This is a function because every element of the domain is assigned to exactly one element in the range. This is not a function because β3 is assigned to multiple values in the range.
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# 26 Subtract 1 from each side of the equation. 13=0.3π+1 β1 β1 0.3 0.3 12=0.3π Divide each side of the equation by 0.3. 40=π 40 miles
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# 27 185=25π₯ +10 Subtract 10 from each side of the equation. β10 β10 175=25π₯ 25 25 Divide each side of the equation by 25. 7=π₯ 7 months
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π. 3 π₯ 3 +2 π₯ 2 β3π₯β2 +( π₯ 2 +5π₯β6) 3 π₯ 3 +3 π₯ 2 +2π₯β8
# 28 Combine Like Termsβ¦ π. 3 π₯ 3 +2 π₯ 2 β3π₯β2 +( π₯ 2 +5π₯β6) 3 π₯ 3 +3 π₯ 2 +2π₯β8 Distribute π π¦ 2 +5π¦β6 β 9 π₯ 2 +3π₯β4 +11 Combine Like Termsβ¦ 12 π¦ 2 +5π¦β6 + β9 π₯ 2 β3π₯+4 +11 12 π¦ 2 β9 π₯ 2 +5π¦β3π₯+9
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π. 6πβ2 2 π 2 β3π+2 π. 2πβ5 3π+6 2 π 2 6π β3π β2 +2 2π 3π β5 +6 12 π 3
# 28 π. 6πβ2 2 π 2 β3π+2 π. 2πβ5 3π+6 2 π 2 6π β3π β2 +2 2π 3π β5 +6 12 π 3 β18 π 2 +12π 6 π 2 β15π β4 π 2 +6π β4 +12π β30 12 π 3 β22 π 2 +18πβ4 6 π 2 β3πβ30
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π. a, b and c are polynomials
# 29 π. (3π₯+5)+(β8π₯β10) π. (3π₯+5) (β8π₯β10) βππβπ π. 3π₯+5 β(β8π₯β10) ππ+π βπ(ππβπ) 3π₯+5 +(+8π₯+10) πππ+ππ π. 3π₯+5 (β8π₯β10) π. a, b and c are polynomials β24π₯ 2 β40π₯ β30π₯ β50 βππ π π βπππβππ
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# 30 π π π =π.ππ(π.ππ+ππ
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π(π 3 ) π(π π₯ ) =π(3 3 +2) =π(2 π₯ 2 ) =π(11) =3 2 π₯ 2 +2 =2 11 2
# 31 π(π 3 ) π(π π₯ ) =π(3 3 +2) =π(2 π₯ 2 ) =π(11) =3 2 π₯ 2 +2 = =6 π₯ 2 +2 =2(121) =242
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# 32 a. There are no games left in the warehouse after 27 shipments. b. He has 9,000 games in the warehouse before making any shipments. c. The slope represents the decrease in the number of games in the warehouse per shipment. d. Domain 0β€π₯β€27 and Range 0β€π¦β€9,000
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a. Domain πβ€πβ€π and Range πβ€πβ€πππ
# 33 192 a. Domain πβ€πβ€π and Range πβ€πβ€πππ b. Both intercepts are 0. This means no cookies can be made with no cups of flour.
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# 34 Elissa uses the function π β =15β to represent the money she makes working at the party store for h hours each week. The function πΆ β =25β+30 represents the money she makes cleaning houses in the neighborhood. Write a function T to describe the total amount of money Elissa makes each week working h hours. π β =40β+30
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# 35 Henry predicts the amount of wheat, in bushels, produced on his farm π‘ years from now can be modeled by the function π π‘ =70π‘ The price, in dollars, per bushel of wheat π‘ years from now can be modeled by the function π π‘ =0.25π‘+6. Write a function πΌ to describe the total income Henry predicts for producing wheat π‘ years from now. πΌ π‘ =π π‘ βπ π‘ =17.5 π‘ π‘+5400
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Slope is the average rate of change for a linear function.
# 36 The graph below shows the average Valentineβs Day spending between and 2012. Slope is the average rate of change for a linear function. (2009, 102) (2004, 100) (2005, 98) (2007, 120) (2010, 103) (2010, 103) π¦ 2 β π¦ 1 π₯ 2 β π₯ 1 ππ π π βπ(π) πβπ π¦ 2 β π¦ 1 π₯ 2 β π₯ 1 ππ π π βπ(π) πβπ π¦ 2 β π¦ 1 π₯ 2 β π₯ 1 ππ π π βπ(π) πβπ 120β β2005 103β β2009 103β β2004 = 22 2 = 1 1 = 3 6 = 1 2 =1 =11 What is the average rate of change in spending between 2005 and 2007? What is the average rate of change in spending between 2004 and 2010? What is the average rate of change in spending between 2009 and 2010? $11 per year $0.50 per year $1 per year
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Graph the line with a y-intercept of 2 and a slope of π π .
# 37 Graph the line with a y-intercept of 2 and a slope of π π . Translate the line 4 units left.
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Graph the line with a y-intercept of β3 and a slope of π π .
# 38 Graph the line with a y-intercept of β3 and a slope of π π . Translate the line 3 units up.
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π. π π =7+(πβ1)(β3) π. π π =β11+(πβ1)(8) Initial value
# 39 Initial value π. π π =7+(πβ1)(β3) Common difference π. π π =β11+(πβ1)(8)
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π= π¦ 2 β π¦ 1 π₯ 2 β π₯ 1 Find the slopeβ¦ Find the y-interceptβ¦ π¦=ππ₯+π
# 40 Find the slopeβ¦ Find the y-interceptβ¦ π= π¦ 2 β π¦ 1 π₯ 2 β π₯ 1 π¦=ππ₯+π 151=6 24 +π 151β79 24β12 = = 6 1 =6 151=144+π β144 β144 7=π The special rate would be $6. The shipping would be $7.
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π= π¦ 2 β π¦ 1 π₯ 2 β π₯ 1 Find the slopeβ¦ Find the y-interceptβ¦ π¦=ππ₯+π
# 41 Find the slopeβ¦ Find the y-interceptβ¦ π= π¦ 2 β π¦ 1 π₯ 2 β π₯ 1 π¦=ππ₯+π 40β30 6β2 = 10 4 = 5 2 30= π 30=5+π β5 β5 25=π He would make 2.5 cards per hour. He started with 25 cards in the box.
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The slope of the line of best fit is π π and itβs y-intercept is 0.
# 42 The slope of the line of best fit is π π and itβs y-intercept is 0. π π = π π π Every 3 days, 1 additional golf cart is repaired
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The slope of the line of best fit is βπ and itβs y-intercept is 90.
# 43 The slope of the line of best fit is βπ and itβs y-intercept is 90. π π =βππ+ππ Every day there are 5 less golf carts sold
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Multiply each term in the first equation by βπ.
# 44 Multiply each term in the first equation by βπ. (βπ) (βπ) (βπ) π. 2π₯+5π¦=20 6π₯+15π¦=15 β6π₯β15π¦=β60 6π₯+15π¦=15 0π₯+0π¦=β45 No Solution
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Infinitely many solutions
# 44 Multiply each term in the first equation by π. π. 9π₯+18π¦=6 12π₯+24π¦=8 Multiply each term in the second equation by βπ. 36π₯+72π¦=24 β36π₯β72π¦=β24 0π₯+0π¦=0 Infinitely many solutions
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π. 5π₯+4π¦=6 6π₯+π¦=13 (19) (19) 30π₯+24 β 29 19 =36 (19) 30π₯+24π¦=36
# 44 Multiply each term in the first equation by π. π. 5π₯+4π¦=6 6π₯+π¦=13 Multiply each term in the second equation by βπ. (19) (19) 30π₯+24 β =36 (19) 30π₯+24π¦=36 β30π₯β5π¦=β65 570π₯β696=684 +696 +696 0π₯+19π¦=β29 19 19 570π₯=1380 570 570 π¦=β 29 19 π₯= 46 19
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28β16=12 blocks π₯+π¦=28 0.38π₯+1.56π¦=24.80 π¦=28βπ₯ 0.38π₯+1.56 28βπ₯ =24.80
# 45 28β16=12 blocks Let x be the number of bricks and y the number of blocks. π₯+π¦=28 0.38π₯+1.56π¦=24.80 π¦=28βπ₯ Substitute 0.38π₯ βπ₯ =24.80 Distribute 0.38π₯+43.68β1.56π₯=24.80 Combine Like Terms β1.18π₯+43.68=24.80 β43.68 β43.68 β1.18 β1.18 Subtract from each side of the equation. β1.18π₯=β18.88 Divide each side of the equation by π₯=16
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300 adult tickets were sold.
# 46 300 adult tickets were sold. 300 can now be substituted into either equation for a. Let a = # adult tickets and c = # of child tickets 20π+10π=15,000 π=3π Use substitution to replace c with 3a. π=3π 20π+10 3π =15,000 π=3(300) 20π+30π=15,000 π=900 50π=15,000 900 child tickets were sold. π=300
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+ Multiply the first equation by βππ.
# 47 Multiply the first equation by βππ. Let S = # of small dogs and L = # of large dogs π+πΏ=22 43π+75πΏ=1234 β43πβ43πΏ=β946 + 43π+75πΏ=1234 Since we are looking for L, letβs eliminate S. Add the second equation. 32πΏ=288 πΏ=9 9 large dogs were groomed.
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# 48 where C is the cost and d is the number of disks 13.75π= π β2.9π β2.9π 10.85π=3255 10.85 10.85 π=300
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π. β2π₯β3π¦=β4 12π₯+3π¦=1 π. β5π₯+35π¦=β40 5π₯+4π¦=20
# 49 Answers will vary. Below are sample answers. π. β2π₯β3π¦=β4 12π₯+3π¦=1 We can multiply the first equation by β 1. π. β5π₯+35π¦=β40 5π₯+4π¦=20 We can multiply the first equation by β 5.
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# 50 π. 7π₯+6π¦>30 β7π₯ β7π₯ 6π¦>β7π₯+30 6 6 6 π¦>β 7 6 π₯+5
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# 50 π. 9π₯+4π¦β₯28 β9π₯ β9π₯ 4π¦β₯β9π₯+28 4 4 4 π¦β₯β 9 4 π₯+7
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# 50 π. 3π₯+7π¦<14 β3π₯ β3π₯ 7π¦<β3π₯+14 7 7 7 π¦<β 3 7 π₯+2
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# 50 π. 7π₯+14π¦β€3 β7π₯ β7π₯ 14π¦β€β7π₯+3 14 14 14 π¦β€β 7 14 π₯+ 3 14 π¦β€β 1 2 π₯+ 3 14
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β5π₯β4π¦β€6 +5π₯ +5π₯ β4π¦β€5π₯+6 β4 β4 β4 π¦β₯β 5 4 π₯β1 1 2 8π₯β16π¦>24 β8π₯
# 51 β5π₯β4π¦β€6 +5π₯ +5π₯ β4π¦β€5π₯+6 β4 β4 β4 π¦β₯β 5 4 π₯β1 1 2 8π₯β16π¦>24 β8π₯ β8π₯ β16π¦>β8π₯+24 β16 β16 β16 π¦< 1 2 π₯β1 1 2
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π₯+π¦>1 βπ₯ βπ₯ π¦>βπ₯+1 5π₯+2π¦<10 β5π₯ β5π₯ 2π¦<β5π₯+10 2 2 2
# 51 π₯+π¦>1 βπ₯ βπ₯ π¦>βπ₯+1 5π₯+2π¦<10 β5π₯ β5π₯ 2π¦<β5π₯+10 2 2 2 π¦<β 5 2 π₯+5
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# 52 175 feet 25 seconds
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# 53 about 4.5 seconds about 26 feet
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What are the factors of βπ that also combine to +π?
# 54 FACTOR π₯ 2 +π₯β6 What are the factors of βπ that also combine to +π? FACTOR π₯ β2 π₯ π₯ 2 β2π₯ +3 3π₯ β6 β6 π₯ 2 π+π
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What are the factors of +ππ that also combine to +ππ?
# 55 FACTOR 2π₯ 2 +13π₯+15 What are the factors of +ππ that also combine to +ππ? FACTOR 2π₯ +3 π₯ 2π₯ 2 3π₯ +5 10π₯ +15 +30 π₯ 2 π+π
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What are the factors of πππ that also combine to +ππ?
# 56 FACTOR 3π₯ +7 π. 3 π₯ 2 +22π₯+35=0 π₯ 3 π₯ 2 7π₯ π₯+5=0 3π₯+7=0 15π₯ +35 β5 β5 β7 β7 +5 3 3 3π₯=β7 π₯=β5 105 π₯ 2 What are the factors of πππ that also combine to +ππ? π₯=β 7 3
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π. 2 π₯ 3 +4 π₯ 2 +2π₯=0 2π₯( π₯ 2 +2π₯+1)=0 2π₯=0 π₯ 2 +2π₯+1=0 π₯=0
# 56 π. 2 π₯ 3 +4 π₯ 2 +2π₯=0 2π₯( π₯ 2 +2π₯+1)=0 2π₯=0 π₯ 2 +2π₯+1=0 π₯=0 (π₯+1)(π₯+1)=0 π₯+1=0 β1 β1 π₯=β1
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π. 45 π₯ 2 β20=0 5 9 π₯ 2 β4 =0 5 3π₯β2 3π₯+2 =0 π. π₯ 2 β6π₯β27=0 3π₯β2=0
# 56 π π₯ 2 β20=0 5 9 π₯ 2 β4 =0 5 3π₯β2 3π₯+2 =0 π. π₯ 2 β6π₯β27=0 3π₯β2=0 3π₯+2=0 +2 +2 β2 β2 π₯β9 π₯+3 =0 3π₯=2 3π₯=β2 π₯β9=0 π₯+3=0 π₯= 2 3 π₯=β 2 3 +9 +9 β3 β3 π₯=9 π₯=β3
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π. 5 π₯ 3 β20π₯=0 π₯(5 π₯ 2 β20)=0 π₯π₯=0 5 π₯ 2 β4 =0 5 π₯β2 π₯+2 =0 π₯β2=0
# 56 π. 5 π₯ 3 β20π₯=0 π₯(5 π₯ 2 β20)=0 π₯π₯=0 5 π₯ 2 β4 =0 5 π₯β2 π₯+2 =0 π₯β2=0 π₯+2=0 +2 +2 β2 β2 π. π₯ 2 +4π₯β5=0 π₯=2 π₯=β2 π₯β1 π₯+5 =0 π₯β1=0 π₯+5=0 +1 +1 β5 β5 π₯=1 π₯=β5
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What are the factors of βπ that also combine to βπ?
# 57 π₯ +2 FACTOR π. 3 π₯ 2 β6π₯β24 π₯ π₯ 2 +2π₯ 3( π₯ 2 β2π₯β8) β4 β4π₯ β8 π(π+π)(πβπ) β8 π₯ 2 What are the factors of βπ that also combine to βπ?
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This is difference of squares.
# 57 Notice that EVERY term is a perfect square and the operation is subtraction. π. 4 π₯ 2 β9 This is difference of squares. (ππβπ)(ππ+π)
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What are the factors of ππ that also combine to π?
# 57 7π₯ +2 FACTOR π. 7 π₯ 2 +9π₯+2 π₯ 7π₯ 2 +2π₯ +1 7π₯ +2 (ππ+π)(π+π) +14 π₯ 2 What are the factors of ππ that also combine to π?
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