Algebra1 Team FAMAT January Regional January Algebra 1 Team Answers 9.A. % & G5

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1 January Algebra 1 Team Answers 1 A. (2% + 3)(2% 3) or (3 + 2%)( 3 + 2%) B. (2% 3)(% + 3) (2% 3)(% 4) D. (2% 3)(% + 1) 2. A. 16 B. G W 4 D. G >.A. % G5 % 12 or 2% 13% 24 B. % + G % + 5 or 4% + 21% + 20 % + 15% D. % 5 5 % + 10 or 3% 23% A. 100 B D A. 1 B D A. % 1 B. % > 6 % 3 D. % 2 5.A. N = 6(% 3) 50 B. (0,4) STU (6,4) - must be written as ordered pairs 1200 D. 2 6.A. % 4N = 8 B. 3% + 4N = 16 % + N = 5.5 D. % + 3N = 26 7.A. (12, 43) must be written as ordered pairs B. ( 1, 11) (15, 41) D. (2, 3) 8.A. $28.64 B. $1.8 $25.25 D. $ A. % = 0, % =, % = 2 2, % = 2 2 B. % = 3, % = 0, % = 1 % = G 5, % = 5 D. % = GG, % = GG 12.A. 2% 5 15% + 14 B. 6% + 17% 5 34% + 30% 12 8% 5 27 D. 125% A. % + B. (;:K)(;:) ;<X ;<5 D. 14.A. 4 B. 4 3 D. 5 ;<> G ; - :> 15.Answers must be in fraction form YZ 1 > K B. I YZ 6 G YZ 4 G > > D. 6 A. G K

2 January Algebra 1 Solutions: 1. A. + 4% = ( [ + \])([ + \]) or (2% + 3)(2% 3) By using the commutative property of multiplication, the factors become: (\] [)(\] + [) B. write so that it is in standard form. 2% + 3% = (\] [)(] + [) 2% 11% + 12 = (\] [)(] ^) D. 2% % 3 = (\] [)(] + _) 2. A. (*+,- )( + ) - = 4*+,- 6( -+ ) = 2 ((5;<)<(;):(>;:)) using the power of a power rule and the quotient rule of exponents. Now work with just the exponent expressions: 3% % (5% 2) = 5% + 2 5% + 2 = 4 This gives: 2 = _` = a45* ,- 28 6b B. 345* , by using the power of a power rule Since the bases are the same; 3, we can pull out the exponents and work with the expressions to simplify the exponents using the multiplication and quotient rules. 3% + ( 1)(2% + 2) ( 5%) = 3% 2% 2 + 5% = 5% + 5% 2 = 2 This gives the following when placed as the exponent on the base of 3. 3 : = G = _ 5 - c = 452* (-+,- 6 *+,8 ( +,8 ) - 4:;< = 4*+, ,- 6 -(+,8) Now we can work with the exponents and simplify the expressions using the multiplication and quotient rules of exponents. 3% ( % + 2) 42(% + 1)6 = 3% + 1 % + 2 (2% + 2) = 2% + 3 2% 2 = 1 This gives: 4 G = ^ > > 2*+ > 2+2-@ = -(2?+2*) = > 2d+2e = >2d+2e >2d+2e = (> -(2*+) )(> -(2+2-) ) (> 2e+ )(> 2-+2? ) > 2e+,(2-+2?) > 2d+2? D. = >2?+2* Now we can work with the exponent expressions and simplify. 8% 6 ( 8% 4) = 8% 6 + 8% + 4 = 2 This gives the following: 5 : = G > - = _ \f 3. A. Use the slope formula to find the slope of the line that passes through (4, 6.5) STU (6.5, 4) :.>.> = 1 for the slope of the perpendicular line, you need the opposite reciprocal of 1 which is _ B. To find the equation of the perpendicular line, use the slope found in part A, and you can do the g:x.> :X.> = X.>: following: = 1 >>>> N 6.5 = % 4 >>>>> N 2.5 = % ;: Solving for x, makes it easy to find the x intercept, which is \. f or ( \. f, h) Using the equation from part B and solving it for y gives: N = % substitute 2 for x and solve. N = = ^. f D. From the slope intercept form of the line from part C, the y intercept is A. 4 3(% + 1) + 7% + 1 :>;<G<GG; 3 >>>> 4 3% 3 + 7% + 1 X;<G % 3% >>>>>> 2 + 4% 3% + 3 >>>>> ] _ B. 3% + 6% + 7% 14% % > 3 5:GI<G :K<G (2 3) + >>>> 16% 15% > K:5 >>>>> % > :X >>>>> % > ( 3) >>>>> % > + 3 >>>> ] > ` 5(>;:5): (L;:5): >>>> G>;:W: GX;:X: >>>>> G>;:GG :(G5:G) : * :(:G) 1 W:L :(:G) (15% 11) 16% + 8 >>>>> 15% % + 8 >>>> ] [ GX;:L :G >>>> D. M 5 L 15% (5 2) 2(2%) (2 + 3)% 15% => + 15% = > + >>>>> 2 15% 3 4% 5% 15% X >>>> 4 15% (3) 4% (5%) >>>> 1 15% 16% 3 >>> ] \

3 5. A. N = 3(% + 2) + (3% 4) multiplying by FOIL gives: N = 3(% + 4% + 4) + (% 24% + 16) N = 3% 12% 12 + % 24% + 16 N = 6% 36% + 4 this is standard form. In order to transform this to vertex form, you need to complete the square. >>>>> N 4 = 6% 36% >>>> N 4 = 6(% 6%) >>>> N 4 + 6() = 6(% 6% + ) = = ( 3) =, but you are really adding 6() to both sides. N = 6(% 3) >>>> N + 50 = 6(% 3) >>>> j = `(] [) \ fh B. To find the points of intersection with the line N = 4, substitute 4 for y and solve for x. 4 = 6(% 3) 50 >>> 54 = 6(% 3) >>> = (% 3) >>>> ±3 = % 3 >>>> 3 = % 3 STU 3 = % 3 >>>> % = 6 STU % = 0 (0,4) (6,4) The discriminant is found using: l 4Sm. From the standard form of the equation found in part A; S = 6, l = 36, m = 4 >>>> ( 36) 4(6)(4) = = _\hh D. With the vertex located at (3, 50) and opening up, it will cross the x axis 2 times or twice. 6. A. Parallel to N = G % + 2 and x-intercept of 8. Since our new line must be parallel to the given line it will have the same slope of G and writing the x intercept as an ordered pair ( 8,0) we have enough information to write the equation of our new line. >>>> % + 8 = 4N >>> ] ^j = n g:i ;:(:L) = G >>>> g ;<L = G >>> 1(% + 8) = 4N B. Perpendicular to N = % + 17 and passes through ( - 8, 2 ). Since our new line must be perpendicular to 5 the given line, it must have the opposite reciprocal of the slope of the given line. Our new line has slope of: using this and the point ( 8,2) we can find the line of our new line. 5 g: = 5, place the negative sign with the 4, so that when you cross multiply x will be positive. ;:(:L) 3(% + 8) = 4(N 2) >>> 3% + 24 = 4N + 8 >>> [] + ^j = _` Our line needs to pass through the points (0, 5.5) STU (4.5, 10). We need to find the slope first. o = :GI:(:>.>).>:I of the line. >>>> = :GI<>.>.> g:(:>.>) ;:I =.> = 1. Now we use the slope and one of the points to write the equation.> = 1 >>> % = N >>> % N = 5.5; multiply both sides of the equation by 1, so that x will be positive. ] + j = f. f D. Has a slope of G and passes through (2, 8). 5 g:l = G >>> % 2 = 3(N 8) >>> % 2 = 3N + 24 >>> ] + [j = \` ;: 5 7. A. Use the elimination method and multiplying the top equation by -1 gives the following system. 2% N = 1 so that when you add the equations together you will get: 5% + N = 17 3% = 36 >>> % = 12; substitute this value into the second equation and solve for y. 5(12) + N = 17 >>> 60 + N = 17 >>> N = 43 Solution: (_\, ^[) % + N = B. ; you can just add the two equations to start. 4% N = 15 6% = 6 >>> % = 1; now substitute the first equation and solve for y. 2( 1) + N = >>> 2 + N = >>> N = 11 Solution: ( _, ) N = 4 3% this system is set up nicely for the substitution method. 5% + 2N = 7

4 5% + 2(4 3%) = 7 >>> 5% + 8 6% = 7 >>> % = 15 >>> % = 15; substituting this value into the first equation gives: N = 4 3(15) >>> N = 4 45 >>> N = 41 Solution: (_f, ^_) N + % = 4 D. R ; this system is set up nicely for elimination; adding the two equations gives: N % = 5 3N = >>> N = 3; substituting this value into to first equation gives: 2( 3) + % = 4 >>> 6 + % = 4 >>> % = 2 Solution: (\, [) 8. A. %.45% = >>>. 55% = >>> % = The original price of jacket was $28.64 B. %.35% = 12. >>>. 65% = 12. >>> % = 1.8 The original price of jacket was $1. %.2% = >>>. 8% = >>> % = The original price of jacket was $25.25 D. %.7% = 47.5 >>>. 3% = 47.5 >>> % = The original price of jacket was $15.83 A. Zeros at % = 5 STU % = 8, this means that the factors are: =% + STU (% 8). By using the FOIL method you will get the equation in standard form. N = =% + (% 8) >>> N = % 8% + 5 % + (8) >> N = % GX % + 5 % + >> j = ]\ _[ ] + _\ or \ 2% 13% 24 B. Zeros at % = 4 STU % = >, this gives factors of: (% + 4) STU =% + By using the FOIL method you will get the equation in standard form. N = (% + 4) =% + >>> N = % + > % + 4% + 4 >>> N = % + > % + GX % + 5 >>> j = ]\ + \_ ^ ] + f or 4% + 21% + 20 Zeros at % = 0 STU % = 15, this gives factors of: % STU (% + 15). By using distribution, you will get the equation in standard form. N = %(% + 15) >>> j = ] \ + _f] D. Zeros at % = > STU % = 6, this gives factors of: =% STU (% 6). By using the FOIL method, you will get the equation in standard form. N = =% (% 6) >>> N = 5 % 6% > % + = ( 6) N = % GL % > % + 5I 5 >>> j = ] \ \[ [ ] + _h or 3% 23% Since there are 4 tests, take the average that they want and multiply by 4 and set it equal to the sum of the 3 tests plus the unknown test grade. A. 4(85) = % >>> 340 = % >>> 100 = % He needs to score 100 on the test. B. 4(80) = % >>> 320 = % >>> 84 = % He needs to score 84 on the test. 4(2) = % >>> 368 = % >>> 3 = % She needs to score 3 on the test. D. 4(75) = % >>> 300 = % >>> 53 = % She needs to score 53 on the test. 11. A. use factoring by grouping to solve. % % 5 8% + 72% = 0 >>> (% % 5 ) + ( 8% + 72%) = 0 >>> % 5 (% ) 8%(% ) = 0 (% 5 8%)(% ) = 0 >>> %(% 8)(% ) = 0 >>> % = 0 YZ % 8 = 0 YZ % = 0 % = 8 YZ % = >>> % = ± 8 = ±2 2 Solutions: ] = h, ] = c, ] = ±\ \ B. 3% 5 + 6% % = 0 >>> 3%(% + 2% 3) = 0 >>> 3%(% + 3)(% 1) = 0 >>> 3% = 0 YZ % + 3 = 0 YZ % 1 = 0 Solutions are: ] = h qr ] = [ qr ] = _ 3% 14% 5 = 0 >>> (3% + 1)(% 5) = 0 >>> 3% + 1 = 0 YZ % 5 = 0 >>> 3% = 1 YZ % = 5 Solutions are: ] = _ [ qr ] = f D. % 121 = 0 >>> (3% + 11)(3% 11) = 0 >>> 3% + 11 = 0 YZ 3% 11 = 0 >>> 3% = 11 YZ 3% = 11 Solutions: ] = qr ] = [ [

5 12. A. (% 2)(2% + 4% 7) >>> = 2% 5 + 4% 7% 4% 8% + 14 = \] [ _f] + _^ B. (2% 3% + 6)( 3% + 4% 2) = 6% + 8% 5 4% + % 5 12% + 6% 18% + 24% 12 = `]^ + _s] [ [^] \ + [h] _\ (2% 3)(4% + 6% + ) = 8% % + 18% 12% 18% 27 = n] [ \s D. (5% + 4)(25% 20% + 16) = 125% 5 100% + 80% + 100% 80% + 64 = _\f] [ + `^ 13. Factor everything you can, then reduce the fractions. A. ;- <L;:W ;:G = (;:G)(;<W) ;:G = ] + cb. (;:K)(;- :) = (;:K)(;<)(;:) = (]:s)(]:\) ; - <L;<G (;<)(;<X) ]<` (;<5) - = (;<5)(;<5) = ]<[ ; - <L;<G> (;<5)(;<>) ]<f D. ;- <> ;? :> = ; - <> (; - :>)(; - <>) = _ ] \ :f 14. A. N = 3 7% 5 + 2%(% 4)(4% + 3% 1) In this function you an exponent of 3, but that is on a term that is not a part of any multiplication. Now, looking at the multiplication part, you only need to worry about the first terms in the parenthesis as these all have variables with exponents. 2%(%)(4% ) = 8% The degree of this function is 4. B. N = 6%(2% + 5) 8% 5 (5% + 3) >>> N = 6% + 30% 40% 24% 5 The degree of this function is 4. N = 3% :G (2% %) + (% 4)(% + 4) >>> N = 6% % 5 + 4% 4% 16 There are two terms with degree 3, so the degree of the function is 3. D. N = (2% + 5)(3% 5 5% + 2) >>> 2% (3% 5 ) = 6% > > degree of the function is A. G + G = G >>> 4u + 3u = 12 >>> 5u = 12 >>> v = _\ = _ f wqxry 5 t s s B. G + G = G >>> 2.5u + 4(2.5) = 4u >>> 2.5u + 10 = 4u >>> 10 = 1.5u >>> 10 = 5 u >>> 10 t.> = u 5 v = \h [ qr ` \ [ wqxry G 5 + G t = G G.K> >>> 1.75u + 3(1.75) = 3u >>> 5.25 = 1.25u >>> u = 4.2 > 4 GI >>> v = ^ _ \_ qr wqxry f f D. G 5 + G ; = G >>> 2u + 6 = 3u >>> v = ` wqxry