. Then 2 3 the circumference is 8π 25 in. 2(1) both of which violate the original assumptions. Therefore, we have no solution.

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Cathleen Golden
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1 WYSE MATH SECTIONAL 04 SOLUTIONS Ans B: (a) and (d) are equivalent Therefore since the determinant of the product of matrices is the product of their determinants, it too cannot be zero If three square matrices of the same dimensions are multiplied together, the result has the same dimensions The only thing which isn t necessarily true is (b), as if there are three matrices with a negative determinant or a set of two with positive determinant and one with negative determinant, the product would have a negative determinant mt r 005 Ans D: Using FV = PV + m and INT = FV PV, we get FV = FV = INT = So the interest is about $88 Ans B: The equilibrant vector to a set of vectors is the opposite of their sum as it serves to balance them The sum of these three vectors is (a) so the equilibrant would be (b) 4 Ans D: Since CGE is 60, then CAE is 0 which represents angle measure of the circle So arc CE represents the circumference of circle A Then arc EC represents circumference of the circle C = ( 6) Then the circumference is 8 5 in A B 5 Ans E: = A(A + C) = B(B + C) A + AC = B + BC B+ C A + C C ± C 4()( B BC) C ± C + 4BC + 4B A + AC B BC = 0 A = A = () C ± (C + B) C ± (C + B) A = A = Giving us solutions A = B and A = C B, both of which violate the original assumptions Therefore, we have no solution 6 6 Ans D: By the Chain Rule, d d sin cos cos d + = + d + = Ans E: gives us + y ( y + + y + y 4 ( + ) y ( y + y + + y + y 4 ( ) 4 ( ) Then ( y + y + y y y y y y + y + y + y + y y 8 Ans C: Use Pythagorean theorem twice, once to find the diagonal of a face, then to find the distance between the opposing vertices The diagonal of a face is, the distance between the opposing vertices is Ans C: r = a + b θ has a distance between successive turns of b

2 0 Ans C: 6s 45 s 9 s = = = Then V = = V 7 Ans C: If P dollars are invested at an interest rate of r (written as a decimal, not a percentage), compounded n times per year for t years, the approimate total in the nt r account after that time is A P = + n In this scenario, P = 6000, r = 09, n = 4 and t = 7 Please note that this would give us (d), but that counts the principal as well as the interest Ans A: Suppose the rd tree is (c) 5 ft Then using Heron s formula we find that s = 05( ) 75 So A = 75( 85)( 5)( 5) But we can t have a negative under the radical so the distance must be greater than 5 ft We then assign s = Then the area is the distance to be ft Here A = 4( 4 4)( 9) But this produces 0 The distance must be greater than ft The distance must be 40 ft With this length the value s = 05( ) 45 The area is A = 45( 6)( 4)( 5) 80 This is the given area The distance is 40 ft Ans C: Simply solve sinθ = 8 and get the angle as 50 degrees 0 a b 4 Ans C: For an ellipse, the eccentricity is e = It s easier to find a and b if the a y ellipse is placed in the form + = (We ll see that the major ais is on the side a a b 4 9 little bit down the road) First, we divide the equation by 5 to get + y = This 5 5 y 5 is equivalent to + = So a 5 5 = 4 and b =, which means e = By multiplying the numerator and denominator of the fraction by 6 and then subsequently dividing each by 5, we get e = = = Ans B: ( )( + ) + ( ) 6 Ans D: Out of all students, 080*070, or 56% both own their own car and have jobs Out of the 60% that have cars, this represents 56/60, or about 9% 7 Ans C: The area of a triangle when two sides and the angle included between them are given is A = ab sinθ

3 8 Ans A: r = 09 and a = 8 Then 8 8 S = Ans D: Let θ be the angle opposite the side with measure 5 Let a be the measure of the altitude On the large triangle, the tangent of θ will be 5 On the small triangle which is created with a hypotenuse of 5, the tangent of θ will be a So 5 5 = and = a a By the Pythagorean theorem on that small triangle, a + = 5 and thus a + a = 5 Then a + a = 5 and a = 5 So a = and a = Ans B: h'(t) = t + 80t tracks speed Solving t + 80 = 0 gives us t = 5 Ans D: There are C(, ) ways to pick the denominations which will be paired and then there are C(4, ) ways to pick the two cards of those denominations There are C(, ) ways to pick the three denominations of which there are single cards chosen, and four ways to choose the single cards from each of those denominations Then by the multiplication principle, there are C(,) C(,) C(4,) 4 = 9,65,480 ways to choose the hand with two pairs in it Ans B: Using Cramer s Rule, numerator is d b c d b c d b c = The determinant of the 4 4 8( 8 + ) ( 0 ) + ( 5 6) = 4 sin Ans E: 6 sin tan = 0 sin = tan sin = sin cos = sin cos 5 Then sin = sin So = 0,,, Only is one of the listed solutions 4 Ans E: The first valve fills the tank at a rate of /50 = 00 tank per minute, the second fills at a rate of /40 = 005 tank per minute, and the drain empties the tank at /5 = 004 tank per minute Since the two valves have a greater combined rate than the drain, the tank will fill when all three are open In the first 0 minutes, the tank fills to 00*0 = 0 tank After 0 minutes, the tank is up to 0 + 0*( ) = 065 tank Solving the equation = *( ) tells us that the tank fills up 70 minutes after the drain was opened, which would be :0 PM

4 5 Ans C: There are! ways to order the ones who aren t the top three and! ways to order the top three The product of! and! is (c) y 6 Ans C: The parametric equation (I) produces = cos( t ) and = sin( t ) Using the identity cos sin y t + ( t) =, by substitution we have + = Similarly the y second parametric equation produces = cos( t ) and = sin( t ) Using the identity cos ( ) sin y t + ( t) =, substitution gives us + = Both give the same equation for an ellipse Finally, making a table of points for both parametric equations we find the following: I II t y t y Using the graph below, we can see the first parametric equation travels a quarter of the ellipse in a clockwise direction from time t = 0 to t = while the second equation, travels half of the ellipse in the same time 7 Ans D: There are! possible arrangements of those keys in a ring There are 0! ways to arrange the non-triplicate keys and! ways to arrange the triplicate keys Thus there are,77,800 ways to arrange the keys in a way that the three are net to one another, so there s a in chance that they re net to one another (and a in chance that they re not all found together) 8 8 Ans A: 4 4 Then 8 4 = 4 = 4 9 Ans A: For a hemisphere, V = r, and for a cone, V = r h The given information tells us r = 5 and h = 65, so V = Ans D: Angle measure, betweenness and collinearity are all preserved Distance is not as size change means that distances will be changed

5 log log5 = log log5 = log5 + 4 = Rewrite as an eponential equation we have 0 = = + 4 = 0 Ans B: Ans D: The radius of the cylinder would be while its height would be 4 So, for the cylinder, the volume would be V = r h = 456cubic inches For the sphere, the 4 volume would be V = r = 04 cubic inches, which is two-thirds of the volume of the cylinder Ans C: The quadrilateral must be a parallelogram due to similar triangles, but none of the other properties are guaranteed simply from having the diagonals bisected 4 Ans C: 00 + i 00 + i 0i + i 7i 5 Ans A: By the Remainder Theorem, when a polynomial P() is divided by a simple linear function Q() = c, P(c) is the remainder P(-7) = Ans D: 9 C 6 C = 60 7 Ans B: The additional cost per item is ( )/( ) = $05 per item For the additional 500 items, it will cost 500*05 = $50 above the $500, or $750 A quick check also reveals that the equation is cost = 05 * items Ans D: We first need to multiply out the left-hand side to get = 6 If we add 6 to both sides, we get = 0 Factoring the left-hand side gives us ( 5) = 0 Then has to be 5 e dy e 9 Ans E: y = ln lne ln ( e + ) y = ln( e + ) = e + d e + e + e = e + e + 40 Ans A: Based on I, II, and III, the nine individual event values must be $0, $0, $0, $40, $50, $60, $70, $80, and $90 The three event totals must be $40, $50, and $60 Based on V, the si individual totals must be $50, $60, $70, $80, $90, and $00 First place who was at what event(s) B was at RS and GT, and E was at RS and FS Since A was also at two events, she must be at FS and GT F must be at FS D must have been at GT with A, which means C was at RS If we call E s FS value, then A s FS value is + 40, and F s FS value is + ( + 40) Adding these up gives us If we set this equal to the three possible totals, = 40 gives = 5, = 50 gives = 75, and = 60 gives = 0, the only acceptable result E s FS value is $0, A s FS value is $60, and F s FS value is $80 This means E s RS value must be $0 To have a gap of $50 between A and D at GT, A s GT value must be $40 and D s must be $90 To get a gap of $40 for B s event values, $0 must be his GT value to make the total $40, leaving $50 for his RS value By default, C s RS value is $70 For the final totals, we have E = $50, B = $60, C = $70, F = $80, D = $90, and A = $00

( 3x 2 y) 6 (6x 3 y 2 ) x 4 y 4 b.

( 3x 2 y) 6 (6x 3 y 2 ) x 4 y 4 b. 1. Simplify 3 x 5 4 64x Algebra Practice Problems for MDPT Pre Calculus a. 1 18x 10 b. 7 18x 7 c. x 6 3x d. 8x 1 x 4. Solve 1 (x 3) + x 3 = 3 4 (x 1) + 1 9 a. 77 51 b. 3 17 c. 3 17 d. 3 51 3. Simplify