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question:For (mathbf{v} = begin{pmatrix} 2 z end{pmatrix}) and (mathbf{w} = begin{pmatrix} 8 4 end{pmatrix}), [text{proj}_{mathbf{w}} mathbf{v} = begin{pmatrix} 12 6 end{pmatrix}.] Find (z).
answer:To solve for (z), we start by calculating the projection of (mathbf{v}) on (mathbf{w}): [ text{proj}_{mathbf{w}} mathbf{v} = frac{mathbf{v} cdot mathbf{w}}{mathbf{w} cdot mathbf{w}} mathbf{w} ] First, calculate ( mathbf{v} cdot mathbf{w} ) and ( mathbf{w} cdot mathbf{w} ): [ mathbf{v} cdot mathbf{w} = 2 times 8 + z times 4 = 16 + 4z ] [ mathbf{w} cdot mathbf{w} = 8^2 + 4^2 = 64 + 16 = 80 ] Now, plug these into the projection formula: [ text{proj}_{mathbf{w}} mathbf{v} = frac{16 + 4z}{80} begin{pmatrix} 8 4 end{pmatrix} = frac{4(4 + z)}{80} begin{pmatrix} 8 4 end{pmatrix} = frac{4 + z}{20} begin{pmatrix} 8 4 end{pmatrix} = begin{pmatrix} frac{8(4 + z)}{20} frac{4(4 + z)}{20} end{pmatrix} = begin{pmatrix} frac{32 + 8z}{20} frac{16 + 4z}{20} end{pmatrix} ] Set each component equal to the corresponding component in (begin{pmatrix} -12 -6 end{pmatrix}): [ frac{32 + 8z}{20} = -12 quad text{and} quad frac{16 + 4z}{20} = -6 ] Solving the first equation: [ 32 + 8z = -240 quad rightarrow quad 8z = -272 quad rightarrow quad z = -34 ] So, (z = boxed{-34}).
question:f a b = 3 and a^2 + b^2 = 29, find the value of ab.
answer:We can solve this problem by using the given equations to express ab in terms of a and b. First, let's square the first equation: (a - b)^2 = 3^2 a^2 - 2ab + b^2 = 9 Now we have an expression that includes a^2, ab, and b^2. We also have another equation that gives us the value of a^2 + b^2: a^2 + b^2 = 29 Let's subtract the second equation from the squared first equation to isolate the term -2ab: (a^2 - 2ab + b^2) - (a^2 + b^2) = 9 - 29 -2ab = -20 Now we can solve for ab: ab = -20 / -2 ab = 10 Therefore, the value of ab is boxed{10} .
question:A boy has the following eight coins in his pocket: 2 pennies, 2 nickels, 2 dimes, 1 quarter, and 1 halfdollar. He takes out two coins, records the sum of their values, and then puts them back with the other coins. He continues to take out two coins, record the sum, and put them back. How many different sums can he record at most?
answer:The following pairs represent the values of the two coins the boy could take from his pocket: begin{array}{ccccc} (1,1) & (1,5) & (1,10) & (1,25) & (1,50) (5,5) & (5,10) & (5,25) & (5,50) & (10,10) & (10,25) & (10,50) & & (25,25) & (25,50) & & & end{array} Each of these pairs generates a unique sum: - Two pennies: 1+1=2 - Penny and nickel: 1+5=6 - Penny and dime: 1+10=11 - Penny and quarter: 1+25=26 - Penny and half-dollar: 1+50=51 - Two nickels: 5+5=10 - Nickel and dime: 5+10=15 - Nickel and quarter: 5+25=30 - Nickel and half-dollar: 5+50=55 - Two dimes: 10+10=20 - Dime and quarter: 10+25=35 - Dime and half-dollar: 10+50=60 - Two quarters: 25+25=50 - Quarter and half-dollar: 25+50=75 Thus, there are boxed{14} different sums.
question:Sunshine Car Rentals rents a basic car at a daily rate of 17.99 dollars plus 0.18 per mile. City Rentals rents a basic car at 18.95 dollars plus 0.16 per mile. For what mileage is the cost the same?
answer:Let's set up an equation to find out at what mileage the cost of renting a car from Sunshine Car Rentals is the same as renting from City Rentals. Let x be the number of miles driven. The cost for Sunshine Car Rentals would be: Cost_Sunshine = 17.99 + 0.18x The cost for City Rentals would be: Cost_City = 18.95 + 0.16x We want to find the mileage at which the costs are the same, so we set the two costs equal to each other: 17.99 + 0.18x = 18.95 + 0.16x Now, we solve for x: 0.18x - 0.16x = 18.95 - 17.99 0.02x = 0.96 Divide both sides by 0.02 to solve for x: x = 0.96 / 0.02 x = 48 So, the cost of renting a car from Sunshine Car Rentals is the same as renting from City Rentals when the mileage is boxed{48} miles.