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Math Skills Review I Subjects include: Algebra I/IIAlgebra I/II GeometryGeometry TrigonometryTrigonometry Christopher Bullard 4/17/2007.

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Presentation on theme: "Math Skills Review I Subjects include: Algebra I/IIAlgebra I/II GeometryGeometry TrigonometryTrigonometry Christopher Bullard 4/17/2007."— Presentation transcript:

1 Math Skills Review I Subjects include: Algebra I/IIAlgebra I/II GeometryGeometry TrigonometryTrigonometry Christopher Bullard 4/17/2007

2 Question #1 John is at a dealership to buy a used car. There are two vehicles he is choosing from (an 86 four door sedan, and an 89 two door coupe). All cars at the dealership have a 10% mark-up of their fair market value. The 89 coupe has a dealership price $500 more than the fair market value of the 86 sedan. The 86 sedans dealership price is $4,950. What is the fair market value of the 89 sedan?

3 Solution to Question #1 First, we must determine the fair market value of the 86 sedan. To do this, we set up the equation : 1.1x=4950 where x represents the 86 sedans fair market value Solving for this yields x=$4,500 Next, we know that the 89 sedans dealership price is $500 more than the 86 sedans fair market value. Therefore, we set up the equation: x+500=1.1y We already know what x is, so solving for y, we find the fair value of the 89 sedan to be $4,955, rounded to the nearest dollar.

4 Question #2 A box has a volume of 2,000 cm 3 (cubic centimeters, where 2.54 cm = 1 inch), and the vertical surface adjacent to A has surface area of 250 cm 2 (10cm x 25cm). Given angle A= 30 degrees, find the length of the line AB.

5 Solution to Question #2 First, our objective is to solve for the hypotenuse of the right triangle formed on the floor of the box (the right triangle with adjacent side of length 25cm, and opposite side of length 8cm). We can either use trig functions (sine, or cosine), or the Pythagorean Theorem. Either way, the result is the square root of 689 (we will leave it in this form until after we have proceeded with the second step). The next step involves solving for the right triangle now formed by this new side (call it AF), and the height of the box (10cm). Solving for this right triangle will yield ABs length of cm (the square root of 789). Note, however, that we could also solve this problem if only given the surface area of the 25cm by 10cm side, and the total volume of the box. We could obtain the length of 8cm by dividing the volume by the surface area, and then obtain the hypotenuse of the floor of the box by the cotangent of 30 degrees, multiplied by division of volume by area.

6 Question #3 Find the value of the variable x from the equation 37 = log x 80 Question #4 Find the value of the variable y from the equation 42 = 100 – 6 log 7 y Question #5 Find the value of the variable z from the equation 91 = ln (13/z)

7 Solution to Question #3 Given log a b = (log b)/(log a) (34/47) = (log 10 80)/ (log 10 x) so… x = 10 [(log 80)/(34/47)] Solution to Question #4 Given log a b = (log b)/(log a) (-58/6) = (log y)/ (log 7) so… y = 10 (-58/6) (log7) Solution to Question #5 Given ln(a/b) = ln(a) – ln (b) ln (13) - 91 = ln (z) So… z = e (ln (13) - 91 )

8 Question #6 Given the following equations (1) 3w – 2x + 7y +z = 108 (2) w + x – 2y +3z = 372 (3) 2w + z = 13 (4) x + y = 7 Solve for w, x, y, and z, listing them in the form (w, x, y, z).

9 Solution to Question #6 We must perform either row operations or substitution with equations (3) and (4) to simplify what we have. For this example, we will begin with row operations to eliminate unnecessary variables. For convenience, we will eliminate either w or z from equation (2) or (1), respectively. This results in… (1) x + y +10z = 1224 (2) w + x – 2y +3z = 372 (3) 2w + z = 13 (4) x + y = 7 We can now use equation (4) to reduce the equation (1) to… (7 – y) + y +10z = 1224 which now lets us determine… z = 1224 where z = Next, since we know z, we can substitute into equation (3) to determine… 2w = 13 – Which results in… w = Finally, given w and z, we can use equations (2) and (4) to determine y, which results in… (2) x – 2y +3(121.7) = 372, or… x – 2y = 372… where we can substitute equation (4) to get… (Continued on next slide!)

10 Solution to Question #6 (continued) Finally, given w and z, we can use equations (2) and (4) to determine y, which results in… (2) x – 2y +3(121.7) = 372, or… x – 2y = 372… where we can substitute equation (4) to get… (7 – y) -2y = 372, or… -3y = 54.25, leaving y = and x = 7 – y, or x = Leaving the solution to the system of equations to be… (-54.35, , , 121.7)

11 End of Math Skills Review I Contact information: Website: AIM: goatman95111


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