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The following problems are to be written up in polished form, showing all relevant work.

2. A manufacturer purchases a key component used in the production of their product from two different suppliers. (They want to keep their flexibility by maintaining a relationship with both suppliers.) The cost for this key component from supplier A is $5.25 per unit. From supplier B, the component cost is $4.80 per unit. The manufacturer needs to 7500 units and has $37,350 available to make the purchases. Set up a system of equations to describe this situation. Solve the system of eqautions to find the solution.
3. On a sunny day, a building and its shadow form the sides of a right triangle. If the hypotenuse is 32 m long and the shadow is 24 m, how tall is the building?
4. An airplane leaves an airport and flies due west 120 miles and then 210 miles in the direction 220°30′. Assuming the earth is flat, how far is the plan from the airport at this time (to the nearest mile)? (Hint: draw a picture of this with north pointing up.)

2007-11-29 04:06:30 · 2 answers · asked by Nicole 4 in Science & Mathematics Mathematics

2 answers

x = number of components from supplier A
y = number of components from supplier B
x+y = 7500 <-- Eq 1
5.25x + 4.8y = 37,350 <-- Eq 2
-5.25 x - 5.25y = -39,375 <-- Eq 3 = -5.25 * Eq 1
-4.5y = -2025 <-- Eq2 +Eq 3
y = 450 = B supplied components
7500-450 = 7050 = A supplied components

height = sqrt(32^2- 24^2) = 8sqrt(7) m

Use law of cosines.
d = sqrt( 120^2+210^2-120*210cos 139.5)

2007-11-29 04:29:29 · answer #1 · answered by ironduke8159 7 · 0 0

2) a) x+y=7500, equation 1
b) 5.25x + 4.8y = 37350, equation 2
c) multiply equation 1 by 5.25 to get:
5.25x+5.25y = 39375, new equation 1
d) subtract equation 2 from new equation 1 to get:
.45y = 2025
e) y = 4500, number of components to buy from vendor y
f) 7500-4500 = 3000, number to buy from vendor x

3) a) a^2 + b^2 = c^2, Pythagorean theorem
b) 24^2 + b^2 = 32^2
c) b^2 = 448
d) b = 21.166

4) a) COS angle c = (a^2 + b^2 - c^2) / 2ab
b) angle c = 130.5°, COS(130.5) = -.64945
c) -.64945 = (120^2 + 210^2 - c^2) / (2(120)(210))
d) -.64945 = (14400 + 44100 - c^2) / 50400
e) -32732.28 = 58500 - c^2
f) -91232.28 = -c^2
g) c = 302.7337 miles

2007-11-29 05:03:46 · answer #2 · answered by Anonymous · 0 0

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