2008 MATH 436 PART C

17) The area of the rectangle is the polynomial 5x² + 38x – 63. The diagonal measures 52 cm. Find the perimeter.

Solution:

5x² + 38x – 63 = (5x – 7)(x + 9), so we set base b = 5x – 7 and height h = x + 9.
Now, b² + h² = 52² which means (5x – 7)² + (x + 9)² = 2704
We expand to get: 25x² – 70x + 49 + x² + 18x + 81 – 2704 = 0
We collect terms: 26x² – 52x – 2574 = 0 (all constants are multiples of 26)
We divide by 26: x² – 2x – 99 = 0 and then we factor (x – 11)(x + 9) = 0
This means x = 11, so the base = 48 and the height = 20.
Perimeter = 2(b + h) = 2(68) = 136 cm.

18) f(x) = – 2/3 x + 6. V is the vertex of the parabola and the x-intercept of f.
We know that f(63) = g(16). Find the rule of correspondence for the parabola (and the coordinates of point P).

Solution:

f(63) = g(16) means that the y-value on the line at x = 63 is the same as the y-value on the parabola when x = 16. Since f(x) = – 2/3 x + 6, f(63) = – 42 + 6 = – 36.
We now know that (16, – 36) is a point on the parabola.
The vertex is the x-intercept of the line so it is (9, 0).
Now we find the rule: g(x) = a (x – 9)² + 0 since (h, k) = (9, 0).
Now substitute (16, – 36) to find a: – 36 = a (16 – 9)²
The rule of correspondence is

19) A print shop gets an order for 212, 000 items that are either newspapers, magazines or books. The job is worth \$61, 650.00 and the shop charges \$0.12/newspaper, \$0.60/magazine and \$3.90 per book. In this job, the number of magazines is equal to the number of books and there is also a different number of newspapers to print. The print machines break down after printing
153, 214 of the newspapers for the job. How many more newspapers must they print to complete that part of the job?

Solution:

Let n = the # of newspapers to print for the job
Let m = the # of magazines = the # of books to print for the job.
Our system of equations is:
n + 2m = 212, 000, for the count
0.12n + 0.60m + 3.90m = \$61, 650, for the value.
We'll substitute n = 212, 000 – 2m into the 2nd equation.
0.12 (212, 000 – 2m) + 0.60m + 3.90m = \$61, 650
4.26m = 36 210 so m = 8500. From n = 212, 000 – 2m, we get n = 195 000.
The number of newspapers left to print is 195 000 – 153, 214 = 41 786.

20) Line 1 (l1) and Line 2 (l2) are parallel to each other while Line 3 (l3) is parallel to the y-axis. The equation of Line 1 is 5x + 4y – 564 = 0. Points D and E are on Line 2 with D as the intersection point of Line 2 and Line 3. The coordinates of D and E are (32, 75) and (72, 25) respectively. Point P is of the way from D to E. Determine if P is closer to Line 1 or Line 3.

Solution:

First, we find the coordinates of P which is 2/5 the way to E.
Change in x = 40, Change in y = 50,
so FP = 2/5(40) = 16, and DF = 2/5(50) = 20.
We add 16 to 32 to get 48 and we subtract 20 from 75 to get 55.
Point P is (48, 55). We already know it is 16 units from Line 3.
Now we find perpendicular distance between P and Line 1.
units from Line 1.
Point P is a teeny weeny bit closer to Line 3 than it is to Line 1.

22) A carpenter is going to cut off the top of a right circular cone with height H = 17.6 dm. and base radius r = 4.8 dm. to use the frustrum or base to build a table. The volume of the frustrum is 106.16 dm³. Find h the height of the small cone he cut from the big one.

Solution:

First, we find the Volume of the whole cone using
Volume (big cone) :
Now we find the Volume of the small cone by subtraction:
Volume (small cone) : 424.64 – 106.16 = 318.48 dm³.
The quotient of 318.48 and 424.64 gives us the level 3 ratio of small to big.
We find k ³ = 0.75 which means the ratio of h to H is the cube root of 0.75 = 0.9085.
This tells us that h = 0.9085 × 17.6 = 16 dm.

25) There was a long story about a mouse and a motion detector. Here's the picture.

Solution:

First, we find angle AEF using the law of Cosines.
We know that ,where a is the side opposite the angle we need.
From the triangle AEF:
Angle AEF = arccos ( – 0.7882) = 142°. This makes angle AED = 180° – 142° = 38°.
In triangle ADE: x = 4 cos 38°, and y = 4 sin 38° so the area of triangle ADE is ½(xy).
Area = ½(4 cos 38°)(4 sin 38°) = 3.88 units².