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pg189
#5

using each of the digits 2 through 5 only once, write 2 two-digit whole numbers whose product is as large as possible.

#6

using each of the digits 3 through 8 only once, write 2 three-digit whole numbers whose product is as large as possible.

#8

the cube of a whole number is 48 more than the square of the same number. what is the number?

pg 193 #1b
solve. use the method of completeing the square.

m^2= -6m-7

2006-10-22 10:37:42 · 6 answers · asked by koalabear 2 in Science & Mathematics Mathematics

6 answers

pg189
#5: 52 and 43
#6: 843 and 765
#8: 4
pg193 #1b: can't use completing the square....use the quadratic formula.

2006-10-22 10:52:18 · answer #1 · answered by D 3 · 0 0

#5. Suppose the two-digit numbers are ab, cd where a,b,c,d are the digits 2 through 5. Then (ab)*(cd) = (10*a+b)*(10*c+d) = 100ac + 10(ad+bc) + bd. Therefore, we want ac as large as possible, so we choose them to be a=5 and c=4. It remains to choose b and d from 2,3. To make ad+bc larger, choose d to be the larger, 3, so b=2. Therefore the numbers are 52 and 43.

#6. Similar idea, let the two numbers be abc and def. Their product is (abc)(def) = (100*a+10*b+c)(100*d+10*e+f) =
10000ad + 1000(ae+bd) + 100(af+cd+be) + 10(bf+ce) + cf. To maximize the product, choose a=8 and d=7. Then you have
560000 + (8000e+7000b) + 100(8f+7c+be) + 10(bf+ce) + cf, where b,c,e,f come from 3,4,5,6. We chose e=6 since it has the largest coefficient, 8000, and choose b=5. Then you have
646000 + 100(8f+7c) + 10(5f+6c) + cf. Then, choose f=4 and c=3, (choosing f larger because it has the larger coefficient in the second term) to get the two numbers 853 and 764.

#8. Let x be the number. Then x^3 = 48 + x^2.
To solve x^3 - x^2 - 48 = 0, use the rational root theorem, which says that rational roots should be factors of 48. Trying such factors, you'll find that x=4 does the job.

#1b. You have m^2 + 6m + 7 = 0. To complete the square, use the fact that (m+a)^2 = m^2 + 2am + a^2. Matching this with m^2 + 6m + 7, you'll see that you want 2a=6, so a=3.

(m+3)^2 = m^2 + 6m + 9 = (m^2 + 6m + 7) + 2. Therefore,
m^2 + 6m + 7 = (m+3)^2 - 2.

To solve (m+3)^2 - 2 = 0, move the 2 to the other side, and take the square root of both sides. You get m+3 = +/-sqrt(2). Therefore m = -3-sqrt(2) and m = -3+sqrt(2) are the solutions.

2006-10-22 11:00:50 · answer #2 · answered by James L 5 · 0 0

5 52 x 43
6 853 x 764
8 4

2006-10-22 10:47:50 · answer #3 · answered by Anonymous · 0 0

#5
52 x 43

#6
853 x 764

#8
4

2006-10-22 10:51:57 · answer #4 · answered by Philip W 7 · 0 0

#5
52 * 43 = 2236
#6
853 * 764 = 651692
#8
x^3 - x^2 = 48
x = 4

2006-10-22 10:48:34 · answer #5 · answered by jacinablackbox 4 · 0 0

#5 is
53*42 = 2226

#6 is
863*754= 642935

#8 is
4

#1b ???????

try it out to see if this is right hope you find this useful>

2006-10-22 10:57:19 · answer #6 · answered by Movie Star 2 · 0 0

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