This is false, -5(3e-2) is not less than -2(6e+7)
Answer: See below
Step-by-step explanation:
27. -(a-3)
28. (b-1)(b+3)
29. (c+4)(c+5)
30. d(d+5)
31. -(3/4)(2e-5)
Sorry - I don't have time to enter the details. Look for areas where the expressions can be factored in a manner that forms as many equivalent expressions in both the numerator and denominator.
For example: In problem 30:
(5d-20)/(d^2+d-20) * [??]/20d = 1/4
Factor:
<u>(5(d-4))</u> <u>d(d+5)</u> = 1/4
(d-4)(d+5<u>)</u> 20d
The (d-4), d+5, and d terms cancel, leaving
5/20 = 1/4
Answer:
22 onces
Step-by-step explanation:
16 ounces in a pound
5-3=2
2x16=32
32-10=22
Answer:
67
Step-by-step explanation: Given the quadratic equation $z^2 + bz + c = 0$, Vieta's formulas tell us the sum of the roots is $-b$, and the product of the roots is $c$. Thus,
\[-b = (-7 + 2i) + (-7 - 2i) = -14,\]so $b = 14.$
Also,
\[c = (-7 + 2i)(-7 - 2i) = (-7)^2 - (2i)^2 = 49 + 4 = 53.\]Therefore, we have $b+c = \boxed{67}$.
There are many other solutions to this problem. You might have started with the factored form $(z - (-7 + 2i))(z - (-7 - 2i)),$ or even thought about the quadratic formula.
This is the aops answer :)
