Answer:
For complex numbers,
a + bi and a - bi
they have the interesting property that if you add them you get the real number 2a
and if you multiply them , because of the difference of square pattern, you get a^2 - b^2 i^2
But since i^2 = -1, we end up with a real number as a product.
e.g. 6 - 5i and 6 + 5i are conjugates of each other
sum = 6-5i + 6+5i = 12
product = 36 - 25i^2
= 36 -(-25) = 61
Your question is even easier, since the denominator is a monomial instead of a binomial, so we just have to multiply by i/i
Also I believe, according to the answer, that you have a typo, and you meant
(-5+i)/(2i)
= (-5+i)/(2i) *i/i
= (-5i + i^2)/2i^2)
= (-5i +i^2)/-2
= (-5i - 1)/-2
= (1 + 5i)/2 or they way they have it: 1/2 + 5i/2
2: Midpoint of segment AC is given as B3: Given4: Knowing that AC~=BC, AC~= DE
1.8, Problem 37: A lidless cardboard box is to be made with a volume of 4 m3
. Find the
dimensions of the box that requires the least amount of cardboard.
Solution: If the dimensions of our box are x, y, and z, then we’re seeking to minimize
A(x, y, z) = xy + 2xz + 2yz subject to the constraint that xyz = 4. Our first step is to make
the first function a function of just 2 variables. From xyz = 4, we see z = 4/xy, and if we substitute
this into A(x, y, z), we obtain a new function A(x, y) = xy + 8/y + 8/x. Since we’re optimizing
something, we want to calculate the critical points, which occur when Ax = Ay = 0 or either Ax
or Ay is undefined. If Ax or Ay is undefined, then x = 0 or y = 0, which means xyz = 4 can’t
hold. So, we calculate when Ax = 0 = Ay. Ax = y − 8/x2 = 0 and Ay = x − 8/y2 = 0. From
these, we obtain x
2y = 8 = xy2
. This forces x = y = 2, which forces z = 1. Calculating second
derivatives and applying the second derivative test, we see that (x, y) = (2, 2) is a local minimum
for A(x, y). To show it’s an absolute minimum, first notice that A(x, y) is defined for all choices
of x and y that are positive (if x and y are arbitrarily large, you can still make z REALLY small
so that xyz = 4 still). Therefore, the domain is NOT a closed and bounded region (it’s neither
closed nor bounded), so you can’t apply the Extreme Value Theorem. However, you can salvage
something: observe what happens to A(x, y) as x → 0, as y → 0, as x → ∞, and y → ∞. In each
of these cases, at least one of the variables must go to ∞, meaning that A(x, y) goes to ∞. Thus,
moving away from (2, 2) forces A(x, y) to increase, and so (2, 2) is an absolute minimum for A(x, y).
Answer:
2.4
Step-by-step explanation:
RSQ is a right anle triangle
tan(Q) = opposite/adjacent
tan(Ql = SR/QS
tan(Q = 12/5 = 2.4
Answer:
c
Step-by-step explanation:
Uh, you got it right.
8,2 and 8,7
The differnece between 7 and 2 is 5