The answer to this one would be B
A.)
Z = kx/y where k is a constant.
32 = 16k/4
k = 128/16
k = 8
z = 8x/y
b.)
Z = 8(21)/6
Z = 28
c.)
4 = 8x/48
x = 192/8
x = 24
Answer:
FALSE, (2, 9) is not a solution to the set of inequalities given.
Step-by-step explanation:
Simply replace x by 2 and y by 9 in the inequalities and see if the inequality is true or not:
irst inequality:

so thi inequality is verified as true since 9 is larger or equal than 8
Now the second inequality:

This is FALSE since 9 is larger than 4 (not smaller)
Therefore the answer to the question is FALSE, (2, 9) is not a solution to the set of inequalities given.
We are going to make the product step by step.
We have:
(6 + 5i) * (3-7i)
Multiplying:
((6) * (3)) + ((6) * (- 7i)) + ((5i) * (3)) + ((5i) * (- 7i))
Rewriting:
18 - 42i + 15i - 35i ^ 2
18 - 42i + 15i - 35 * (- 1)
18 - 42i + 15i + 35
53 - 27i
Answer:
the product of and (6 + 5i) and (3-7i) is:
53-27i