For this case, the first thing we must do is define a variable.
We have then:
x: additional amount of weight that Li can add.
We write now the inequality that represents the problem.
We have then:
Answer:
An inequality that can be used to determine how much more weight can be added to the suitcase without going over the 50-pound weight limit is:
1. Suppose that BC is hypotenuse, then by Pythagorean theorem:

.
2. If BC is not hypotenuse, then BC is a leg and AC is hypotenuse (because AC>AB). Hence by Pythagorean theorem:

.
X< -2, means all numbers less than -2,
-3,-4,-5 and so on
Correct answer is -3
Assuming the ratio of 4:5 represents the ratio of width:length then we can set up a simple proportion to find the length...

L = 5(28)÷4
L = 35
Now we can find the perimeter and area of the rectangle:
A = LW
A = 28(35) = 980
P = 2L + 2W
P = 2(35) + 2(28) = 126
Complex numbers are written in the generic way:
a + bi
The complex numbers are composed of the following parts:
a = real part
bi = imaginary part
where,
i = root (-1)
Answer:
Every complex number in the form a + bi is composed of:
real and imaginary