How to tell if a complex solution is real or imaginary?

Mathematics

2 answers:

skad [1K]3 years ago

8 0

Answer:

If a parabola has a descrimant is less than zero its solution is imaginary, if it is equal to 0 it has one solution and if it is greater than 0 it has 2 solutions. The discramanant can be written as b^2 - 4ac and it is the area under the square root in the quadratic equation.

Step-by-step explanation:

fgiga [73]3 years ago

4 0

Answer:

The expression b2 − 4ac is called the discriminant, and can be used to determine whether the solutions are real, repeated, or complex: 1) If the discriminant is less than zero, the equation has two complex solution(s). 2) If the discriminant is equal to zero, the equation has one repeated real solution(s).

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The perimeter of a triangle is 73 cm. the longest side is 5 cm less than the sum of the other two sides. twice the shortest side

LiRa [457]

a+b+c = 73 (a is shortest side, b is second shortest side, c is longest side)

c = a+b-5

c = 2a+8

b = a+13

2a+8+a+13+a = 73

4a+21=73

4a = 52

a = 13

b = 26

c = 34

However when you plug that back in, you don't get a triangle, so I think you are missing something

4 0

3 years ago

Let X be a random variable with probability density function fX(x) = ( c(1 − x 2 ) if − 1 < x < 1 0 otherwise. a) What is

xeze [42]

Answer:

a) c=3/4

b) for -1<x<1

F(x) = (3/4) ( x + (2-x^3)/3 )

E(x)=0

Var(x) = 1/5

Step-by-step explanation:

Hi!

In order to f(x) be a probability density function it must be normalized, which means that its integral must be equal to 1:

$1 = \int\limits^{\infty}_{-\infty} {f(x)} \, dx$

Since f(x) is zero for x>1 and x<-1

$1 = \int\limits^1_{-1} {f(x)} \, dx = \int \limits^1_{-1} {c(1-x^2)}\, dx \\\\\\1 = c ((1-\frac{1}{3}) - (-1 - \frac{-1}{3})) = c(\frac{2}{3} - \frac{-2}{3} ) = c\frac{4}{3}$