Answer:
I got -64
Step-by-step explanation:
The begging multiplication is 0 and subtract -7 is becomes just -7 then add 3 and it is -4 to the third power is -64
A' (3(-5),3(-4))= (-15,-12)
B' (3(2),3(6))=(6,18)
C'(3(4),3(-3))=(12,-9)
Discriminant D is given by:
D=b²-4ac
Implication of discriminant is as follows:
D<0 two zeros that are complex conjugate
D=0 one real zero of multiplicity 2
D>0 two distinct real zers
D= (+ve perfet square) two distinct rational zeros
From:
12x^2+10x+5=0
plugging in the equation we get:
10²-4×12×5
=100-240
=-140
thus
D<0
Answer is:
<span>A two irrational solutions </span>
Answer: Variables
F and C are placeholders for numbers. They are called variables because they are allowed to vary, or change. If you change C then it affects F, and vice versa. If the values for the placeholders is not allowed to change, yet it holds a number, then it is considered a constant. In this case, we don't have constants or else the formula isn't too useful.
For this case we must find the solution set of the given inequalities:
Inequality 1:

Applying distributive property on the left side of inequality:

Subtracting 3 from both sides of the inequality:

Dividing by 6 on both sides of the inequality:

Thus, the solution is given by all the values of "x" greater than 3.
Inequality 2:

Subtracting 3x from both sides of the inequality:

Subtracting 3 from both sides of the inequality:

Thus, the solution is given by all values of x less than 4.
The solution set is given by the union of the two solutions, that is, all real numbers.
Answer:
All real numbers