Substitute x = -5
|2* -5 -1| + 10
|-10-1| + 10
21
Answer:
A direct variation is when x and y (or f(x) and x) are directly proportional to each other... For example, if you have a chart that says x and y, and in the x column is 1, 2 and 3, and the y column says 2, 4 and 6... then you know it's proportional because for each x, y increases by 2
Step-by-step explanation:
Sec(x/2) = 1/cos(x/2)
sec(x/2)=cos(x/2) ----> cos^2(x/2)=1 ---> cos(x/2) = -1 and cos(x/2) = 1
Cos(x/2)=1 --- > x/2 = 0, only. x = 0;
cos(x/2)=-1 ----> x/2 = pi -> x = 2pi. But the statement says [0,2pi), so 2pi can not be chosen.
Only x = 0.
In fact, your equation is equivalent to sec(x)=cos(x), for x in [ 0, pi), so yes, only x = 0 .
Answer:
D) 0 = 2(x + 5)(x + 3)
Step-by-step explanation:
Which of the following quadratic equations has no solution?
We have to solve the Quadratic equation for all the options in other to get a positive value as a solution for x.
A) 0 = −2(x − 5)2 + 3
0 = -2(x - 5) × 5
0 = (-2x + 10) × 5
0 = -10x + 50
10x = 50
x = 50/10
x = 5
Option A has a solution of 5
B) 0 = −2(x − 5)(x + 3)
Take each of the factors and equate them to zero
-2 = 0
= 0
x - 5 = 0
x = 5
x + 3 = 0
x = -3
Option B has a solution by one of its factors as a positive value of 5
C) 0 = 2(x − 5)2 + 3
0 = 2(x - 5) × 5
0 = (2x -10) × 5
0 = 10x -50
-10x = -50
x = -50/-10
x = 5
Option C has a solution of 5
D) 0 = 2(x + 5)(x + 3)
Take each of the factors and equate to zero
0 = 2
= 0
x + 5 = 0
x = -5
x + 3 = 0
x = -3
For option D, all the values of x are 0, or negative values of -5 and -3.
Therefore the Quadratic Equation for option D has no solution.
