Answer:
x = 9/4
Step-by-step explanation:
Solve for x. Start by adding 6x to both sides, which results in 3 + 4x = 12.
Combine the constants, obtaining 4x = 9.
Dividing both sides by 4, we get x = 9/4.
The answer is C.
x^2 + z = 103x^2 + 22 = 103*Subtract both sides by 22*x^2 + 22 - 22 = 103 - 22x^2 = 81*square root both sides to find x*√x^2 = <span>√81</span>
<span>x = +- 9</span>
Answer:
The complement of the given set in interval notation is
. It can we written as (-inf,5]U(6,inf).
Step-by-step explanation:
The given set in interval notation is
(−5,6]
It means the set is defined as
![A=\{x|x\in R,-5](https://tex.z-dn.net/?f=A%3D%5C%7Bx%7Cx%5Cin%20R%2C-5%3Cx%5Cleq%206%5C%7D)
If B is a set and U is a universal set, then complement of set B contains the elements of universal set but not the elements of set B.
Here, universal set is R, the set set of all real numbers.
![U=\{x|x\in R\}](https://tex.z-dn.net/?f=U%3D%5C%7Bx%7Cx%5Cin%20R%5C%7D)
The complement of the given set is
![A^c=U-A](https://tex.z-dn.net/?f=A%5Ec%3DU-A)
![A^c=\{x|x\in R,-\infty](https://tex.z-dn.net/?f=A%5Ec%3D%5C%7Bx%7Cx%5Cin%20R%2C-%5Cinfty%3Cx%5Cleq%20-5%2C6%3Cx%3C%5Cinfty%5C%7D)
Complement of the given set in interval notation is
![A^c=(-\infty,-5]\cup(6,\infty)](https://tex.z-dn.net/?f=A%5Ec%3D%28-%5Cinfty%2C-5%5D%5Ccup%286%2C%5Cinfty%29)
Therefore the complement of the given set in interval notation is
. It can we written as (-inf,5]U(6,inf).
A 96 because 25 percent of 400 is 100 and 24 percent is 96