
the domain:

the equation:

4 is in the domain
-2 is not in the domain
The answer:

***
Also, the answer in 13 is {8}. -3 is not in the domain. Replace x with -3 and you'll see:

It's not true so -3 isn't a solution to this equation.
Answer:

Step-by-step explanation:
We know:

We have

Use 
![\left(\dfrac{1}{2}\right)^2+\cos^2\theta=1\\\\\dfrac{1}{4}+\cos^2\theta=1\qquad\text{subtract}\ \dfrac{1}{4}\ \text{from both sides}\\\\\cos^2\theta=\dfrac{4}{4}-\dfrac{1}{4}\\\\\cos^2\theta=\dfrac{3}{4}\to\cos\theta=\pm\sqrt{\dfrac{3}{4}}\to\cos\theta=\pm\dfrac{\sqrt3}{\sqrt4}\to\cos\theta=\pm\dfrac{\sqrt3}{2}\\\\\theta\in[0^o,\ 90^o],\ \text{therefore all functions have positive values or equal 0.}\\\\\cos\theta=\dfrac{\sqrt3}{2}](https://tex.z-dn.net/?f=%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E2%2B%5Ccos%5E2%5Ctheta%3D1%5C%5C%5C%5C%5Cdfrac%7B1%7D%7B4%7D%2B%5Ccos%5E2%5Ctheta%3D1%5Cqquad%5Ctext%7Bsubtract%7D%5C%20%5Cdfrac%7B1%7D%7B4%7D%5C%20%5Ctext%7Bfrom%20both%20sides%7D%5C%5C%5C%5C%5Ccos%5E2%5Ctheta%3D%5Cdfrac%7B4%7D%7B4%7D-%5Cdfrac%7B1%7D%7B4%7D%5C%5C%5C%5C%5Ccos%5E2%5Ctheta%3D%5Cdfrac%7B3%7D%7B4%7D%5Cto%5Ccos%5Ctheta%3D%5Cpm%5Csqrt%7B%5Cdfrac%7B3%7D%7B4%7D%7D%5Cto%5Ccos%5Ctheta%3D%5Cpm%5Cdfrac%7B%5Csqrt3%7D%7B%5Csqrt4%7D%5Cto%5Ccos%5Ctheta%3D%5Cpm%5Cdfrac%7B%5Csqrt3%7D%7B2%7D%5C%5C%5C%5C%5Ctheta%5Cin%5B0%5Eo%2C%5C%2090%5Eo%5D%2C%5C%20%5Ctext%7Btherefore%20all%20functions%20have%20positive%20values%20or%20equal%200.%7D%5C%5C%5C%5C%5Ccos%5Ctheta%3D%5Cdfrac%7B%5Csqrt3%7D%7B2%7D)

Answer:
all positive numbers and their negatives are opposites. I suppose you could say infinite and zero are also opposites.
Step-by-step explanation:
Answer:
6(7a + 9b) = 3(14a + 18b)
Step-by-step explanation:
We need to find the equivalent expression for the given expression i.e. 6(7a + 9b).
We can use the distributive property here such that,
a(b+c) = ab + ac
We have, a = 6, b = 7 and c = 9
6(7a + 9b) = 6(7a) + 6(9b)
= 42a + 54b
or taking 3 common from the above expression.
= 3(14a+18b)
Hence, the correct option is (a).
Since, the coordinates of the midpoint of line GH are M
.
The coordinates of endpoint G are (-4,1)
We have to determine the coordinates of endpoint H.
The midpoint of the line segment joining the points
and
is given by the formula
.
Here, The endpoint G is (-4,1) So, 
Let the endpoint H be 
The midpoint coordinate M is
.
So, 



Now, 


So, the other endpoint H is (-9,-13).