A^2 - b^2 = (a - b)( a +b);
<span>cos^4x-sin^4x =( cos^2x + sin^2x )( cos^2x - sin^2x) = 1* cos2x = co2x.</span>
True. No imaginary number is considered "whole", all whole numbers are rational and real.
Answer:
-3r + 15 ---> answer
Step-by-step explanation:
r < 5
You are going to multiply both sides with 3. The reason being is that 3 is a positive number and the equality sign will not change if you use +3.
3r < 15
Now, subtract 15 from both sides, you will get this:
3r < 15
-15 -15
-------------
3r — 15 < 0
Lastly, using the Modulus function, we are going to add a negative sign to the content of our previous step because it's already negative.
So, -3r + 15 is the final solution if r < 5 in the given equation of l3r-15l
The answer to the question is -24
