Answer:
Remove backets
Group and Evaluate like terms
6a. 1 - 2sin(x)² - 2cos(x)² = 1 - 2(sin(x)² +cos(x)²) = 1 - 2·1 = -1
6c. tan(x) + sin(x)/cos(x) = tan(x) + tan(x) = 2tan(x)
6e. 3sin(x) + tan(x)cos(x) = 3sin(x) + (sin(x)/cos(x))cos(x) = 3sin(x) +sin(x) = 4sin(x)
6g. 1 - cos(x)²tan(x)² = 1 - cos(x)²·(sin(x)²)/cos(x)²) = 1 -sin(x)² = cos(x)²
Answer: 4
Step-by-step explanation:
The degree is the highest exponent , in which in this problem it is 4. The 3x counts as an exponent of 1 because the variable x is 1, and the 2 counts as an exponent of zero. Which means the degree is 4.
Hey there!
In order for you to find the one that is equivalent to the given expression is to COMBINE YOUR LIKE TERMS then work from there!
12q + q^2 - 8 + q = ?
(q^2) + (12q + q) + (-8)
q^2 = q^2 since it doesn't have any like term(s)
12q + q = 12q + 1q = 13q
-8 = -8 since it doesn't have any like term(s)
Put all the numbers we solved for in An equation and Thats your answer!
If you did it correctly you should have: q^2 + 13q - 8
Answer: q^2 + 13q - 8 ✅
Good luck on your assignment and enjoy your day!
~LoveYourselfFirst:)
Constant is the number that cannot change the value
Im not sure..but you can copy it
