The answer is D) -r-6
if you distribute the -2 to the 5r and 3 you get -10r and -6.
-10r + 9r = -r (then just add on the negative 6)
-r-6
Given and (say).
Then,
From the above 3 equations,
From the equations, we get
Since , the negative value is rejected.
No, it is not a solution, as 15 is not less-than fifteen.<span />
Answer:
Step-by-step explanation:
We're going to leave the right side alone and work on the left side. In other words we are going to use a series of substitutions for these trig idenitites and get the left side manipulated to look like the right side.
Begin with the fact that sin(2x) = 2sin(x)cos(x) and make that first substitution:
2sin(x)cos(x) - tan(x) = right side
Now use the fact that the tangent is the same as the sin over the cos:
= right side
Now find a common denominator of cos(x) by multiplying the 2sin(x)cos(x) by cos(x) and writing the whole mess over that common denominator:
= right side
Now factor out a sin(x):
= right side
If we "split" that up and simplify at the same time, we'll see that sin(x) ovr cos(x) is the same as the tan(x), and that 2cos^2 - 1 is the same as cos(2x):
and that the left side now is the same as the right side. You MUST learn to recongize these identities. I'll attach a copy that I made and give to my pre-calculus and calculus classes every year, if I am able to.
It should be y=2x-3 since it’s touching -3 on the y axis and the it’s going 2 units up and 1 to the right, making the slope = 2.