Simply substitute x for -3 and y for -5, creating the equation shown below:
-6 (-3 - 2(-5) )
Now that you have your values set up, now you can do the math by starting off with Multiplying -2 by -5, which will get you 10.
Now that you have the parenthesis simplified in the parenthesis, you can use the distributive property by multiplying all the numbers with -6, like so:
-6(-3+10)= 18 - 60, which simplifies to -42, in which (A) is your answer.
<u>Answer:</u>
The correct answer option is B. (x - 1).
<u>Step-by-step explanation:</u>
We are to determine the factor of the polynomial function
which is graphed on the given coordinate plane.
We know that if the zeros (or intercepts) of an equation are
and
then the factors for this equation will be
and
.
From the graph, we can see that it intercepts the x axis at
and
. So the roots will be
and
.
Therefore, the correct answer option is B. (x - 1).
Answer: distributive
Step-by-step explanation:
You distribute the 12
So you can do this multiple ways, I'll do this the way that I think makes sense the l most easily.
Cos (0) = 1
Cos (pi/2)=0
Cos (pi) =-1
Cos (3pi/2)=0
Cos (2pi)=1
Now if you multiply the inside by 4, the graph oscillates more violently (goes up and down more in a shorter period).
But you can always reduce it.
Cos (0)= 1
Cos (4pi/2) = cos (2pi)=1
Cos (4pi) =Cos (2pi) =1 (Any multiple of 2pi ==1)
etc...
the pattern is that every half pi increase is now a full period as apposed to just a quarter of one. That's in theory.
Now that you know that, the identities of Cosine are another beast, but mathematically.
You have.
Cos (2×2t) = Cos^2 (2t)-Sin^2 (2t)
Sin^2 (t)=-Cos^2 (t)+1..... (all A^2+B^2=C^2)
Cos (2×2t) = Cos^2 (t)-(-Cos^2 (t)+1)
Cos (2×2t)= 2Cos^2 (2t) - 1
2Cos^2 (2t) -1= 2 (Cos^2(t)-Sin^2(t))^2 -1
(same thing as above but done twice because it's cos ^2 now)
convert sin^2
2Cos^2 (2t)-1 =2 (Cos^2 (t)+Cos^2 (t)-1)^2 -1
2 (2Cos^2(t)-1)^2 -1
2 (2Cos^2 (t)-1)(2Cos^2 (t)-1)-1
2 (4Cos^4 (t) - 2 (2Cos^2 (t))+1)-1
Distribute
8Cos^4 (t) -8Cos^2 (t) +1
Cos (4t) =8Cos^4-8Cos^2 (t)+-1
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