Factorization of x²+6x+5
1 answer:
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Answer:
Part a) The company can not build 12 chairs and 5 tables
Part b) The company can build 18 chairs and 3 tables
Part c) The company can build 12,13,14,15,16 or 17 chairs
Step-by-step explanation:
we know that
If a ordered pair is a solution, then the ordered pair must be belong to the feasible region of the graph
Part a) Can the company build 12 chairs and 5 tables?
In this problem we have the ordered pair A(12,5)
Plot and verify if the point belong to the feasible region of the graph
see the attached figure to better understand the problem
The point A is not on the feasible region
therefore
The company can not build 12 chairs and 5 tables
Part b) Can the company build 18 chairs and 3 tables?
In this problem we have the ordered pair B(18,3)
Plot and verify if the point belong to the feasible region of the graph
see the attached figure to better understand the problem
The point B is on the feasible region
therefore
The company can build 18 chairs and 3 tables
Part c) One week, the company decides to build 4 tables. how many chairs can the company build?
For y=4
Find the value of x in the graph
The solutions point for y=4 tables are
(12,4), (13,4),(14,4),(15,4),(16,4) and (17,4)
therefore
The company can build 12,13,14,15,16 or 17 chairs
Answer:
These type of questions are super tricky b/c you have to remember all the different versions of the identities, and then they put the question in some odd form, I feel like this should land math professors in jail , for dishonesty , b/c it's really a form of "how tricky can I make a question and still have a way to solve it" anyway,
Step-by-step explanation:
a)
next the question asks 1-cos 2A and this is total abuse of notation. the way this should be written is 1- cos( 2A) so we know that the A is part of the cosine functions input... btw.. in any computer program, it would never ever let you get away with that top form of the expression. :/ anyway... I keep ranting.. huh... sorry :P
1-cos(2A) is an odd form of the identity 1/2(1-cos(2A) = (A) the 1/2 is missing but we can add that pretty easy, we just have to remember to take it out too. I usually forget to do that. and my professor marks me off completely, totally wrong, but I just miss one small thing :/ anyway....
our 1-cos(2A) needs the 1/2 added to it. or if we move that 1/2 to the other side it looks like 2*(A) = 1-cos(2A) and this is that "odd" from of the identity that I was talking about.
next let's deal with sin(2A) it has an identity of 2 sin(A)cos(A) which is really nice for us b/c it will cancel out the 2 in then numerator for us, nice !
now our fraction looks like [2* (A)] / 2 sin(A)cos(A)
so cancel out one of the sines
2*sin(A) / 2 cos(A)
cancel the 2s
Sin(A) / Cos(A) = Tan(A)
nice it worked out :P
b)
by the above that we just worked out, then
Tan(15) = Sin(15) / Cos(15)
I had to look up what sin of 15 is b/c it's not one of those special angles but it does have an exact form of
Sin(15) = (√3 - 1) / 2√2
Cos(A) = (√3 + 1) / 2√2
you can use rule of Cos(A-B) = Cos(A)Cos(B)+Sin(A)Sin(B) to get the above and a similar rule for Sin(A-B)
back to our problem, the 2√2 will cancel out
then we have
Tan(15) = (√3 - 1) /(√3 + 1)
in the form that is above that's exact, the roots could be approximated but i'll just leave that in the form that is exact. Most math professors like that form.