Let's do

Release k for some while
If
So
So vertical asymptote is at origin now
It mentioned that it's at x=-5 so we need to change x


- Vertical asymptote at x=-5
Now
- for k=0 horizontal asymptote at origin
But it's given
Same put y=12 in place of k

Graph attached for verification
<u>the correct question is</u>
The denarius was a unit of currency in ancient rome. Suppose it costs the roman government 10 denarii per day to support 4 legionaries and 4 archers. It only costs 5 denarii per day to support 2 legionaries and 2 archers. Use a system of linear equations in two variables. Can we solve for a unique cost for each soldier?
Let
x-------> the cost to support a legionary per day
y-------> the cost to support an archer per day
we know that
4x+4y=10 ---------> equation 1
2x+2y=5 ---------> equation 2
If you multiply equation 1 by 2
2*(2x+2y)=2*5-----------> 4x+4y=10
so
equation 1 and equation 2 are the same
The system has infinite solutions-------> Is a consistent dependent system
therefore
<u>the answer is</u>
We cannot solve for a unique cost for each soldier, because there are infinite solutions.
4+4+4+6+6=24 Hours
15+15+20+20+20= 1 Hr 30 Min
24+1= 25 Hours + 30 Minutes=
Final answer: C) 25 hours and 30 minutes
Answer:
The work is in the explanation.
Step-by-step explanation:
The sine addition identity is:
.
The sine difference identity is:
.
The cosine addition identity is:
.
The cosine difference identity is:
.
We need to find a way to put some or all of these together to get:
.
So I do notice on the right hand side the
and the
.
Let's start there then.
There is a plus sign in between them so let's add those together:

![=[\sin(a+b)]+[\sin(a-b)]](https://tex.z-dn.net/?f=%3D%5B%5Csin%28a%2Bb%29%5D%2B%5B%5Csin%28a-b%29%5D)
![=[\sin(a)\cos(b)+\cos(a)\sin(b)]+[\sin(a)\cos(b)-\cos(a)\sin(b)]](https://tex.z-dn.net/?f=%3D%5B%5Csin%28a%29%5Ccos%28b%29%2B%5Ccos%28a%29%5Csin%28b%29%5D%2B%5B%5Csin%28a%29%5Ccos%28b%29-%5Ccos%28a%29%5Csin%28b%29%5D)
There are two pairs of like terms. I will gather them together so you can see it more clearly:
![=[\sin(a)\cos(b)+\sin(a)\cos(b)]+[\cos(a)\sin(b)-\cos(a)\sin(b)]](https://tex.z-dn.net/?f=%3D%5B%5Csin%28a%29%5Ccos%28b%29%2B%5Csin%28a%29%5Ccos%28b%29%5D%2B%5B%5Ccos%28a%29%5Csin%28b%29-%5Ccos%28a%29%5Csin%28b%29%5D)


So this implies:

Divide both sides by 2:

By the symmetric property we can write:
