Answer:
A
Step-by-step explanation:
A is the procedure
Answer:
they would be intersecting lines
Step-by-step explanation:
brainly crown apricated if I'm right
Answer:
Hotdog: $3.00
Hamburger: $4.00
Step-by-step explanation:
For the first time that Bob buys food, we can make an equation to find how much a single hotdog and a single hamburger costs, where:
x = cost of a hotdog
y = cost of a hamburger
He bought 2 hotdogs and 1 hamburger for $10, so the equation for his first time buying food is:
2x + y = 10
For the second time buying food, he bought 1 hotdog and 3 hamburgers for $15, so his equation would be:
x + 3y = 15
To find the value for x and y we need to solve this system of equations using the two equations we just came up with. We can do this multiple ways, but I'll be demonstrating the substitution method.
Using the second equation, we can solve for x by simply subtracting 3y from both sides:
x = 15 - 3y
We can then insert this value of x into the first equation so that way we are only dealing with one variable to solve - y:
2(15-3y) + y = 10
Distribute out the 2 into the paratheses, combine like terms, and then solve for y:
30 - 6y + y = 10
30 - 5y = 10
-5y = -20
y = 4
This means the cost for one hamburger is $4. But we still need to find the price of one hotdog, so we can insert this value of y into the equation we came up with earlier for x, and then solve for x:
x = 15 - 3y
x = 15 - 3(4)
x = 15 - 12
x = 3
So the price of one hotdog is $3 and the price of one hamburger is $4. Hope this helps.
(√3 - <em>i </em>) / (√3 + <em>i</em> ) × (√3 - <em>i</em> ) / (√3 - <em>i</em> ) = (√3 - <em>i</em> )² / ((√3)² - <em>i</em> ²)
… = ((√3)² - 2√3 <em>i</em> + <em>i</em> ²) / (3 - <em>i</em> ²)
… = (3 - 2√3 <em>i</em> - 1) / (3 - (-1))
… = (2 - 2√3 <em>i</em> ) / 4
… = 1/2 - √3/2 <em>i</em>
… = √((1/2)² + (-√3/2)²) exp(<em>i</em> arctan((-√3/2)/(1/2))
… = exp(<em>i</em> arctan(-√3))
… = exp(-<em>i</em> arctan(√3))
… = exp(-<em>iπ</em>/3)
By DeMoivre's theorem,
[(√3 - <em>i </em>) / (√3 + <em>i</em> )]⁶ = exp(-6<em>iπ</em>/3) = exp(-2<em>iπ</em>) = 1
