9514 1404 393
Answer:
obtuse
Step-by-step explanation:
The law of cosines tells you ...
b² = a² +c² -2ac·cos(B)
Substituting for a²+c² using the given equation, we have ...
b² = b²·cos(B)² -2ac·cos(B)
We can subtract b² to get a quadratic in standard form for cos(B).
b²·cos(B)² -2ac·cos(B) -b² = 0
Solving this using the quadratic formula gives ...
The fraction ac/b² is always positive, so the term on the right (the square root) is always greater than 1. The value of cos(B) cannot be greater than 1, so the only viable value for cos(B) is ...
The value of the radical is necessarily greater than ac/b², so cos(B) is necessarily negative. When cos(B) < 0, B > 90°. The triangle is obtuse.
Answer:
c
Step-by-step explanation:
because they are in large sizes but small amount of time
Answer:
see below
Step-by-step explanation:
(3x + 8) units by (6x + 5) units.
A = l*w
= (3x+8)(6x+5)
Distribute
3x*(6x+5) + 8(6x+5)
18x^2 +15x + 48x +40
Combine terms
18x^2+63x+40 units^2
The degree is 2 ( the highest power of x)
This is a trinomial ( it has 3 terms)
Multiplication is closed for polynomial multiplication
We started with polynomials and we ended up with a polynomial.
Problem 1
<h3>Answer: False</h3>
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Explanation:
The notation (f o g)(x) means f( g(x) ). Here g(x) is the inner function.
So,
f(x) = x+1
f( g(x) ) = g(x) + 1 .... replace every x with g(x)
f( g(x) ) = 6x+1 ... plug in g(x) = 6x
(f o g)(x) = 6x+1
Now let's flip things around
g(x) = 6x
g( f(x) ) = 6*( f(x) ) .... replace every x with f(x)
g( f(x) ) = 6(x+1) .... plug in f(x) = x+1
g( f(x) ) = 6x+6
(g o f)(x) = 6x+6
This shows that (f o g)(x) = (g o f)(x) is a false equation for the given f(x) and g(x) functions.
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Problem 2
<h3>Answer: True</h3>
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Explanation:
Let's say that g(x) produced a number that wasn't in the domain of f(x). This would mean that f( g(x) ) would be undefined.
For example, let
f(x) = 1/(x+2)
g(x) = -2
The g(x) function will always produce the output -2 regardless of what the input x is. Feeding that -2 output into f(x) leads to 1/(x+2) = 1/(-2+2) = 1/0 which is undefined.
So it's important that the outputs of g(x) line up with the domain of f(x). Outputs of g(x) must be valid inputs of f(x).
Answer:
a square what
Step-by-step explanation:
