The answer is L = 59 + 2(d). If we plug in our numbers, we will get L = 59 + 2(32). 2*32 is 64, and if you add 59 you get 123 miles.
This isn’t apart of the question, but 59 would be the y-intercept. 2 miles per day (2/1) is the slope.
Maybe because you're finding the "difference" between the two numbers. We could also question why the answer to an addition problem is called sum.
Well, 'sum'body said that's what they should be called a long time ago, that's why we're using that term to call them now.
Answer:
In order for me to help you I need to either see a photo or know more about the other angle.
Step-by-step explanation:
Answer:
Remember,
and the range of g must be in the domain of f.
a)


The domain of f(g(x)) and g(f(x)) is the set of reals.
b)


The domain of f(g(x)) is the set of nonnegative reals and the domain of g(f(x)) is the set of number such that 
c)


The domain of f(g(x)) is the set of reals except the 1 and the domain of g(f(x)) is the set of reals except the 1 and -1
d)


The domain of f(g(x)) is the set of reals except 2, and the domain of g(f(x)) is the set of reals except -1.
e)


The domain of f(g(x)) is the set of nonnegative reals except -3. The domain of g(f(x)) is the set of nonnegative reals except -2.
Answer:
A∪B = {0,1,3,5,6,7,9}
A∩B = {3}
Step-by-step explanation:
A = {1,3,5,7}
B = {0,3,6,9}
Now
A∪B is a union (unity) of both sets
A∪B = {0,1,3,5,6,7,9}
Now
A∩B is the intersection(common elements) of both sets
A∩B = {3}