Answer:
Here we have the relation:
m = 140*h
Where m is the distance in miles, and h is time in hours.
And we want to complete a table like:
![\left[\begin{array}{ccc}in, h&out, m\\&\\&\\&\\&\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Din%2C%20h%26out%2C%20m%5C%5C%26%5C%5C%26%5C%5C%26%5C%5C%26%5Cend%7Barray%7D%5Cright%5D)
The way to complete this, is to evaluate the function:
m = 140*h
in different values of h, and then record both values of h and m in the table.
Let's use values of h that increase by 0.5, then:
if h = 0.5
m = 140*0.5 = 70
We have the pair: h = 0.5, m = 70
if h = 1
m = 140*1 = 140
We have the pair: h = 1, m = 140
if h = 1.5
m = 140*1.5 = 210
Then we have the pair h = 1.5, m = 210
if h = 2
m = 140*2 = 280
We have the pair: h = 2, m = 280
Now we can complete the table, and it will be:
![\left[\begin{array}{ccc}in, h&out, m\\0.5&70\\1&140\\1.5&210\\2&280\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Din%2C%20h%26out%2C%20m%5C%5C0.5%2670%5C%5C1%26140%5C%5C1.5%26210%5C%5C2%26280%5Cend%7Barray%7D%5Cright%5D)
I sort of know the metric system
Given that the associative property of multiplication says that you can group the numbers in any combination, we can reorganize the expression as: 7·(10·3)=210. So the answer would be the first option.
Answer:
x = 21/5 (4 1/5
Solve:
x + 4/5 - 4/5 = 5 - 4/5
Seee:
why do you have so many tabs open
Well as x can never actually be -1 because I'm in the denominator -1 + 1 = 0 and we cannot divide by zero. But we can look at what number it approaches and i assume that is the relative value. sometimes functions will have asymptotes and others will have holes in the graph. this one would have an asymptote going down at a rapid rate. the asymptote would go on forever getting infinitely close to -1 but never touching. So I would say since the asymptote goes down forever that the graph approaches negative infinity
