Answer:
Michael is incorrect.
if we take the steps Michael says and move the ABCDE up 8 and right 10, not all of the points line up.
Step by step explanation:
In order to get the shapes to line up, we need to flip (reflect) ABCDE onto the 2nd Quadrant, then we can move (translate) it to the right by 10.
Answer:
31 over 3
Step-by-step explanation:
So basically, let's look at the given. In the problem, Leon is charged a fee of 75. The interest rate would be 12.5 % if he doesn't pay his due amount. So in order to get the total amount that he has to pay if he doesn't pay on time, you just need to multiply 75 with 12.5% or simply 0.125. If you do this, you will get 9.375. After that, you add 9.375 to the principal amount which is 75 and you will get 84.375. So Leon has to now pay a total of 84.375.
I hope this HELP! :)
It is the first answer. Notice that if you multiply it out you get 6x + 30, and they have factored out the 6.
Answer:
<em>f(0)>f(1) and f(1)<f(2), therefore a critical point exists at f(1). As the value is greater before the critical point and is greater after as well, thus there exists a local minima at x=1.</em>
<em>f(3) <f(4) and f(4)>f(5), therefore a critical point exists at f(4). As the value is less before the critical point and is less after as well, thus there exists a local maxima at x=4.</em>
Step-by-step explanation:
As the data table is missing in the question, a similar question is found, which is as attached here with.
From the data of table
x | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
----------------------------------------------------
y=f(x) | -3 | -5 | -4 | -1 | 2 | 1 | -1 | -3 | -4 | -6 | -7 |
From the graph attached the critical points are as given below
As
<em>f(0)>f(1) and f(1)<f(2), therefore a critical point exists at f(1). As the value is greater before the critical point and is greater after as well, thus there exists a local minima at x=1.</em>
<em>f(3) <f(4) and f(4)>f(5), therefore a critical point exists at f(4). As the value is less before the critical point and is less after as well, thus there exists a local maxima at x=4.</em>