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>
The 7 in 27,026 means 7,000. 7,000*10=70,000, so the answer is 70,000.
Answer:
Step-by-step explanation:
Let the number be 'x'
x*8 + 6 = 7
8x + 6 = 7
8x = 7 - 6
8x = 1
x = 1/8
The base's length is 8 and it's width is 7, so it's perimeter will be
. To find the lateral area you shoul multiply this with it's height. So,
(h is height).
It tells you how many places you have to move the decimal to right in order to get the proper quotient(answer).