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>
1 yard=36 inches 1 foot=12 inches and 1/3 yard 1 inch= 1/36 yard and 1/12 foot
Answer:
3/50
Step-by-step explanation:
P( 1st ball is a black ball) = number of black balls/ total balls
= 2/10 = 1/5
We put the ball back so we still have the same number of balls
P( 2nd ball is a red ball) = number of red balls/ total balls
= 3/10
P (1st ball black, 2nd red) = P( 1st ball is a black ball)* P( 2nd ball is a red ball)
=1/5* 3/10
= 3/50
<u>Answer:</u>
<u>Step-by-step explanation:</u>
We are given the following expression which are to solve for a and give the answer in radical form:
To solve this, we will multiply both the sides by to get:
Simplify it to get: