The first one, WX is perpendicular to ZY and WX is parallel to AB
Answer:
17 feet
Step-by-step explanation:
Let x represent the shorter leg of the triangle. Then the longer leg is (2x-1) and the area is ...
A = (1/2)bh = (1/2)(x)(2x-1) = 60
2x^2 -x = 120 . . . . multiply by 2 and simplify
2x^2 -x -120 = 0 . . . . subtract 120
(x -8)(2x +15) = 0 . . . . factor. The positive solution is x=8.
The square of the hypotenuse is ...
hypotenuse^2 = x^2 + (2x -1)^2 = 8^2 + 15^2 = 289
hypotenuse = √289 = 17
The length of the hypotenuse is 17 feet.
Wow, that is really eassy. It is 4! What did you thin k it would be? 4 is the correct answer! unless you ment 2x2 and that is 4 too
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>
Find the mean of all values ... use it to work out distances ... then find the mean of those distances!
In three steps:
1. Find the mean of all values
2. Find the distance of each value from that mean (subtract the mean from each value, ignore minus signs)
3. Then find the mean of those distances