The absolute value functions that contains the points are:
- f(x) = |x| + 2
- f(x) = |-x| + 2
<h3>Which could be the function represented by this graph?</h3>
Here we have the points (-3, 5), (0, 2), and (3, 5). We want to see which ones of the given functions have that points.
To check that, we need to evaluate the functions in the first value of each point and see if the outcome is the second value of the correspondent point.
For example, for the first equation:
- f(-3) = |-3| + 2 = 5 so it has the point (-3, 5)
- f(0) = |0| + 2 = 2 so it has the point (0, 2)
- f(3) = |3| + 2 = 5 so it has the point (3, 5).
The other option that also contains these 3 points is:
f(x) = |-x| + 2
- f(-3) = |3| + 2 = 5 so it has the point (-3, 5)
- f(0) = |-0| + 2 = 2 so it has the point (0, 2)
- f(3) = |-3| + 2 = 5 so it has the point (3, 5).
And all the other options can be trivially discarded (by evaluating them).
So the two correct options are:
- f(x) = |x| + 2
- f(x) = |-x| + 2
If you want to learn more about absolute value functions:
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Answer: 90
Step-by-step explanation:
c = 2πr
r = 15
d = 30
30π = 90
Answer: x = 9.6
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Explanation:
We have two smaller right triangles that are glued together so to speak.
The base of the smaller triangle on the left is 5 while the height is h.
Let's use the tangent rule to find the value of h
tan(angle) = opposite/adjacent
tan(55) = h/5
5*tan(55) = h
h = 5*tan(55)
h = 7.14074003371058
Make sure your calculator is in degree mode. That value of h above is approximate.
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Now focus on the smaller triangle on the right
It has the same height value h. This side is the adjacent side while x is the hypotenuse.
We'll use the cosine ratio
cos(angle) = adjacent/hypotenuse
cos(42) = h/x
cos(42) = 7.14074003371058/x
x*cos(42) = 7.14074003371058
x = 7.14074003371058/cos(42)
x = 9.6088135029715
x = 9.6
Perimiter=6 times legnth of one side
P=6(x+1/3y)
P=6x+6/3y
P=6x+2y
if x=5 and y=6
P=6(5)+2(6)
P=30+12
P=42
A. P=6x+2y
B. 42 units is the perimiter
Answer:
sometimes
Step-by-step explanation:
Similar figures can also be a different size, as long as they're the same shape. So both rigid transformations and dilations can create similar figures.