The classifications of the functions are
- A vertical stretch --- p(x) = 4f(x)
- A vertical compression --- g(x) = 0.65f(x)
- A horizontal stretch --- k(x) = f(0.5x)
- A horizontal compression --- h(x) = f(14x)
<h3>How to classify each function accordingly?</h3>
The categories of the functions are given as
- A vertical stretch
- A vertical compression
- A horizontal stretch
- A horizontal compression
The general rules of the above definitions are:
- A vertical stretch --- g(x) = a f(x) if |a| > 1
- A vertical compression --- g(x) = a f(x) if 0 < |a| < 1
- A horizontal stretch --- g(x) = f(bx) if 0 < |b| < 1
- A horizontal compression --- g(x) = f(bx) if |b| > 1
Using the above rules and highlights, we have the classifications of the functions to be
- A vertical stretch --- p(x) = 4f(x)
- A vertical compression --- g(x) = 0.65f(x)
- A horizontal stretch --- k(x) = f(0.5x)
- A horizontal compression --- h(x) = f(14x)
Read more about transformation at
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Answer:
|x|=6 we have x1=-6 x2=6
|x|=3.2 we have x1=-3.2 x2=3.2
|x|=0 x=0
2+8-z=-24
^
10-z=-24
-10 -10
——————-
-z=14
———
-z -z
Z=14
Answer:
Kindly check explanation
Step-by-step explanation:
Given the data:
Temperature (degrees Celsius) : 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Percent heat loss from beak : 35 36 38 28 41 43 55 46 39 54 45 58 60 56 62 67
Using an online regression calculator ; the regression equation obtained is :
ŷ = 2.0927X + 0.6029
X = independent variable
Y = predicted variable
2.0927 = slope
0.6029 = intercept
B.) temperature = 25
ŷ = 2.0927(25) + 0.6029
= 52.9204
C.) the explained variance is the value of the coefficient of determination (R²) which is the square of the correlation Coefficient
0.8785² = 0.7718
D.) the correlation Coefficient r is 0.8785 using the Coefficient of regression calculator
