The given function is:
x^2 - 1
We want to calculate the limit of this function as x approaches zero. To do so, we will use direct substitution.
We will substitute the x with 0 in the given function to calculate its limit as follows:
Limit as x approaches 0 = (x)^2 - 1 = -1
Therefore, the correct choice is:
-1
f(x) = x2 - 2x + 2
g(x) = x - 3
(f • g)(x) = (x2 - 2x + 2)(x - 3)
(f • g)(x) = x2(x - 3) - 2x(x - 3) + 2(x - 3)
(f • g)(x) = x3 - 3x2 - 2x2 + 6x + 2x - 6
(f • g)(x) = x3 - 5x2 + 8x - 6
The answer is B.
Answer:
760,000
Step-by-step explanation:
759,993
we round 993 up to 1,000
AV = AX
2x + 1 = x + 3
x + 1 = 3
x = 2
BV = (2/3)ZV
3x + 2 = (2/3)ZV
3(2) + 2 = (2/3)ZV
6 + 2 = (2/3)ZV
8 = (2/3)ZV
24 = 2ZV
ZV = 12
Answer:
A)Predicted value = 103.5
B)Actual value = 102
C)So, these are not same
Step-by-step explanation:
The researcher used the line
to model the data.
A)When the researcher substituted the value of x = 65 into this equation, what is the resulting y value?
Substitute value x=65 in equation :
y = -0.1 x +110
y=-0.1(65)+110
y=103.5
Predicted value = 103.5
B)Based on the coordinates of the given data points, what is the actual actual vision score when x=65?
Use the given graph
When x = 65
y = 102
So, The actual actual vision score when x=65 is 102
C)When the researcher substituted x = 65 into the line of equation, is the resulting y value the predicted value or the exit value for the vision score of a 65-year-old? Use your results for parts a and b to answer this question.
Predicted value = 103.5
Actual value = 102
So, these are not same