1st term = a(1) = -7(1)-1 = -7-1 = -8
2nd term = a(2) = -7(2)-1 = -14-1 = -15
3rd term = a(3) = -7(3)-1 = -21-1 = -22
term would be -8, -15, -22
B. slope = -1/2
C. slope = -2
Answer:
-1/2
I hope that helps
Step-by-step explanation:
y = 15x, in 9 years he will have saved $13,500
y = x + 15, in 9 years he will have saved $2,400
y = 6x, in 9 years he will have saved $5,400
y = x + 90, in 9 years he will have saved $9,900
Answer:it’s a, 13,500. I took the test
Step-by-step explanation:
7 less than -2 times a number X is greater than or equal to 41
To find the instantaneous rate of change of the function f(x,y) = x^2 + ln(y) at (3,1) to (1,2), we can use the partial derivatives with respect to x and y:
fx(x,y) = 2x
fy(x,y) = 1/y
Then, we can use the gradient vector to find the direction of maximum increase:
∇f(x,y) = <fx(x,y), fy(x,y)> = <2x, 1/y>
At point (3,1), the gradient vector is:
∇f(3,1) = <6, 1>
At point (1,2), the gradient vector is:
∇f(1,2) = <2, 1/2>
To find the instantaneous rate of change from (3,1) to (1,2), we can use the formula for directional derivative:
Dv(f) = ∇f(x,y) · v
where v is the unit vector in the direction from (3,1) to (1,2). The direction vector v is given by:
v = <1, 2> - <3, 1> = <-2, 1>
To make v a unit vector, we need to normalize it by dividing it by its length:
|v| = sqrt((-2)^2 + 1^2) = sqrt(5)
u = v/|v| = <-2/sqrt(5), 1/sqrt(5)>
Then, the instantaneous rate of change from (3,1) to (1,2) is:
Dv(f) = ∇f(3,1) · u = <6, 1> · <-2/sqrt(5), 1/sqrt(5)> = (-12/sqrt(5)) + (1/sqrt(5)) = -11/sqrt(5)
Therefore, the instantaneous rate of change of the function f(x,y) = x^2 + ln(y) from (3,1) to (1,2) is -11/sqrt(5).
To learn more about instantaneous rate of change refer below:
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