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krek1111 [17]
3 years ago
14

Evaquate-18÷3×8(-8)\-5×-2+(-2)​

Mathematics
2 answers:
MArishka [77]3 years ago
4 0

Answer:

-12/5 - 2

Step-by-step explanation:

-18÷3×8(-8)/-5×-2+(-2)​ =

-6×8(-8)/-5×-2+(-2)​ =

-48(-8)/-5×-2+(-2)​ =

6/-5×-2+(-2)​ =

-12/5 - 2

svetlana [45]3 years ago
3 0

Answer:

Step-by-step explanation:

-18÷3×8(-8)/-5×-2+(-2)

-6×8(-8)/10+(-2)

-48×(-8)/8

40/8

5

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Step-by-step explanation:

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Given f(x) = (lnx)^3 find the line tangent to f at x = 3
kirill [66]
Explanation

We must the tangent line at x = 3 of the function:

f(x)=(\ln x)^3.

The tangent line is given by:

y=m*(x-h)+k.

Where:

• m is the slope of the tangent line of f(x) at x = h,

,

• k = f(h) is the value of the function at x = h.

In this case, we have h = 3.

1) First, we compute the derivative of f(x):

f^{\prime}(x)=\frac{d}{dx}((\ln x)^3)=3*(\ln x)^2*\frac{d}{dx}(\ln x)=3*(\ln x)^2*\frac{1}{x}=\frac{3(\ln x)^2}{x}.

2) By evaluating the result of f'(x) at x = h = 3, we get:

m=f^{\prime}(3)=\frac{3}{3}*(\ln3)^2=(\ln3)^2.

3) The value of k is:

k=f(3)=(\ln3)^3

4) Replacing the values of m, h and k in the general equation of the tangent line, we get:

y=(\ln3)^2*(x-3)+(\ln3)^3.

Plotting the function f(x) and the tangent line we verify that our result is correct:

Answer

The equation of the tangent line to f(x) and x = 3 is:

y=(\ln3)^2*(x-3)+(\ln3)^3

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