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Dmitrij [34]
3 years ago
12

To make six cakes Fred uses four and a half cups of sugar. How much sugar does he need for one cake?

Mathematics
1 answer:
kramer3 years ago
6 0

Answer:

1.333 cups

Step-by-step explanation:

if you divide 6 by 4.5 you get 1.333

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Can someone explain HOW to simplify this algebraic fraction, as in tell me why and what I need to do for each step : 1-n/n^2-1
Andreyy89

- I’m assuming the expression implies the following:

(1-n)/(n^2 - 1)

The steps goes as follows:

= -(n-1)/(n^2-1) (factored a negative out)
= -(n-1)/[(n+1)(n-1)] (Factored denominator into a difference of squares, but don’t just blindly accept the fact, expand it and you should get the original denominator)

At this point, the (n-1) terms cancel, as you should know: a real number, besides 0, over itself is 1.

= -1/(n+1) , which is the answer

Note: You could also see it as 1/-(n+1) or 1/(-n-1)
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2 years ago
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3 years ago
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

6 0
1 year ago
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masha68 [24]

Answer:

So sorry , but the pic isn't very seen well

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