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Salsk061 [2.6K]
3 years ago
10

B. Write a part-to-part ratio using colon notation.

Mathematics
1 answer:
UNO [17]3 years ago
4 0

Answer:

3:5, 4:9, 6:8 those are examples

Step-by-step explanation:

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A frost has caused a temporary spike in the price of orange juice. the price is now $10 a gallon. before the frost, the price wa
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It is $6 more now than before...maybe you were looking for percent increase?

%change=100(final-initial)/initial

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Someone please help !! I don’t know what I’m doing with this !!
dimulka [17.4K]

Answer:

  a) d(sinh(f(x)))/dx = cosh(f(x))·df(x)/dx

  b) d(cosh(f(x))/dx = sinh(f(x))·df(x)/dx

  c) d(tanh(f(x))/dx = sech(f(x))²·df(x)/dx

  d) d(sech(4x+2))/dx = -4sech(4x+2)tanh(4x+2)

Step-by-step explanation:

To do these, you need to be familiar with the derivatives of hyperbolic functions and with the chain rule.

The chain rule tells you that ...

  (f(g(x)))' = f'(g(x))g'(x) . . . . where the prime indicates the derivative

The attached table tells you the derivatives of the hyperbolic trig functions, so you can answer the first three easily.

__

a) sinh(u)' = sinh'(u)·u' = cosh(u)·u'

For u = f(x), this becomes ...

  sinh(f(x))' = cosh(f(x))·f'(x)

__

b) After the same pattern as in (a), ...

  cosh(f(x))' = sinh(f(x))·f'(x)

__

c) Similarly, ...

  tanh(f(x))' = sech(f(x))²·f'(x)

__

d) For this one, we need the derivative of sech(x) = 1/cosh(x). The power rule applies, so we have ...

  sech(x)' = (cosh(x)^-1)' = -1/cosh(x)²·cosh'(x) = -sinh(x)/cosh(x)²

  sech(x)' = -sech(x)·tanh(x) . . . . . basic formula

Now, we will use this as above.

  sech(4x+2)' = -sech(4x+2)·tanh(4x+2)·(4x+2)'

  sech(4x+2)' = -4·sech(4x+2)·tanh(4x+2)

_____

Here we have used the "prime" notation rather than d( )/dx to indicate the derivative with respect to x. You need to use the notation expected by your grader.

__

<em>Additional comment on notation</em>

Some places we have used fun(x)' and others we have used fun'(x). These are essentially interchangeable when the argument is x. When the argument is some function of x, we mean fun(u)' to be the derivative of the function after it has been evaluated with u as an argument. We mean fun'(u) to be the derivative of the function, which is then evaluated with u as an argument. This distinction makes it possible to write the chain rule as ...

  f(u)' = f'(u)u'

without getting involved in infinite recursion.

7 0
3 years ago
The Big O notation of the composite function 5N^3 O(N^2) is _____. a. O(5N^3) b. O(N^3 N^2) c. O(N^3) d. O(N^5)
stira [4]

The Big O notation of the composite function 5N^3 O(N^2) is c. O(N^3).

A composite function of two capabilities combines the given two functions inside the given order. i.e., for any given functions f(x) and g(x), there may be four composite functions: f(g(x)) that is substituting g(x) into f(x) g(f(x)) that's substituting f(x) into g(x).

A composite characteristic is a function that relies upon any other feature. A composite characteristic is created while one characteristic is substituted for another feature. For instance, f(g(x)) is the composite function that is fashioned whilst g(x) is substituted for x in f(x). f(g(x)) is studied as “f of g of x”.

In Maths, the composition of a character is an operation wherein two capabilities say f and g generate a brand new feature say h in one of these manners that h(x) = g(f(x)). It method right here feature g is implemented to the function of x. So, basically, a feature is carried out to the result of every other feature.

Learn more about composite functions here brainly.com/question/10687170

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Answer:

I guess 36

Step-by-step explanation:

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