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Bogdan [553]
2 years ago
10

= 28.16

Mathematics
1 answer:
frosja888 [35]2 years ago
8 0

Answer:

20 m

Step-by-step explanation:

4(3+2) = 20

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The slope of a line parallel to the graph of 4x-5y=12
Leya [2.2K]
Parallel lines have the same slope.

4x-5y=12 \\ -5y=-4x+12 \\ y= \frac{4}{5}x- \frac{12}{5}   \\  \\ slope:\frac{4}{5}
6 0
3 years ago
I need help on this :/ <br><br><br>Ignore my answer I accidentally clicked it
Ivanshal [37]
The answer is -4. I solved it on a calculator and got -4
6 0
3 years ago
What are the domain and range of the function f(x) = 4(3/81)*?
Ulleksa [173]

Answer:108

Step-by-step explanation: divided 81 and 3. the answer is 27. multiply 27 x 4 you get 108.

8 0
3 years ago
7 freshmen, 9 sophomores, 8 juniors, and 8 seniors are eligible to be on a committee.
Setler79 [48]

Answer:

i)32C16

ii)1185408

<em><u>Explanation</u></em><em><u>:</u></em>

i)Total number of selected/eligible is 7+9+8+8=32

Total ways of selecting dance committee of 16 is

<em><u>3</u></em><em><u>2</u></em><em><u>C</u></em><em><u>1</u></em><em><u>6</u></em>

ii)Total ways of selecting 3 seniors from 8 is 8C3

and Total ways of selecting 6 juniors from 8 is 8C6

ways of selecting 2 sopho from 9 is 9C2

ways of selecting 5 freshman from 7 is 7C5

now, total way of selection come to be

8C3×8C6×9C2×7C5

=56×28×36×21

=1185408

✌️

6 0
3 years ago
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
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