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kolbaska11 [484]
3 years ago
9

What is the slope of a line that is perpendicular to the line whose equation is 2y = 3x - 1? -3 - -

Mathematics
1 answer:
Shalnov [3]3 years ago
8 0

Answer: 2/3x

Step-by-step explanation:

First, you need to simplify your original line to point slope formula. Dividing the whole thing by 2 to get y by itself leaves you with

y = 3/2x - 1/2

Any perpendicular line has the opposite slope to the line it is perpendicular to, so you flip the slope to get 2/3x

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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
1. Identify the base and the exponent in the expression 10^15
JulsSmile [24]

Answer:

Base is 10, exponent is 15

Step-by-step explanation:


3 0
3 years ago
A. - 8/5<br><br>B. 5/8<br><br>C. - 5/8 <br><br>D. 8/5​
IrinaK [193]

Answer:

5/8

Step-by-step explanation:

To find the slope, use the formula

m = (y2-y1)/(x2-x1)

   = (2 - -3)/(6 - -2)

   = (2+3)/(6+2)

   = 5/8

5 0
3 years ago
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pentagon [3]
A.) wooden blue, wooden green.....
4 0
2 years ago
Please help, thank you :)​
BARSIC [14]

Answer:

30

Step-by-step explanation:

You already have 40 points and every movie is two points. 100-40=60. 60 divided by 2 is 30, so you still need to watch 30 more movies.

8 0
2 years ago
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