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Aneli [31]
3 years ago
8

50 POINTS AND I WILL MARK BRAINLIEST Find the volume please help

Mathematics
2 answers:
ANEK [815]3 years ago
8 0

Answer:348

Step-by-step explanation:

For composite prisms, where the bases are a composite shape, the area of the bases is the sum of the areas of the parts it is made of. The volume of the composite prism is then given by multiplying the whole base area by the height of the prism.In math, volume is the amount of space in a certain 3D object. For instance, a fish tank has 3 feet in length, 1 foot in width and two feet in height. To find the volume, you multiply length times width times height, which is 3x1x2, which equals six. So the volume of the fish tank is 6 cubic feet.

Fed [463]3 years ago
3 0

Answer:

1,205.8\:  {ft}^{3}

Step-by-step explanation:

Given object is made up of cylindrical and conical parts.

Fir cylindrical part:

r = 16/2 = 8 ft

h = 4 ft

For conical part:

r = 16/2 = 8 ft

h = 6 ft

Volume of the object = Volume of cylinderical part + Volume of the conical part

= \pi {r}^{2} h +  \frac{1}{3} \pi {r}^{2} h \\  \\  =\pi \times  {8}^{2}  \times 4 +  \frac{1}{3}  \times \pi \times  {8}^{2}  \times 6 \\  \\  = 64\pi \times 4 +  \frac{1}{ \cancel3}  \times 64\pi \times  \cancel6 \\  \\  = 64\pi(4 + 2) \\  \\  = 64\pi \times 6 \\  \\  = 384\pi \\  \\  = 384 \times 3.14 \\  \\  = 1,205.8 \:  {ft}^{3}

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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
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