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zzz [600]
3 years ago
11

What is the volume of the sphere given below? if necessary, round your answer to two decimal places.

Mathematics
1 answer:
Marrrta [24]3 years ago
6 0
The formula for volume of a sphere is V=(4pi*r^3)/3.
Plugging in 12 for r we get
V=(4pi*(12)^3)/3
V=(4pi*1728)/3
V=(6912pi)/3
V=2304pi
Round to 2 decimal places: 7238.23
Final Answer:
7238.23 cubic units.
Hope I helped :)
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What value of x makes this equation true 12x-15=6-3x
Alex

Answer:

x = 1.4 (or 21/15)

Step-by-step explanation:

12x - 15 = 6 - 3x

+3x              +3x

-----------------------

15x - 15 = 6

       +15  +15

-----------------------

15x = 21

/15   /15

----------------------

x = 21/15  or

x = 1.4

6 0
3 years ago
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Identify the volume of the composite figure rounded to the nearest tenth. HELP PLEASE!!
nirvana33 [79]

Answer:

V = 115.3 ft³

Step-by-step explanation:

The left part of the figure shows a cube of side length 4.2 ft.  The volume of a cube is V = s³, where s is the side length.  Hence, the volume of this particular cube is V = (4.2 ft)³ = 74.088.

The volume of a pyramid is V = (1/3)(base area)(height).

Here V = (1/3)(4.2 ft)²(7 ft) = 41.16 ft³.

Summing up the two distinct areas, we get V = 41.16 ft³ + 74.088 ft³, or

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3 0
3 years ago
In the derivation of Newton’s method, to determine the formula for xi+1, the function f(x) is approximated using a first-order T
dimaraw [331]

Answer:

Part A.

Let f(x) = 0;

suppose x= a+h

such that f(x) =f(a+h) = 0

By second order Taylor approximation, we get

f(a) + hf'±(a) + \frac{h^{2} }{2!}f''(a) = 0

h = \frac{-f'(a) }{f''(a)} ± \frac{\sqrt[]{(f'(a))^{2}-2f(a)f''(a) } }{f''(a)}

So, we get the succeeding equation for Newton's method as

x_{i+1} = x_{i} + \frac{1}{f''x_{i}}  [-f'(x_{i}) ± \sqrt{f(x_{i})^{2}-2fx_{i}f''x_{i} } ]

Part B.

It is evident that Newton's method fails in two cases, as:

1.  if f''(x) = 0

2. if f'(x)² is less than 2f(x)f''(x)    

Part C.

In case  x_{i+1} is close to x_{i}, the choice that shouldbe made instead of ± in part A is:

f'(x) = \sqrt{f'(x)^{2} - 2f(x)f''(x)}  ⇔ x_{i+1} = x_{i}

Part D.

As given x_{i+1} = x_{i} = h

or                 h = x_{i+1} - x_{i}

We get,

f(a) + hf'(a) +(h²/2)f''(a) = 0

or h² = -hf(a)/f'(a)

Also,             (x_{i+1}-x_{i})² = -(x_{i+1}-x_{i})(f(x_{i})/f'(x_{i}))

So,                f(a) + hf'(a) - (f''(a)/2)(hf(a)/f'(a)) = 0

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Also,             x_{i+1} = x_{i} -f(x_{i})/f'(x_{i}) + [(x_{i+1} - x_{i})f''(x_{i})f(x_{i})]/[2(f'(x_{i}))²]

6 0
3 years ago
The profits in hundreds of dollars, p(c), that a company can make from a product is modeled by a function of the price, c, they
Fiesta28 [93]

Answer:

640,000

Step-by-step explanation:

took test

7 0
2 years ago
I need help pls help
Black_prince [1.1K]

Answer:

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Step-by-step explanation:

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so 1000 - 550 =450

3 0
2 years ago
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