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Gennadij [26K]
3 years ago
10

Find the volume of the sphere shown to the right. Give an answer in terms of pi and rounded to the nearest cubic unit.

Mathematics
1 answer:
Vladimir [108]3 years ago
5 0
The correct answers are:
Exact volume = V = \frac{32}{3} \pi in^3
Approximate volume = V = 33.51 in^3 

Explanation:

The volume of the sphere is: V = \frac{4}{3} \pi r^3

Since r = 2 in

Hence
V = \frac{4}{3} \pi (2)^3 \\ V =  \frac{32}{3} \pi in^3 (exact)
V = 33.51 in^3 (approximate)
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An unused roll of paper towels is shown. What is the volume of the unused roll?
vlada-n [284]

Answer:

11762.12

-hope this helped have a good rest of your day :)

Step-by-step explanation:

Calculate the area of the base (which is a circle) by using thee equation πr² where r is the radius of the circle. Then, multiply the area of the base by the height of the cylinder to find the volume. ... Since the volume V of a right circular cylinder is given by V = πr²h.

8 0
3 years ago
Read 2 more answers
How is 1/x + 2 - 4/2x +1 equivalent to - 2x + 7 / (x + 2)(2x + 1)
DanielleElmas [232]

We have,


\frac{1}{x + 2}  -  \frac{4}{2x + 1}  \\

Since the denominator isn't same we cannot directly perform any arithmetic operation on the terms, in order to be able to do that we must make the denominator same,

It's like the way we have,
\frac{2}{3}  +  \frac{4}{9}   \\

You know you cannot just add the terms and write the answer as,
\frac{8}{12}  \\
It completely wrong.

each of the terms have different values when when added doesn't gives 8/12.

Let me prove that to you by converting them into decimal values,

so
\frac{2}{3}  = 0.67 \\  \\ and \\  \\  \frac{4}{9}  = 0.43
Adding these two gives, 1.1

and
The result we had earlier equates to,
\frac{8}{12}  =  \frac{4}{3}  = 1.33 \\

You see the results are not equal.

So there must be something wrong.

And the wrong this here was not making the denominators equal,
To do that you basically multiply 2/3 with a 3/3 which may look stange at first but it does gives you same denominator as 4/9 in the form of 6/9.

Now you can add the numerators up and leave the denominator as it is which will give 10/9 = 1.1

Back to you question to show that the terms sperated by = are equal we must show one is identical to another.


Let's do this using the Right hand side,

\frac{( - 2x + 7)}{(x + 2)(2x + 1)}  \\  \\  =  \frac{(2x + 1) -4 (x + 2)}{(x + 2)(2x + 1)}  \\  \\  =    \frac{(2x + 1)}{(x + 2)(2x + 1)}  -  \frac{4(x + 2)}{(x + 2)(2x + 1)}  \\  \\  \frac{1}{x + 2}  -  \frac{4}{2x + 1}




Now this value is clearly equal to the the one on the left hand side of "=" .


3 0
3 years ago
I need the answers for questions 1 through 5
SOVA2 [1]

Answer: Sorry but we need the questions before I can give you answers.

Step-by-step explanation:

3 0
3 years ago
Write an expression in simplest form that represents the total amount. Suppose you buy 3 shirts that each cost s dollars, a pair
charle [14.2K]

Answer:

d

Step-by-step explanation:

3 shirts that each costs s dollars would be 3s

jeans is $30, pair of shoes is $50, add $30+$50 and you'll end up with $80

so, an expression in the simplest form that represents the total cost would be:

d. 3s+80

5 0
3 years ago
Which statement describes the inverse of m(x) = x2 – 17x?
stealth61 [152]

Answer:

The correct option is;

The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}

Step-by-step explanation:

The given information is that m(x) = x² - 17·x

The above equation can be written in the form;

y = x² - 17·x

Therefore;

0 = x² - 17·x - y

From the general solution of a quadratic equation, 0 = a·x² + b·x + c we have;

x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}

By comparison to the equation,0 = x² - 17·x - y, we have;

a = 1, b = -17, and c = -y

Substituting the values of a, b and c into the formula for the general solution of a quadratic equation, we have;

x = \dfrac{-(-17)\pm \sqrt{(-17)^{2}-4\times (1) \times (-y)}}{2\times (1)} = \dfrac{17\pm \sqrt{289+4\cdot y}}{2}

Which can be simplified as follows;

x =  \dfrac{17\pm \sqrt{289+4\cdot y}}{2}= \dfrac{17}{2} \pm \dfrac{1}{2}  \times \sqrt{289+4\cdot y}} = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +\dfrac{4\cdot y}{4} }}

And further simplified as follows;

x = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +y }} = \dfrac{17}{2} \pm \sqrt{y + \dfrac{289}{4} }}

Interchanging x and y in the function of the inverse, m⁻¹(x), we have;

m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}

We note that the maximum or minimum point of the function, m(x) = x² - 17·x found by differentiating the function and equating the result to zero, gives;

m'(x) = 2·x - 17 = 0

x = 17/2

Similarly, the second derivative is taken to determine if the given point is a maximum or minimum point as follows;

m''(x) = 2 > 0, therefore, the point is a minimum point on the graph

Therefore, as x increases past the minimum point of 17/2, m⁻¹(x) increases to give;

The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }} to increase m⁻¹(x) above the minimum.

8 0
3 years ago
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