All categories

3 years ago

The volume in cubic centimeters of a sphere whose diameter is 6 centimes is ?

Mathematics

1 answer:

aivan3 [116]3 years ago

5 0

Answer:

Volume of the sphere is 36π (or) 113.097 cubic centimetre.

Step-by-step explanation:

$given \\ diameter \: of \: sphere = 6 \: cm \\ radius \: of \: sphere \: = \frac{diameter}{2} \\ radius \: = \frac{6}{2} \\ radius \: (r) = \: 3 \: cm \\ volume \: of \: sphere \: = \frac{4}{3} \pi \: {r}^{3} \\ here \: r \: = 3 \: cm \\ on \: substituting \: r \: = 3 \: \: in \: formula \\ volume \: of \: sphere \: = \frac{4}{3} \pi \: {3}^{3} \\ \: \: = \frac{4}{3} \pi \: 27 \\ \: = \: 4 \pi \times 9 \\ \: \: = 36\pi \\ \: \: = 113.097 \: {cm}^{3}$

HAVE A NICE DAY!

THANKS FOR GIVING ME THE OPPORTUNITY TO ANSWER YOUR QUESTION. 

You might be interested in

A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$

$6 = ln (\frac{820}{820-p_0})$

Using exponentials we got:

$e^6 = \frac{820}{820-p_0}$

$(820-p_0) e^6 = 820$

$820-p_0 = \frac{820}{e^6}$

$p_0 = 820-\frac{820}{e^6}$

8 0

3 years ago

. A special safe lets you choose from 9 symbols for a 3-symbol-long pass code. You may enter a symbol any number of times. How m

Lady_Fox [76]

The question is an illustration of combination and there are 729 potential pass codes available

<h3>How to determine the number of potential pass codes?</h3>

The given parameters are

Symbols available = 9

Length of pass code = 3

From the question, we understand that a symbol may be entered any number of times.

This means that each of the 9 available symbols can be used three times

So, the number of potential pass codes is

Passcodes = 9 * 9 * 9

Evaluate the product

Passcodes = 729

Hence, there are 729 potential pass codes available

Read more about combination at:

brainly.com/question/11732255

#SPJ1

4 0

1 year ago

Chose parameters h and k such that the system has a) a unique solution, b) many solutions, and c) no solution. X1 + 3x2 = 4 2x1

AysviL [449]

Answer:

a) The system has a unique solution for $k\neq 6$ and any value of $h$ , and we say the system is consisted

b) The system has infinite solutions for $k=6$ and $h=8$

c) The system has no solution for $k=6$ and $h\neq 8$

Step-by-step explanation:

Since we need to base the solutions of the system on one of the independent terms ( $h$ ), the determinant method is not suitable and therefore we use the Gauss elimination method.

The first step is to write our system in the augmented matrix form:

$\left[\begin{array}{cc|c}1&3&4\\2&k&h\end{array}\right]$

The we can use the transformation $r_0\rightarrow r_0 -2r_1$ , obtaining:

$\left[\begin{array}{cc|c}1&3&4\\0&k-6&h-8\end{array}\right]$ .

Now we can start the analysis:

If $k\neq 6$ then, the system has a unique solution for any value of $k$ , meaning that the last row will transform back to the equation as:

$(k-6)x_2=h-8\\x_2=h-8/(k-6)$

from where we can see that only in the case of $k=6$ the value of $x_2$ can not be determined.

if $k=6$ and $h=8$ the system has infinite solutions: this is very simple to see by substituting these values in the equation resulting from the last row:

$(k-6)x_2=h-8\\0=0$ which means that the second equation is a linear combination of the first one. Therefore, we can solve the first equation to get $x_1$ as a function of $x_2$ o viceversa. Thus, $x_2$ ( $x_1$ ) is called a parameter since there are no constraints on what values they can take on.

if $k=6$ and $h\neq 8$ the system has no solution. Again by substituting in the equation resulting from the last row:

$(k-6)x_2=h-8\\0=h-8$ which is false for all values of $h\neq 8$ and since we have something that is not possible $(0\neq h-8,\ \forall \ h\neq 8)$ the system has no solution

6 0

3 years ago

José is blocking off several rooms in a hotel for guest coming to his wedding. The hotel can reserve small rooms that can hold 3

miv72 [106K]

Answer:

$x=2y\,,\,3x+4y=70$ where x denotes number of small rooms and y denotes number of large rooms

Step-by-step explanation:

Given: Small rooms can hold 3 people, and large rooms can hold 4 people.

Number of small rooms is twice the number of large rooms.

Small rooms and large rooms altogether can accommodate 70 guests.

To find: system of equations that could be used to determine the number of small rooms reserved and the number of large rooms reserved

Solution:

Let x denotes number of small rooms and y denotes number of large rooms.

As number of small rooms is twice the number of large rooms,

$x=2y$

As small rooms and large rooms altogether can accommodate 70 guests such that small rooms can hold 3 people, and large rooms can hold 4 people,

$3x+4y=70$

4 0

2 years ago

Sheena baked 112 cookies she kept 16 cookies at home for her family and shared the rest evenly with her 8 friends how many cooki

sesenic [268]

Answer:

Each friend would get 12 cookies.

Step-by-step explanation:

Given:

Number of cookies she baked = 112 cookies

Number of cookies kept aside for family = 16

Number of friends she need to distribute cookies evenly = 8

We need to find number of cookies each friend would get;

Solution:

We will first find the number of cookies which needs to be distributed.

So we can say that;

number of cookies which needs to be distributed can be calculated by subtracting Number of cookies kept aside for family from Number of cookies she baked.

framing in equation form we get;

number of cookies which needs to be distributed = $112-16=96$

Now we will find number of cookies each friend would get.

number of cookies each friend would get can be calculated by dividing number of cookies which needs to be distributed from Number of friends she need to distribute cookies evenly.

framing in equation form we get;

number of cookies each friend = $\frac{96}{8}=12$

Hence Each friend would get 12 cookies.

7 0

3 years ago