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3 years ago

If h = 9 units and r = 5 units, then what is the volume of the cone shown above?

Mathematics

1 answer:

dexar [7]3 years ago

6 0

Answer:

Answer = 235.5

Step-by-step explanation:

Cone volume formula is V = 1/3 * pi * r^2 * h

V = 1/3 * 3.14 * 5^2 * 9

V = 235.5

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Sara has two puppies, Sonny & Cher. Every month she purchases enough food to feed her puppies for the month. Sonny eats 8.25

Ipatiy [6.2K]

The amount of Cute Puppy food needed for the year is 1140 cans.

<h3>Determining the total number of cans that the dogs eat in a year</h3>

The first step in determining the total number of cans needed in year is to calculate the sum of dog food consumed in a week.

8.25 + 15.5 = 23.75 cans

The second step is to determine the total cans the dogs consume in a month

23.75 x 4 = 95

The third second step is to determine the total cans the dogs consume in a year:

95 x 12 = 1140 cans

<h3>Cost of cute puppy food needed in a year</h3>

1140 x ($21 /40) = $598.50

To learn more about addition, please check: brainly.com/question/19628082

3 0

2 years ago

In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$

This gives the infinite many solution.

5 0

3 years ago

17. Scott cuts his birthday cake into nine equal pieces and eats two of them.

Roman55 [17]

Answer:

A. 0.2222

B. Repeat indefinitely

Step-by-step explanation:

A. We have that Scott has 9 pieces in total and he eats 2 of them. That means he eats 2 <em>out of </em>the 9, so we have 2 ÷ 9. (see attachment)

The decimal form is: 0.2222.

B. Clearly, we can see that throughout the long division, we keep getting 20 - 18 = 2, which makes another 20 and so on. So, this will repeat indefinitely.