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beks73 [17]
1 year ago
13

A candy store called "Sugar" built a giant hollow sugar cube out of wood to hang above the entrance to their store. It took 13.5

\text{ m}^213.5 m
2
13, point, 5, start text, space, m, end text, squared of material to build the cube.
What is the volume inside the giant sugar cube?
Give an exact answer (do not round).
Mathematics
1 answer:
Liula [17]1 year ago
7 0

Answer: 3.375

Step-by-step explanation: I got it wrong to see the answer

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Catherine has $54. She plans to spend more than $20 of the money for a painting canvas. The rest will go toward paints. Each tub
kakasveta [241]

Answer:

4. 54- 8.5x>20

Step-by-step explanation:

Catherine only has $54, so she cannot spend more than that.

The canvas will cost at least $20, but we don't know how much exactly.

The tubes cost $8.50 each.

So, she starts with a total budget of $54, out of which she will buy paints (8.5x) and she wants to have at least $20 left for canvas.

So, we transpose those facts into the inequity:

54 - 8.5x > 20

8 0
2 years ago
Read 2 more answers
What is the 185th digit in the following pattern 12345678910111213141516
Delicious77 [7]

Answer:

The 185th digit following that pattern of '12345678910111213141516' would be the number '5'

Step-by-step explanation:

The reason the 185th digit is '5' is because the pattern increases it's standard number by the following number (ex, 1,2,3,4,5, etc), and eventually it reached 185 as one of it's numbers, '5' would be the digit because the '18' part of 185 would only be the 184th digit, hence you are needed to include the '5' to complete the sequence of that number to reach the 185th digit.

5 0
2 years ago
Choose the option that best completes the statement below. In finding the number of permutations for a given number of items, __
vodomira [7]
Let’s look at the permutations of the letters “ABC.” We can write the letters in any of the following ways:
ABC
ACB
BAC
BCA
CBA
CAB
Since there are 3 choices for the first spot, two for the next and 1 for the last we end up with (3)(2)(1) = 6 permutations. Using the symbolism of permutations we have: 3 P_{3}=(3)(2)(1)=6. Note that the first 3 should also be small and low like the second one but I couldn’t get that to look right.

Now let’s see how this changes if the letters are AAB. Since the two As are identical, we end up with fewer permutations.
AAB
ABA
BAA
To make the point a bit better let’s think of one A are regular and one as bold A.
A
BA and ABA look different now because we used bold for one of the As but if we don’t do this we see that these are actual the same. If they represented a word they would be the same exact word.

So in this case the formula would be \frac{3 P_{3} }{2!}= \frac{(3)(2)(1)}{(2)(1)}= \frac{6}{2}=3. We use 2! In the denominator because there are 2 repeating letters. If there were three we would use 3!


Hopefully, this is enough to let you see that the answer is A. The number of permutations is limited by the number of items that are identical.



7 0
2 years ago
Find all x in set of real numbers R Superscript 4 that are mapped into the zero vector by the transformation Bold x maps to Uppe
sukhopar [10]

Answer:

 x_3 = \left[\begin{array}{c}4&3&1\\0\end{array}\right]

Step-by-step explanation:

According to the given situation, The computation of all x in a set of a real number is shown below:

First we have to determine the \bar x so that A \bar x = 0

\left[\begin{array}{cccc}1&-3&5&-5\\0&1&-3&5\\2&-4&4&-4\end{array}\right]

Now the augmented matrix is

\left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\2&-4&4&-4\ |\ 0\end{array}\right]

After this, we decrease this to reduce the formation of the row echelon

R_3 = R_3 -2R_1 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\0&2&-6&6\ |\ 0\end{array}\right]

R_3 = R_3 -2R_2 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\0&0&0&-4\ |\ 0\end{array}\right]

R_2 = 4R_2 +5R_3 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&4&-12&0\ |\ 0\\0&0&0&-4\ |\ 0\end{array}\right]

R_2 = \frac{R_2}{4},  R_3 = \frac{R_3}{-4}  \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&1\ |\ 0\end{array}\right]

R_1 = R_1 +3 R_2 \rightarrow \left[\begin{array}{cccc}1&0&-4&-5\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&-1\ |\ 0\end{array}\right]

R_1 = R_1 +5 R_3 \rightarrow \left[\begin{array}{cccc}1&0&-4&0\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&-1\ |\ 0\end{array}\right]

= x_1 - 4x_3 = 0\\\\x_1 = 4x_3\\\\x_2 - 3x_3 = 0\\\\ x_2 = 3x_3\\\\x_4 = 0

x = \left[\begin{array}{c}4x_3&3x_3&x_3\\0\end{array}\right] \\\\ x_3 = \left[\begin{array}{c}4&3&1\\0\end{array}\right]

By applying the above matrix, we can easily reach an answer

5 0
2 years ago
A mail clerk found that the total weight of 155 packages was 815 pounds. if each of the packages weighed either 3 pounds or 8 po
8090 [49]
For this on you need to set up two equations
x + y = 155 total number of packages
3x+8y=815 total weight
Then multiply both sides of the first equation by 3: 3x +3y = 465
Then combine the two equations by subtracting the first one from the second one which gets you 5y=350 so y = 70 so we have 70 packages weighting 8 pounds. 
3 0
1 year ago
Read 2 more answers
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