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raketka [301]
2 years ago
8

2. A store is to pack pencils in boxes of 10 and 12 pieces. What is the smallest number of pencils that can be packed using the

boxes?​
Mathematics
1 answer:
Taya2010 [7]2 years ago
3 0

Answer:

2

Step-by-step explanation:

Obtain the greatest common factor of both 12 and 10

Factors of 10 = 1, 2, 5, 10

Factors of 12 = 1, 2, 3, 4, 6, 12

From the above, the greatest common factor of 10 and 12 is 2

Hence, smallest number of pencils using the boxes is 2

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Justin is constructing a line through point Q that is perpendicular to line n. He has already constructed the arcs shown. A line
xz_007 [3.2K]

I believe that this problem has the following choices:

It must be equal to BQ .<span>
It must be wider than when he constructed the arc centered at point A.
It must be equal to AB .
It must be the same as when he constructed the arc centered at point A.</span>

 

The correct answer is the last one:

It must be the same as when he constructed the arc centered at point A.

<span> </span>

4 0
2 years ago
Read 2 more answers
One semester in a chemistry class, 14 students failed due to poor attendance, 23 failed due to not studying, 15 failed because t
BlackZzzverrR [31]

Answer:

(a) 16 students failed for exactly two of the three reasons.

(b) 7 students failed because of poor attendance and not studying but not because of not turning in assignments.

(c) 14 students failed because of exactly one of the three reasons.

(d) 3 students failed because of poor attendance and not turning in assignments but not because of not studying.

Step-by-step explanation:

We are given that one semester in a chemistry class, 14 students failed due to poor attendance, 23 failed due to not studying, 15 failed because they did not turn in assignments, 9 failed because of poor attendance and not studying, 8 failed because of not studying and not turning in assignments, 5 failed because of poor attendance and not turning in assignments, and 2 failed because of all three of these reasons.

As shown in the diagram attached below, a Venn diagram represents the situation given in the question;

In the diagram below, 2 in represents purple color represents the number of students failed because of all three of these reasons.

7 represents the number of students who failed because of poor attendance and not studying but not due to not turning in assignments.

6 represents the number of students who failed because of not studying and not turning in assignments but not due to poor attendance.

3 represents the number of students who failed because of poor attendance and not turning in assignments but not due to not studying.

2 in red color represents the number of students who failed only due to poor attendance.

8 in red color represents the number of students who failed only due to not studying.

4 in red color represents the number of students who failed only due to not turning in assignments.

(a) The number of students who failed because of exactly two of the three reasons = 7 + 6 + 3 = 16 students.

(b) The number of students who failed because of poor attendance and not studying but not because of not turning in assignments = 9 - 2 = 7 students.

(c) The number of students who failed because of exactly one of the three reasons = 2 + 8 + 4 = 14 students.

(d) The number of students who failed because of poor attendance and not turning in assignments but not because of not studying = 5 - 2 = 3 students.

7 0
2 years ago
A line contains the points (82, −96) and (87, −86).
Alik [6]

Answer:

<h2>The answer is 2</h2>

Step-by-step explanation:

The slope of a line given two points can be found by using the formula

m =  \frac{ y_2 - y _ 1}{x_ 2 - x_ 1} \\

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

(82, −96) and (87, −86)

We have

m =  \frac{ - 86 -  - 96}{87 - 82}  =  \frac{ - 86 + 96}{5}  =  \frac{10}{5}  = 2 \\

We have the final answer as

<h3>2 </h3>

Hope this helps you

8 0
2 years ago
Enrollment in a school has grown exponentially since the school opened. A graph depicting this growth is shown. Determine the pe
DedPeter [7]

Answer:

Your correct answer is A. 0.6%

8 0
2 years ago
Read 2 more answers
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
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