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Firlakuza [10]
3 years ago
7

Is the square root of 15 an irrational number

Mathematics
1 answer:
Mkey [24]3 years ago
6 0
Yes, because nothing multiplied by itself is 15.. it will lead you to a number with a decimal
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Suppose there are 200 lockers and 200 students. Kayla reasons that since there are 10
Rus_ich [418]

Answer:

Kayla's reasoning is not correct.

Step-by-step explanation:

The locker problem is as follows:

Imagine you are at a school that has student lockers. There are 200 lockers, all shut and unlocked, and 200 students. Suppose the first student goes along the row and opens every locker. The second student then goes along and shuts every other locker beginning with number 2. The third student changes the state of every third locker beginning with number 3. (If the locker is open the student shuts it, and if the locker is closed the student opens it). The fourth student changes the state of every fourth locker beginning with number 4. Imagine that this continues until the 200 students have followed the pattern with the 200 lockers. At the end, which lockers will be open and which will be closed? Why?

Solution:

So from the information we know that the first student goes along the row and opens every locker.

Then the second student shuts every other locker, i.e. locker numbers 2, 4, 6, 8, 10, ..., 196, 198 and 200.

Then the third students changes the state of every third locker, i.e. he/she closes an open locker and opens a closed locker.

So the open lockers are: 1, 5, 6, 12,...

Then the fourth students changes the state of every fourth locker.

So the open lockers are: 1, 4, 5, 6, 8,....

So, on we will observe that the open lockers have a perfect square number such as, 1, 4, 9, 16,....

Consider that the pattern is as follows:

Student 1 opens the locker, Student 2 closes it, Student 3 opens it, person 4 Student and so on.

This is because the square numbers always have an odd number of factors, which leads them to be open at the end.

Take any locker number, 40, for example. Its state (open or closed) is changed for every student whose number in line is a factor of the locker number.

Student      Locker 40 status

     1                    Open

     2                   Close

     4                   Open

     5                   Close

     8                   Open

     10                  Close

    20                  Open

    40                  Close

Like all other lockers numbered with non-square numbers, it ends up closed after all the students have gone through the line because it has an even number of factors.

Consider the locker number 16:

Student      Locker 16 status

     1                    Open

     2                   Close

     4                   Open

     8                   Close

     16                  Open

Thus, we can conclude that all the doors with square numbers on them will remain open because all square numbers have an odd number of factors and the doors with non-square numbers on them will remain close because they have even number of factors.

There will be a total of 14 lockers open.

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169 and 196

So, if there are 10 lockers open in the first 100 lockers there must be only 4 other lockers opened in the next 100.

5 0
3 years ago
The rectangle shown is dilated by a scale factor of 3.
Law Incorporation [45]
Given that the image was dilated by scale factor 3 then:
a] Length of the dilated side will be:
length/width=(original length/width)*(scale factor)
length=3*10=30 in
Width=3*6=18 in
thus the dimensions of the new image will be:
30 in by 18 in

b. The diagram below represent s the rectangle. 1 cm represent 10 inches

7 0
3 years ago
Answer this question here :))
Studentka2010 [4]
Company A it should be lmk if im wrong
3 0
3 years ago
100 Points PLEASE ANSWER ASP! Will Give BRAINLIST!
VikaD [51]

Let's work to solve this system of equations:

y = 2x ~~~~~~~~\gray{\text{Equation 1}}y=2x        Equation 1

x + y = 24 ~~~~~~~~\gray{\text{Equation 2}}x+y=24        Equation 2

The tricky thing is that there are two variables, xx and yy. If only we could get rid of one of the variables...

Here's an idea! Equation 11 tells us that \goldD{2x}2x and \goldD yy are equal. So let's plug in \goldD{2x}2x for \goldD yy in Equation 22 to get rid of the yy variable in that equation:

\begin{aligned} x + \goldD y &= 24 &\gray{\text{Equation 2}} \\\\ x + \goldD{2x} &= 24 &\gray{\text{Substitute 2x for y}}\end{aligned}  

x+y

x+2x

​    

=24

=24

​    

Equation 2

Substitute 2x for y

​  

Brilliant! Now we have an equation with just the xx variable that we know how to solve:

x+2x3x 3x3x=24=24=243=8Divide each side by 3

Nice! So we know that xx equals 88. But remember that we are looking for an ordered pair. We need a yy value as well. Let's use the first equation to find yy when xx equals 88:

\begin{aligned} y &= 2\blueD x &\gray{\text{Equation 1}} \\\\ y &= 2(\blueD8) &\gray{\text{Substitute 8 for x}}\\\\ \greenD y &\greenD= \greenD{16}\end{aligned}  

y

y

y

​    

=2x

=2(8)

=16

​    

Equation 1

Substitute 8 for x

​  

Sweet! So the solution to the system of equations is (\blueD8, \greenD{16})(8,16). It's always a good idea to check the solution back in the original equations just to be sure.

Let's check the first equation:

\begin{aligned} y &= 2x \\\\ \greenD{16} &\stackrel?= 2(\blueD{8}) &\gray{\text{Plug in x = 8 and y = 16}}\\\\ 16 &= 16 &\gray{\text{Yes!}}\end{aligned}  

y

16

16

​    

=2x

=

?

2(8)

=16

​    

Plug in x = 8 and y = 16

Yes!

​  

Let's check the second equation:

\begin{aligned} x +y &= 24 \\\\ \blueD{8} + \greenD{16} &\stackrel?= 24 &\gray{\text{Plug in x = 8 and y = 16}}\\\\ 24 &= 24 &\gray{\text{Yes!}}\end{aligned}  

x+y

8+16

24

​    

=24

=

?

24

=24

​    

Plug in x = 8 and y = 16

Yes!

​  

Great! (\blueD8, \greenD{16})(8,16) is indeed a solution. We must not have made any mistakes.

Your turn to solve a system of equations using substitution.

Use substitution to solve the following system of equations.

4x + y = 284x+y=28

y = 3xy=3x

4 0
3 years ago
Read 2 more answers
Algebra help needed. Overwhelmed with other papers. See attached
MA_775_DIABLO [31]

Answer:

Step-by-step explanation:

whitch answer how do you want us to answer

5 0
3 years ago
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