GYanggg

Adding y and subtracting 5 to both sides on the top equation, we have:

y = 5x - 5 

Substituting y = 5x - 5 into the bottom equation, we have:

10x - 10 - x^2 = 11

x^2 -10x + 21 = 0.

Factoring:

( x - 3 ) ( x - 7 ) = 0

x = 3, x = 7

Solving for y, we have: y = 30, y = 10.

You can also graph the functions on desmos.com and then find the intersection points. 

I hope this helped, 

Gavin.

The second to last digit must be odd. 

1, 3, 5, 7, and 9.

We can write the number in the form (10x + 6)^2 or (10x + 4)^2.

We then expand it.

100x^2 + 120x + 36 and 100x^2 + 80x + 16.

We don't need to 100x^2.

Then we factor into the form 10A + 6.

120x + 30 + 6 and 80x + 10 + 6. 

Factoring, we have:

30(4x+1)+6 and 10(8x+1)+6.

The second to last digit of the perfect square is the last digit of 4x + 1 or 8x + 1

Since 8x is even, 8x + 1 must be odd. 

Therefore the tens digit possible values or 1, 3, 5, 7, and 9. 

Note: 0 is even, not odd. 

I hope this helped, 

Gavin.

We can factor each of these numbers.

1! + 2! = 1! ( 1 + 2 ) 

2! + 3! = 2! ( 1+ 3 )

3! + 4! = 3! ( 1 + 4 )

4! + 5! = 4! ( 1 + 5 )

5! + 6! = 5! ( 1 + 6 )

6! + 7! = 6! ( 1 + 7 )

7! + 8! = 7! ( 1 + 8 )

8! + 9! = 8! ( 1 + 9 )

Once again:

1! + 2! = 1! ( 3 ) 

2! + 3! = 2! ( 4 )

3! + 4! = 3! ( 5 )

4! + 5! = 4! ( 6 )

5! + 6! = 5! ( 7 )

6! + 7! = 6! ( 8 )

7! + 8! = 7! ( 9 )

8! + 9! = 8! ( 10 )

We can split the number into two parts, a! * b

The LCM of 1!, 2!, 3!, ... , 8! is 8!.

The LCM of 3, 4, 5, ... , 10 is 2520

The LCM is 2520 * 8! 

a = 2520

b = 8

a + b = 2528

I hope this helped,

Gavin

Since 24% of 1200 students have a part-time job and on the honor roll,

there are \( 0.24 \cdot 1200= 288\) students that have a part-time job and on the honor roll.

there are \(0.62\cdot1200=744\) students that have a part-time job, including the ones on the honor roll.

Since the problem states that they selected a student with a part time job, it must be one of the 744 students.

Out of the 744, we need the student to be one of the 288 people, so the probabilty of this happening is:

\(288\div744\approx0.3871\)

Rounding this to the nearest percent, we have 39%.

your final answer will be b.

I hope this helped, 

Gavin.

Once again, I do not understand the method you are asking me to solve the problem with. 

Here is my method. 

Since the problem is asking for the percentage of students that pass edthe test, we first need to know how many students actually did passed the test. 

We need to find all the numbers above 60, from this list 76, 55, 73, 93, 29, 74, 57, 34, 54.

We have 76, 73, 93, and 74, the rest are all below 60.

Since there are a total of 9 students, and only 4 passed the test. 

The percentage of students that passed the test is 4/9, or \(0.4\overline4%\)%

I hope this helped,

Gavin.

I do not understand the method you want me to solve the problem with. 

This is mine:

We can first find the median of this list of data, then find the numbers that are most of.

First, we can arrange it from least to greatest:

\(29, 34 , 54, 55, 57, 73, 74, 76, 93\)

The median, or middle number is 57.

The numbers farthest away from 57 are:

\( |57 - 29| = 28 \)

\(|57-93|=-36\)

\(|57-34|=23\)

Therefore, the scores that are most unusual are:

29, 93, and 34.

I hope this helped,

Gavin

Here is my solutions.

4. To "light" up all the squares with a knight going from A1 to H8, the knight must travel 63 times.

Everytime a knight moves, it goes from black to white square and vise versa. 

To go from white to white, the knight must move an even amount of times. 

Therefore this is not possible. 

Wow! Well done!

Hi Severoth!

To standard equation for a circle with its radius \( r\), and center \((h,k)\) is:

\((x - h)^2 + (y - k)^2 = r^2\)

Since your circle has a center \((-1,2)\) and a radius \(3\), 

The equation will be:

\((x+1)^2+(y-2)^2=9\)

I hope this helps, 

Gavin.

If l1 is horizontal to (1,1), then the equation of the line is y = 1.

If l2 is horizontal to (2,2) then the equation of the line is x = 2.

the intersection must be 2,1.

i hope this helped.

gavin