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Sorry, I made a mistake on the y=mx+b formula so I changed it. 

      $1    $2$5$10

1  = 20   0   0   0  
2  = 18   1   0   0  
3  = 16   2   0   0  
4  = 14   3   0   0  
5  = 12   4   0   0  
6  = 10   5   0   0  
7  = 8   6   0   0  
8  = 6   7   0   0  
9  = 4   8   0   0  
10  = 2   9   0   0  
11  = 0   10   0   0  
12  = 15   0   1   0  
13  = 13   1   1   0  
14  = 11   2   1   0  
15  = 9   3   1   0  
16  = 7   4   1   0  
17  = 5   5   1   0  
18  = 3   6   1   0  
19  = 1   7   1   0  
20  = 10   0   2   0  
21  = 8   1   2   0  
22  = 6   2   2   0  
23  = 4   3   2   0  
24  = 2   4   2   0  
25  = 0   5   2   0  
26  = 5   0   3   0  
27  = 3   1   3   0  
28  = 1   2   3   0  
29  = 0   0   4   0  
30  = 10   0   0   1  
31  = 8   1   0   1  
32  = 6   2   0   1  
33  = 4   3   0   1  
34  = 2   4   0   1  
35  = 0   5   0   1  
36  = 5   0   1   1  
37  = 3   1   1   1  
38  = 1   2   1   1  
39  = 0   0   2   1  
40  = 0   0   0   2  
 

It's actually really easy to do it yourself. Try it.

:)

Sorry if it's a bit of a bother but could you put it in standard form? like Ax + By=C? Thank you!

x (x + 9) = 112

1. Let's find two random points on the graph. (-3, 3) and (-1, -1)

2. Slope=(y2-y1)/(x2-x1)

In this case, slope=((3)-(-1))/((-3)-(-1))=4/-2=-2.

Slope=-2

3. The y-intercept is -3, so the equation will be x=-2y-3.

Answer: y=-2x-3 

Joshua rolls two dice to form a two-digitinteger. If the number on the first die represents the tens digit and the number on the second die represents the units digit, what is the probability that the integer formed is divisible by 8?  

There are a possible 36 combinations. Since 5 of those combinations satisfy the criterion, the probability is 5/36. 

1,1   1,2   1,3   1,4   1,5   1,6  

2,1   2,2   2,3   2,4   2,5   2,6  

3,1   3,2   3,3   3,4   3,5   3,6  

4,1   4,2   4,3   4,4   4,5   4,6  

5,1   5,2   5,3   5,4   5,5   5,6  

6,1   6,2   6,3   6,4   6,5   6,6

_.

(2x + y)/(3x - 2y) = 2/7.

1. You can make 6 unique pairs with 1, 4, 5, and 8.

Pairs: 1+4, 1+5, 1+8, 4+5, 4+8, 5+8

Sums: 5, 6, 9, 9, 12, 13

2. After adding the pairs together, we get 6 sums. Out of these 6 sums, 4 out of 6 are greater than 7.

3. The probability will be written as 4/6. Simplify, then we get 2/3.

Answer: 2/3 

https://latex.artofproblemsolving.com/7/7/c/77c7d3326de037aba782d38a9ee2a6b2f0a03c83.pngHeres the graph btw

Just corrected my answer....check again!

:)

Sorry, the answer is 4! Sillly me, I need to learn how to count...

yeah im struggling on it too...

I put in 2 as the answer but it said it was wrong. Are you sure it is right? It asked for the number of values for c.

onehundredtwentysix

Replied on original post.

:)

This question is asking for how many times the y axis is equal to -1. We see that there are 2 times.

:)

In trapezoid PQRS, Base PQ is parallel to base RS. Let point X  be the intersection of diagonals PR and QS. The area of triangle PQX is 20 and the area of triangle RSX is 45 Find the area of trapezoid PQRS.

Since the type of trapezoid is not given, we may use isosceles trapezoid.

Bases are 10 and 18, and the height is 9. 

Sigh, brothers these days...

um this was blocked by the moderator...

Since f(x) is divisible by x^3, f(x) is of the form ax^5 + bx^4 + cx^3.

You then want ax^5 + bx^4 + cx^3 + 2 to be divisible by (x + 1)^3.  Using long division, you get the equations

-10a  + 6b - 3c = 0

4a - 3b + 2c = 0

-a + b - c + 2 = 0

==> a = 6, b = 16, c = 12

So f(x) = 6x^5 + 16x^4 + 12x^3.