catmg

-2(s + 7) > 4s + 8

s + 7 < -2s - 4 (when we divide by a negative number we flip the sign)

3s < -11

s < -11/3

In interval notation, that is (-infinity, -11/3).

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Since there are 5 green marbles and 10 total marbles. The chance I draw a green marble at first is 5/10. Then, there are 4 green marbles and 9 total marbles. This means that the chance of drawing 2 green marbles is 5/10*4/9 = 2/9. 

Using the same method, solve for the chance of drawing 2 blue marbles and red marbles. 

Then, add all three up for the possibility of drawing two marbles of the same colour. 

:))

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A parabola with a vertex of (2, 4) with a vertical axis of symmetry can be written in the form a(x-2)^2 + 4.

Since the parabola contains the point (1, -11) -11 = a(1-2)^2 + 4.

Solve for a and expand the equation for equation of the parabola. 

I hope this was helpeful. :))

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q(x) = ax + b

p(x) = cx + d

p(2) = 2c + d = 3

p(q(x)) = c(ax + b) + d = acx + bc + d = 4x + 7

Now we have a system of equations, where we're looking for q(-1) = -a + b.

Can you take it from here?

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a^3 = x

1/(x+7) - 7 =- x/(x+7) + 6

(1 + x)/(x + 7) = 13

1 + x = 13x + 91

-90 = 12x

x = -15/2

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I'm gonna try giving problem 2 a shot. 

Part a: without diagonal lines 

Note that vertical lines and horizontal lines are the same, so we only need to count for one and multiply it by 2. 

There are 2 cases.

Case 1: a line of triangles, a line of squares, and a line of both triangles and squares. 

For the line of both triangles are squares, there are 6 possible arrangements. (s, t, t) (t, s, t) (t, t, s) (t, s, s) (s, t, s) (s, s, t)

There are also 6 possible arragements of the 3 lines (3!). 

Therefore, there are 36 possible options. 

Case 2: a line of triangles, a line of squares, and a line containing all triangles or all squares. 

For the line containing all triangles or all squares, there are 2 possible arragements. (s, s, s) (t, t, t)

Finally, for each option there are 3 possible arrangements for the 3 lines. (3C2)

Therefore, there are 6 possible options. 

Adding up both the cases we have 42 options. 

However, we have to account for the fact that we can use both horizontal and diagonal lines, so we have a final answer of 84.

Part b: without diagonal lines

I don't think it is possible to have a grid with a diagonal line of squares or triangles since it would make having another line for the other shapes impossible. 

Therefore, the answer is the same as without diagonal lines, 84. 

My answer seems to be different from yours, do you mind explaining your logic?

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x^2 + 10x + 25 = (x+5)^2

(x+5)^2 = 18

(x+5) = +/-sqrt(18)

Can you take it from here?

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That is a very cool way to visualize the problem. 

One thing I love about math is how many different ways there are to solve the question. :))

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Everything about your solution seems right to me. 

However, at the very end I got the answer 9/28 which is about 32.14%

6C2 = 15

4C2 = 6

9C3 = 84

6C3 = 20

3! = 6

(6*15*6)/(84*20) = (3*3*3)/(84) = 9/28

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Nice job. :DDD

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