answer:This is a problem about units and ratios, so I will try to use some algebra to express the relationships given in the problem. Let t be the weight of one treek, s be the weight of one squig, and g be the weight of one goolee. Then I can write the following equations based on the problem: 10t = 3s + g 2t + g = s I want to find out how many treeks equals one squig, so I need to isolate s in one equation and then substitute it into the other. I will start with the second equation and solve for s: s = 2t + g Now I will plug this into the first equation and simplify: 10t = 3(2t + g) + g 10t = 6t + 3g + g 4t = 4g t = g This tells me that one treek and one goolee have the same weight. That means I can eliminate g from the second equation and get: s = 2t + t s = 3t This is the answer I was looking for. One squig is equal to the weight of three treeks.

answer:To find b, I need to plug in q(1) into f(x) and set it equal to -3. First, I'll find q(1) by substituting 1 for x in q(x). That gives me q(1) = b(1) + 1 = b + 1. Next, I'll plug in b + 1 for x in f(x). That gives me f(b + 1) = (b + 1) - 3 = b - 2. Now, I'll set f(b + 1) equal to -3 and solve for b. I have b - 2 = -3. Adding 2 to both sides, I get b = -3 + 2 = -1. Therefore, b = -1.

answer:To find the absolute value of the difference of the two roots, I need to first find the roots themselves. I can use the quadratic formula to find the roots: x = frac{-5 pm sqrt{5^2 - 4a(-3)}}{2a}. The difference of the two roots is then frac{-5 + sqrt{5^2 - 4a(-3)}}{2a} - frac{-5 - sqrt{5^2 - 4a(-3)}}{2a} = frac{2sqrt{5^2 - 4a(-3)}}{2a}. The absolute value of this difference is frac{sqrt{5^2 - 4a(-3)}}{a}. I want this to equal frac{sqrt{61}}{3}, so I can set up an equation: frac{sqrt{5^2 - 4a(-3)}}{a} = frac{sqrt{61}}{3}. To solve for a, I can cross-multiply and square both sides: (sqrt{5^2 - 4a(-3)})^2 = (frac{sqrt{61}}{3}a)^2. Simplifying, I get 25 + 12a = frac{61}{9}a^2. Multiplying both sides by 9, I get 225 + 108a = 61a^2. Rearranging, I get 61a^2 - 108a - 225 = 0. This is a quadratic equation in a, which I can solve by factoring: (61a + 75)(a - 3) = 0. The solutions are a = -frac{75}{61} and a = 3. However, the problem stated that a is positive, so I can ignore the negative solution and only keep a = 3.

answer:I need to find the coordinates of point A, which is the point where the two lines meet. I can do this by solving the system of equations 3y = x and 2x + 5y = 11. One way to do this is to substitute x = 3y from the first equation into the second equation, and then solve for y. This gives me 2(3y) + 5y = 11, which simplifies to 11y = 11. Therefore, y = 1. Now that I have y, I can plug it back into either equation to find x. I'll use the first one, since it's simpler. This gives me x = 3y = 3(1) = 3. So, the coordinates of point A are (3,1). The sum of the coordinates of point A is 3 + 1 = 4.