so we know the terminal point is at (9, -3), now, let's notice that's the IV Quadrant
![\bf (\stackrel{x}{9}~~,~~\stackrel{y}{-3})\impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{9^2+(-3)^2}\implies c=\sqrt{81+9}\implies c=\sqrt{90} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%28%5Cstackrel%7Bx%7D%7B9%7D~~%2C~~%5Cstackrel%7By%7D%7B-3%7D%29%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Bhypotenuse%7D%7D%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20c%3D%5Csqrt%7Ba%5E2%2Bb%5E2%7D%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20c%3D%5Csqrt%7B9%5E2%2B%28-3%29%5E2%7D%5Cimplies%20c%3D%5Csqrt%7B81%2B9%7D%5Cimplies%20c%3D%5Csqrt%7B90%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
Answer:
k = 8
Step-by-step explanation:
Since the roots of the equation 3x² + 2x + k are x₁ and x₂ and 2x₁ = 3x₂.
By the roots of an equation, ax² + bx + c , where its roots are x₁ and x₂. It follows that sum of roots are x₁ + x₂ = -b/a and product of roots are x₁x₂ = c/a.
Comparing both equations, a = 3 b = 2 and c = k.
So, x₁ + x₂ = -b/a = -2/3 and x₁x₂ = c/a = k/3
x₁ + x₂ = -2/3 and x₁x₂ = k/3
3x₁ + 3x₂ = -2 (1) and 3x₁x₂ = k (2)
Since 2x₁ = -3x₂, and -2x₁ = 3x₂substituting this into equations (1) and (2) above, we have
3x₁ + 3x₂ = -2 (1) and 3x₁x₂ = k (2)
3x₁ + (-2x₁) = -2 and x₁(3x₂) = k
3x₁ - 2x₁ = -2 and x₁(3x₂) = k
x₁ = -2 and x₁(-2x₁) = k
x₁ = -2 and -2x₁² = k
Substituting x₁ = -2 into -2x₁² = k, we have
-2x₁² = k
-2(-2)² = k
2(4) = k
8 = k
So, k = 8
Answer:
45
Step-by-step explanation:
Answer: y = 7- 5x
Step-by Step Explanation: The variable x is multiplied by a larger value here; it's multiplied by 5. So I should expect that my y-values will grow fairly quickly. This means that I should expect a fairly "tall" graph.
First I'll do the T-chart.
T-chart
This equation is an example of a situation in which you will probably want to be particular about the x-values you pick. Because the x is multiplied by a relatively large value, the y-values grow quickly. For instance, you probably wouldn't want to use x = 10 or x = –7 as inputs. You could pick larger x-values if you wished, but your graph would very quickly get awfully tall.
I can see, from my T-chart, that my y-values are getting pretty big on either end (that is, in the positive numbers above the horizontal axis, and in the negative numbers below). I don't want to waste time computing points that will only serve to make my graph ridiculously large, so I'll quit with what I've got so far. But I'm glad I plotted more than just two points, because lines that start edging close to vertical can easily go wrong, if I'm not neat in my work.
Here's my graph:
y = 7 - 5x
25 percentage chance of getting heads three times in a row
