If [math]x + y = 2 \text{ and } x^2 + y^2 = 2[/math], what is the value of [math]xy[/math] ? Could you break it down for me?
If it doesn't specify that x cannot equal y, than this can be resolved several waysFirst a system of equations,[math]x+y=2 [/math][math]x^2+y^2=2 [/math]Since there are two variables in both, we want to use one of the equations to isolate either x or y in terms of the other variable so that you obtain an equation y= or x=. For this example I shall use [math]x+y=2[/math] and subtract the x from both sides yielding [math]y=2-x[/math] (another way to write it is in linear form [math]y=-x+2[/math]Now you see we have an equation which gives us the value of y, in terms of x. This is important because by substituting the [math]-x+2[/math] (which is equivalent to y) into the second equation we get x^2+(-x+2)^2=2. Factoring out (-x+2)^2 → (x^2–4x+4) (but don't forget to add the rest) leaving you with the equation x^2+x^2–4x+4=2the first two x^2’s can be combined into 2x^2…—->(2x^2–4x+4)=2.to factor this out you need to subtract the 2 from both sides so your equation is set to 0, → 2x^2–4x+2=0. because the equation is set to zero you can divide the left side by 2 (since 0/2=0) thus getting x^2–2x+1=0factor that out into (x-1)(x-1) which means the only value for x can be one. Plug this back into the non-substituted equation x+y=2 |x=1 —-> 1+y=2, thus y=1. they must both be one. I thought this may have been a trick question and did it geometrically by graphing the circle x^2+y^2=2 and the line y=-x+2 and the point of intersection was tangental at exactly (1,1). If there is a specification that x and y have to have different value… 0.o then i can't even see how thats possible. However I am not versed in any advanced mathematics.The geometric representation is quite nice but I don't know how to present it on quora but your circle would be of radius sqrt[2] which happens to intersect the line y=-x+2 at exactly (1,1)
Write an equation relating variables x and y?
y= 24x? erm, i dont remember how to do this, and i took regular math so no summer assignment. you probably dont need to know it. the answer is : y= piX + Q bananas
How do we tell if a function is strictly increasing or strictly decreasing?
The easiest way to check if a function f(x) is increasing or decreasing - inspect the first derivative i.e. f’(x).If f’(x) > 0 (for all values of x) , this means f(x) will always be increasing.Or, if f’(x) < 0 (for all values of x) , this means f(x) will always be decreasing.For example, f(x) = ln(x) which exists for x > 0, f’(x) = 1/x. As we can see, f’(x) > 0 for x > 0. Therefore, it is an increasing function.As in your case, given that f’(x) = 7x^6 + 24x^2+4 and f’(x) = 24x^2, therefore the original equation would be something like this -x^7 + 8x^3 + 4x + c =0, where c is a constant.Now, f’(x) = 7x^6 + 24x^2+4.Since x has even powers in f’(x), and as shown in image below,f’(x) > 4, for all values of x.Hence your function will always be increasing.
Math word problem help?
Linear equation, substitution & elimination: x + y + x - y = 9 + 17 2x = 26 x = 13 13 + y = 9 y = - 4 Answer: x = 13, y = - 4 Check 2nd equation: = 13 - (- 4) = 13 + 4 = 17
Show 2 ways to evaluate x(x+y) for x=0.5 and y=0.6?
You could do it with Order of Operations by first doing x+y then multiplying by x, or, use distributive property, x*x + x*y.
Show that the curves y=x^4-2 and y=kx^2 intersect for all values of k?
Hi Jeff, This is a nice problem. Okay, so we have f: y = x^4 - 2; and g: y = kx^2. We have to show that they intersect for all values of k. Intersect means that cross at least once, and so we are going to have to show that for each value of k, f and g intersect at least once. Let's start by seeing where there is an intersection between f and g, if they do intersect. We do this by setting them equal: x^4 - 2 = kx^2; and simplifying gives us h: x^4 - kx^2 - 2 = 0, giving us a quadratic equation in x^2. Now we could use the quadratic formula here to actually try to generate a solution, in which you'd find that x^2 = (k +/- sqrt(k^2 + 8))/2; but Descartes rule of signs is probably a little more straight forward in this case, since we do not need to actually show what the solution is to h(x). Here we have h(x) = h(-x) = x^4 - kx^2 - 2, which has 1 sign change in the positive case and 1 sign change in the negative. Thus h will always have 1 positive real root, and 1 negative real root, and this is independent of the value of k. Therefore, we know that there are always real roots for h, meaning that f and g always intersect, regardless of the value of k. I've searched and found a webpage and a video tutorial that address problems similar to this one, and I thought they might be helpful to you. I've listed them below. As always, if you need more help, please clarify where you are in the process and what's giving you trouble. I'd be more than happy to continue to assist. If we've been helpful in answering your question, please consider being one of the first to stop by our brand new Facebook page at www.facebook.com/protutorcompany.
Can you help me answer some questions about a parabola?
y = x^2 - 4x - 32 1. I think this parabola opens up. Am I correct? Can you explain how you know weather it opens up or down? You are correct. Positive leading coefficient opens upward while negative opens downward. 2. Can you tell me what the axis of symmetry and the vertex of this parabola are? x = -b/2a = -(-4)/2(1) = 2 <---this is the axis of symmetry, x=8, and the x-portion of the vertex f(2) = (2)^2 -4(2) - 32 = -36 So vertex is at (2, -36) 3. Can you tell me the minimum? MIN when positive leading coefficient -- MAX when negative, both are same as vertex. In this case MIN at (2, -36) 4. Could you show me how to find the x-intercepts by factoring? x^2 - 4x - 32 = 0 (x-8)(x+4) = 0 x-8=0 x = 8 x+4=0 x = -4 5. How does the value of the discriminant (what I know to be b^2 - 4ac) support the conclusion in the previous question? Let's see: (-4)^2 - 4(1)(-32) = 144 Positive number means two real roots, which is exactly what we found in the previous question. 6. How does the axis of symmetry relate to the x-intercepts? The x-intercepts are equal distances from the axis of symmetry since the axis of symmetry cuts the parabola exactly in half. :)
In the integral of function f(x) dx, what does dx indicate?
Assuming that you know differentiation, let me explain you what is dx in the term.Let y = f(x).first lets differentiate this, what do we get?dy/dx = f’(x)Now we have obtained the derivative of f(x) with respect to x.remember y = f(x)Now we want f(x) again, what we are gonna do?…. inegrate.Integrate what?simple, we want value of y from the expression,dy/dx = f’(x)So we will integrate dy.Therefore, dy = f’(x)dx.This is what you get as a question in integration, you readily get expression f(x)dx to inegrate but actually its the value of dy. Remember, in differentiation you find a small change in quantity. In integration, you integrate the same quantity to get the real value. So basically both are the same things.Thanks for viewing ! If you have more questions on calculus, algebra, trigonometry etc please refer me.Leave an upvote please!
What are the X & Y intercepts of x^2 + 2x + 1 when the discriminate = 0?
Given x^2 + 2x + 1, I know that when I use the formula to calculate the discriminate, it = 0 . This tells me that the parabola touches the X-axis, here's where I get confused: I know how to get the X & Y intercepts when the discriminant > 0 but when it = 0 all this gives me is the vertex (-1,0) because it touches only at that point. The answer is (0,1) & (-2,1). How did they get this? I think the 1's or Y-intercepts came from substituting '0' for 'x' or (x=0) in the quadratic equation above, but where did the 0 & -2 come from? Please help me understand this. THANKS!