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Assuming you are familiar with all other basic geometry and algebra (which is needed first to even ask the question), it simply means inserting different values of V into the formula and calculating U, then marking the coordinate pairs that are formed in a Cartesian graph (or other appropriate graph form).
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Solve u plus v times u-v when you equals 5i plus j and when v equals i-6j?
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What is the derivative of the position function?
the velocity function v= at + v(initial)
How do you do chain rule?
The chain rule: given two functions of x, u & v, and let their respective derivatives with respect to x denote by u' and v', the derivative with respect to x of (u*v) is: v*u' + u*v' For division of two functions (u/v), derivative of (u/v) = [v*u' - u*v']/(v2) If you cannot remember if you've got it right, try this simple check: take f(x) = x3 but let u = x and v = x2, so u*v = x3. Derivative of x3 = 3*x2. Now try it with the chain rule: d(x*x2) = x2*1 + x*(2*x2)= 3*x2 I think the product rule is easier to remember, but sometimes cannot remember the quotient rule, so I'll try a simple one to make sure I got it right. Take u = x3 and let v = x, so u/v = x2, which the derivative = 2*x So for [v*u' - u*v']/(v2) we have: [x*3*x2 - x3*1]/(x2) = [2*x3]/(x2) = 2*x, so I did it correctly. I'll show one example of the product rule for more complex function, take sin(x)*ex --> u = sin(x), v=ex, so u' = cos(x) & v' = ex d(u*v) = ex * sin(x) + ex * cos(x) = (sin(x) + cos(x)) * ex
How can you use a discriminant to write an equation that has two solutions?
If the discriminant is greater than zero (b^2 - 4ac) > 0, then the equation have two roots that are real and unequal. Further, the roots are rational if and only if (b^2 - 4ac) is a perfect square, otherwise the roots are irrational.Example:Find the equation whose roots are x = u/v and x = v/uSolution:x = u/vx - u/v = 0x = v/ux - v/u = 0Therefore:(x - u/v)(x - v/u) = (0)(0) or(x - u/v)(x - v/u) = 0Let c = u/v and d = v/u. We can write this equation in equation in the form of:(x - c)(x - d) = 0x^2 - cx - dx + CD = 0 orx^2 - (c +d)x + CD = 0The sum of the roots is:c + d = u/v + v/u = (u)(u)/(v)(u) + (v)(v)/(u)(v) = u^2/uv + v^2/uv = (u^2 + v^2)/uvThe product of the roots is:(c)(d) = (u/v)(v/u) = uv/vu = uv/uv = 1Substitute the sum and the product of the roots into the formula, and we'll have:x^2 - (c +d)x + CD = 0x^2 - [(u^2 + v^2)/uv]x + 1 = 0 Multiply both sides of the equation by uv(uv)[x^2 - ((u^2 + v^2)/uv))x + 1] = (uv)(0)(uv)x^2 - (u^2 + v^2)x + uv = 0 which is the equatiopn whose roots are u/v, v/u
How do you make u the subject of the formula 1 over f equals 1 over u plus 1 over v?
1/f = 1/u+1/v Subtract 1/v from both sides: 1/f-1/v = 1/u Multiply all terms by fv: fv/f - fv/v = fv/u => v-f = fv/u Multiply all terms by u: u(v-f) = fv Divide both sides by v-f which will then make u the subject of the formula: u = fv/v-f
