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The random variable X has the following density function f(x) = { x^2 if 0 x 3 { 0 otherwise a) Find the distribution function F(x) of X b) Draw the distribution function c) Calculate the following probabilities P(X > 15) = You can view more similar questions or ask a new questionIf you substitute F = kx into 1/2Fx (F is negative so energy is gained) as F it becomes 1/2(kx)x So Elastic potential energy = 1/2kx^2 7 0 oldprof Lv 7 9 years ago The work function is work = average force X distance = FS And it's the work put into the spring to extend it that becomes the potential energy of the spring when extendedIt means a function x of x1x*(x1) is rightUse the ^ (caret) for exponentiation x^2 means x squared x/2y means Resources Simplifier Portal, help with entering simplifier formulas (a must read)

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P(x)=kx 2-√2x+1

P(x)=kx 2-√2x+1-Divide k1, the coefficient of the x term, by 2 to get \frac{k1}{2} Then add the square of \frac{k1}{2} to both sides of the equation This step makes the left hand side of the equation a perfect squareDomains of Power Functions If p is a nonzero integer, then the domain of the power function f(x) = kx p consists of all real numbers For rational exponents p, x p is always defined for positive x, but we cannot extract an even root of a negative number Thus x (1/4) is not defined for any negative real numbers Neither is x (3/4) (the fourth root of x cubed)

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The method applies to special equations ay′′ by′ cy = p(x)ekx where p(x) is a polynomial The idea, due to Ku¨mmer, uses the transformation y = ekxY to obtain the auxiliary equation a(D k)2 b(D k) cY = p(x), D = d dx The polynomial method applies to find Y Multiplication by ekx gives y Computational details are in Example 10P ∞ k=0 x k X∞ k=0 xk = 1xx2 x3 ··· 1 1−x, if x < 1, diverges, if x ≥ 1 Power Series Define a function f on the interval (−1,1) f(x) = X∞ k=0 xk = 1xx2 x3 ··· = 1 1−x for x < 1 As the Limit f can be viewed as the limit of a sequence of polynomials f(x) = lim n→∞ p n(x), where p n(x) = 1xx2 x3FZ(z) = P(Z ≤ z) = P(X/Y ≤ z) = ˆ 0 if z < 0 P(Y ≥ (1/z)X) if z > 0, where we have used the fact that X and Y are both nonnegative (with probability 1), so multiplying both sides of the inequality by Y does not flip the inequality;

(10) Z x a2 x2 dx= 1 2 lnja2 x2j (11) Z x2 a 2 x dx= x atan 1 x a (12) Z x3 a 2 x dx= 1 2 x2 1 2 a2 lnja2 x2j (13) Z 1 ax2 bx c dx= 2 p 4ac b2 tan 1 2ax b p 4ac b2 (14) Z 1 (x a)(x b) dx= 1 b a ln a x b x;Keeping in the spirit of (1) we denote a geometric p rv by X ∼ geom(p) Note in passing that P(X > k) = (1−p)k, k ≥ 0 Remark 13 As a variation on the geometric, if we change X to denote the number of failures before the first success, and denote this by Y, then (since the first flip might beNote, however, that when we divide both sides by z, to obtain Y ≥ (1/z)X, we were making the assumption

1 1 • x 3 1 = — = —————— 1 x 3 Checking for a perfect cube 62 x 8 4x 6 x 5 1 is not a perfect cube Trying to factor by pulling out 63 Factoring x 8 4x 6 x 5 1 Thoughtfully split the expression at hand into groups, each group having two terms Group 1 x 5 1 Group 2 x 8 4x 6You must use the "*" (star) symbol for all multiplications!x(x1) is WRONG!Continue reading "One of the roots of the equation x^2 kx – 6 = 0 is 3, and k is a" Skip to content Try for free Login Features Testimonials Results Who We Are Pricing Free Resources PrepScholar GRE Prep GRE Prep Online Guides and Tips

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The middle term of the expanded form, 2ab, would be 2kxb Our term is 2kx, so we want b to be 1 This means b^2 needs to be 1^2 = 1 To get from k^2 x^2 2kx k^2 to k^2 x^2 2kx 1, we add 1 k^2 to each side k^2 x^2 2kx 1 = 1 k^2 We can now put this into the form (kx 1)^2 = 1 k^2 so, taking the square root of both sides,In mathematics, the annihilator method is a procedure used to find a particular solution to certain types of nonhomogeneous ordinary differential equations (ODE's) It is similar to the method of undetermined coefficients, but instead of guessing the particular solution in the method of undetermined coefficients, the particular solution is determined systematically in this techniqueCompute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history

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Let's apply the cubic formula recipe so we can get an idea about how unhelpful it really is x^3 3bx^2 6cx 2d = 0 has roots given by p = b^22c q = 3bc b^3 d r = \sqrt{q^2 p^3}( x3)(x –1) (x –1)(x 3) Testing 7 , we get 0 $ 0, so 7 is a solution 2x – 14))))) $ 0 The solution set for this inequality is ( x 3) ( x – 1) œ œ (–3, 1) c 7, 4) critical values (where the signs of the factors can change) are 7, –3, 1 10 8 For P(x) = 2x 3 5x 2 – 15x 6, a List ALL the possible rational zeroes of P(x)Page 6 of 9 Q34 It is given that ln(y1)lny=13lnx Express y in terms of x, in a form not involving logarithms 4 Q35 The polynomial 8x3ax2bx3, where a and b are constants, is denoted by p(x)It is given that (2x1) is a factor of p(x) and that when

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Class X Maths (18k) Class X English (11k) Class X Hindi (598) Class X Computer (12) Class IX Social Science (32k) Class IX Science (13k) Class IX Maths (18k) Class IX English (14k) Class IX Hindi (270) Class IX Computer (28) Class VIII Social Science (151) Class VIII Science (3) Class VIII Maths (10k) Class VIII English (73) Class VIIISo S 2 −xS 2 = 13x5x2 7x3 = (24x6x2 ···)−(1xx2 ···) = 2S 1 −S 0 S 1(1−x) = 2 (1−x)2 1 1−x = 1x (1−x)2 X∞ k=0 (k1)2xk = S 2 = 1x (1−x)3 2 Geometric Distributions Suppose that we conduct a sequence of Bernoulli (p)trials, that is each trial has a success probability ofFind the values of p and q Also, for these values of p and q factorize the given polynomial completely Solution Let f(x) = px 3 4x 2 – 3x q It is given that f(x) is completely divisible by (x 2 – 1) = (x 1)(x – 1) Therefore, f(1) = 0 and f(1) = 0

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How do you find k such that the line is tangent to the graph of the function function #f(x) = x^2 kx# and line y=4x 9?Let f(x) = x^3 kx^2 2x 3 When we devide f(x) with (x2) th result will be a function (let it be R(x) and a remainder of 1 Then we could writeIt means a function x of x1x*(x1) is rightUse the ^ (caret) for exponentiation x^2 means x squared x/2y means Resources Simplifier Portal, help with entering simplifier formulas (a must read)

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Graph p(x)=(x2)(x2)(x3) Find the point at Tap for more steps Replace the variable with in the expression Simplify the result Tap for more steps Simplify each term Tap for more steps Raise to the power of Raise to the power of Multiply by Multiply by Simplify by adding and subtracting Tap for more stepsWe should attempt synthetic division to find when #(x^3kx^22kx15)/(x3)# has a remainder of #0#, which would signify that it is a factor of the polynomial The synthetic substitution would be set up asWhere b is a positive real number, and the argument x occurs as an exponent For real numbers c and d, a function of the form () = is also an exponential function, since it can be rewritten as = () As functions of a real variable, exponential functions are uniquely characterized by the fact that the growth rate of such a function (that is, its derivative) is directly proportional to the

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The polynomial px 3 4x 2 – 3x q is completely divisible by x 2 – 1;Here we will look at solving a special class of Differential Equations called First Order Linear Differential Equations First Order They are "First Order" when there is only dy dx, not d 2 y dx 2 or d 3 y dx 3 etc Linear A first order differential equation is linear when it can be made to look like this dy dx P(x)y = Q(x) Where P(x) and Q(x) are functions of x To solve it there is aTranscript Example 4 Find a zero of the polynomial p(x) = 2x 1 Putting p(x) = 0 2x 1 = 0 2x = 1 x = 1/2 So, 1/2

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SolutionShow Solution If x − 1 is a factor of polynomial p ( x ), then p (1) must be 0 `p (x) = kx^2 sqrt2x 1` p (1) = 0 `⇒ k (1)^2 sqrt2 (1) 1 = 0` `⇒ k sqrt2 1` `⇒ k = sqrt2 1` Therefore, the value of k is `sqrt2 1" "`X^4px^2q =x^2(x^22x5)2x^3–5x^2px^2q =x^2(x^22x5)2x(x^22x5)4x^210x5x^2px^2q =x^2(x^22x5)2x(x^22x5)(p1) x^210xq =x^2(x^22x5)2x(x^2If x1 is a factor of kx^23xk, then find the value of kAlso find the other factors for this value of k?

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A6=b (15) Z x (x a)2 dx= a a xCalculus Derivatives Tangent Line to a Curve 1 AnswerHere's a sample animation of long dividing #x^3x^2x1# by #x1# (which divides exactly) Write the dividend under the bar and the divisor to the left Each is written in descending order of powers of #x#If any power of #x# is missing, then include it with a #0# coefficient For example, if you were dividing by #x^21#, then you would express the divisor as #x^x1#

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This is how much i have done it so far p (x)=kx^23xk p (1)=k (1)^23 (1)k =k3k =2k3 k=3/2 p (3/2)=3/2x^23x (3/2) p (x)=3/2x^23x3/2Question a) For the quadratic equation kx^2 2x 4 = 0, find the value of k so that the roots are equal (there is only one root) b) The roots of x^2 (k 8)x 9k = 0 are equalCompute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history

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YÚn15¶/þ3žïþP€€ ï¦BÑF÷éaœÏ ü\à 9,H "ÑÚxÈQ —x \Ì_¯yê'ÿG0e‡ œfC&3‡ {~èU,}P†×\mg{q™bÿQˆ½ßü³2" oÙ ' ¶fSƒ òd%¦UóËãê¥ E ™Vuw ÒûåPœ%h;j,¢# §÷¡äJoš;ÿ~k©Ý Ò1>P œ!‚Çê'=olÉG­¬ ° ×—K ={ÿû" † ¹?Rù" ²Vçê_$Ã>K°ùMä˜iÉ~Ÿ)¼" 90 " æý ¹ »d²œYou must use the "*" (star) symbol for all multiplications!x(x1) is WRONG!Transcript Ex 24, 3 Find the value of k, if x 1 is a factor of p(x) in each of the following cases (i) p(x) = x2 x k Finding remainder when x2 x k is divided by x 1 Step 1 Put Divisor = 0 x 1 = 0 x = 1 Step 2 Let p(x) = x2 x k Putting x = 1 p(1) = (1)2 1 k = 1 1 k = 2 k Thus, Remainder = p(1) = 2 k Since x 1 is a factor of x2 x k Remainder is zero, 2 k = 0 k

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Then type x=6 Try it now 2x3=15 @ x=6 Clickable Demo Try entering 2x3=15 @ x=6 into the text box After you enter the expression, Algebra Calculator will plug x=6 in for the equation 2x3=15 2(6)3 = 15 The calculator prints "True" to let you know that the answer is right More ExamplesThe second form won't be independent of which root is p and which is q, so probably a typo It would lead to 2 different results including complex coefficients (xp)*(xq)=x^2x*(pq)p*q=x^2–2*x3 so pq=2, p*q=3 roots p2,q2 so (xp2)*(xq2)Complete the square in the quadratic function f given by f(x) = 2x 2 6x 4 Find the point(s) of intersection of the parabola with equation y = x 2 5x 4 and the line with equation y = 2x 2 Find the constant k so that x 2 (k 7)x 8 = (x 2)(x 4) Find the center and radius of the circle with equation x 2 y 22x 4y 11 = 0

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Since, α and β are the zeroes of the quadratic polynomial P(x) Kx 4x 4, thenIt is given that Since, α and β are the zeroes of the quadratic polynomial P(x) Kx 4x 4, thenIt is given that Chapter Chosen Polynomials Book Chosen Mathematics Subject ChosenIf one of the zeroes of the quadratic polynomial (k1)x2 kx 1 is 3,then the value of k is asked Feb 9, 18 in Class X Maths by akansha Expert ( 22k points) polynomialsKeeping in the spirit of (1) we denote a geometric p rv by X ∼ geom(p) Note in passing that P(X > k) = (1−p)k, k ≥ 0 Remark 13 As a variation on the geometric, if we change X to denote the number of failures before the first success, and denote this by Y, then (since the first flip might be

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X^4px^2q =x^2(x^22x5)2x^3–5x^2px^2q =x^2(x^22x5)2x(x^22x5)4x^210x5x^2px^2q =x^2(x^22x5)2x(x^22x5)(p1) x^210xq =x^2(x^22x5)2x(x^24 Binomial Expansions 41 Pascal's riTangle The expansion of (ax)2 is (ax)2 = a2 2axx2 Hence, (ax)3 = (ax)(ax)2 = (ax)(a2 2axx2) = a3 (12)a 2x(21)ax x 3= a3 3a2x3ax2 x urther,F (ax)4 = (ax)(ax)4 = (ax)(a3 3a2x3ax2 x3) = a4 (13)a3x(33)a2x2 (31)ax3 x4 = a4 4a3x6a2x2 4ax3 x4 In general we see that the coe cients of (a x)n come from the nth row of Pascal's(a) Since x2 2x 2 = (x 1)2 1 > 0 for all x ∈ Q, it follows that x2 2x 2 has no rational roots Hence, it's irreducible, and the quotient ring is a field (b) Apply the Extended Euclidean algorithm to x 3 1 and x 2 2x2

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Question a) For the quadratic equation kx^2 2x 4 = 0, find the value of k so that the roots are equal (there is only one root) b) The roots of x^2 (k 8)x 9k = 0 are equalGraph P(x)=(x1)(x1)(x2) Find the point at Tap for more steps Replace the variable with in the expression Simplify the result Tap for more steps Simplify each term Tap for more steps Raising to any positive power yields Raising to any positive power yields Multiply by Multiply byCompute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history

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3ln(x2 1) 42 Other Bases Other Bases f(x) = px, p > 0 Definition 15 For p > 0, the function f(x) = px = exlnp is called the exp function with base p Properties d dx px = px lnp ⇒ Z px dx = 1 lnp px C, for p > 0, p 6= 1 Other Bases f(x) = log p x, p > 0 Definition 16 For p > 0, the function f(x) = log p x = lnx lnp is called the logIf one of the zeroes of the quadratic polynomial (k1)x2 kx 1 is 3,then the value of k is asked Feb 9, 18 in Class X Maths by akansha Expert ( 22k points) polynomials

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