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What are the possible values of k when kx2 plus x2 plus kx plus k plus 1 equals 0 has repeated roots?
Equation: kx^2 +x^2 +kx +k +1 = 0 Using the discriminant: K^2 -4*(k +1)*(k +1) = 0 Expanding brackets: k^2 -4k^2 -8k -4 = 0 Collecting like terms: -3^2 -8k -4 = 0 Dividing all terms by -1: 3k^2 +8k +4 = 0 Factorizing the above: (3k +2)(k +2) = 0 meaning k = -2/3 or -2 Therefore possible values of k are either: -2/3 or -2
What are the possible values of k when kx squared plus x squared plus kx plus k plus 1 equals 0 has equal roots?
Taking all terms and conditions into consideration a quadratic equation can be finally formed as such that 3k^2 +8k +4 = 0 whose solutions are k = -2/3 or k = -2 Check: 3(-2/3)^2 +8(-2/3) +4 = 0 Check: 3(-2)^2 +8(-2) +4 = 0
How can you prove that the straight line of y equals x plus k is a tangent to the curve of x squared plus y squared equals 4 and only when k squared equals 8?
You don't.Furthermore, how would you prove your not a virgin? Either way your fu'd.Improved Answer:-y = x+kx2+y2 = 4Substitute y = x+k into the bottom equation:x2+(x+k)(x+k) = 4x2+x2+2kx+k2 = 4k2 = 8 so therefore:2x2+2kx+8-4 = 0 => 2x2+2kx+4 = 0For the straight line to be a tangent to the curve the discriminant b2-4ac of the quadratic equation must equal zero:Hence:-(2*k)2-4*2*4 = 0k2 = 8So: 4*8-4*2*4 = 0 => 32-32 = 0Therefore the straight line y = x+k is a tangent to the curve x2+y2 = 4
What is a number divisible by 2 3 4 5 6 and has a remainder of 1?
If LCM(2, 3, 4, 5, 6) = L, then the required number is l*k + 1 where k is an integer.
What are the possible values of k when the curve of kx squared plus x squared plus kx plus k plus 1 equals 0 has equal roots?
If the equation has equal roots then the discriminant of b^2 -4ac = 0:- Equation: kx^2 +x^2 +kx +k +1 = 0 Discriminant: k^2 -4(k+1)(k+1) = 0 Multiplying out brackets: k^2 -4k^2 -8k -4 = 0 Collecting like terms: -3k^2 -8k -4 = 0 Divide all terms by -1: 3k^2 +8k +4 = 0 Factorizing: (3k +2)(k +2) = 0 => k = -2/3 or k = -2 Therefore possible values of k are -2/3 or -2
Is it a straight if you have 10 J Q K A?
Yes, that is a straight. A straight consists of five cards of consecutive rank. The Ace can be used as either the lowest or the highest card, (i.e., A, 2, 3, 4, 5 is a straight, as well as 10, J, Q, K, A)
Is it a straight if you have J Q K A 2?
No, a straight are five cards in sequence. Aces are high, deuces are low. They do not wrap around. Examples: A, K, Q, J, 10; A, 2, 3, 4, 5; 5, 6, 7, 8, 9.
What is the factor of 6k2-k-12?
6k2 - k - 12 (2k - 3)(3k + 4) k = 3/2 and -4/3 1.5 and -1.33 are the factors
What is the equation of the quadratic graph with a focus of (3 6) and a directrix of y 4?
For a parabola with a y=... directrix, it is of the form: (x - h)^2 = 4p(y - k) with vertex (h, k), focus (h, k + p) and directrix y = k - p With a focus of (3, 6) and a directrix of y = 4, this means: (h, k + p) = (3, 6) → k + p = 6 y = k - p = 4 → k = 5, p = 1 (solving the simultaneous equations) → vertex is (3, 5) → parabola is (x - 3)^2 = 4(y - 5) which can be rearranged into y = 1/4 x^2 - 3/2 x + 29/4
What are the possible values of k when kx2 plus x2 plus kx plus k plus 1 equals 0 has repeated roots?
Equation: kx^2 +x^2 +kx +k +1 = 0 Using the discriminant: K^2 -4*(k +1)*(k +1) = 0 Expanding brackets: k^2 -4k^2 -8k -4 = 0 Collecting like terms: -3^2 -8k -4 = 0 Dividing all terms by -1: 3k^2 +8k +4 = 0 Factorizing the above: (3k +2)(k +2) = 0 meaning k = -2/3 or -2 Therefore possible values of k are either: -2/3 or -2
What are the possible values of k when kx squared plus x squared plus kx plus k plus 1 equals 0 has equal roots?
Taking all terms and conditions into consideration a quadratic equation can be finally formed as such that 3k^2 +8k +4 = 0 whose solutions are k = -2/3 or k = -2 Check: 3(-2/3)^2 +8(-2/3) +4 = 0 Check: 3(-2)^2 +8(-2) +4 = 0
Which value of k makes 5 - k + 12 = 16 a true statement?
K=1
How can you prove that the straight line of y equals x plus k is a tangent to the curve of x squared plus y squared equals 4 and only when k squared equals 8?
You don't.Furthermore, how would you prove your not a virgin? Either way your fu'd.Improved Answer:-y = x+kx2+y2 = 4Substitute y = x+k into the bottom equation:x2+(x+k)(x+k) = 4x2+x2+2kx+k2 = 4k2 = 8 so therefore:2x2+2kx+8-4 = 0 => 2x2+2kx+4 = 0For the straight line to be a tangent to the curve the discriminant b2-4ac of the quadratic equation must equal zero:Hence:-(2*k)2-4*2*4 = 0k2 = 8So: 4*8-4*2*4 = 0 => 32-32 = 0Therefore the straight line y = x+k is a tangent to the curve x2+y2 = 4
Show that n factorial is greater than or equals to 2 the power of n for n greater than or equal to 4?
For n = 4, 4! = 4*3*2*1 = 24 and 24 = 2*2*2*2 = 16 So the statement is true for n = 4. Suppose it is true for n = k, that is, k! > 2k Then (k+1)! = (k+1)*k! > (k+1)*2k (since k! > 2k) > 2*2k (since k >= 4 > 2) = 2k+1 So if the statement is true for n = k then it must be true for n = k+1. Therefore, since it is true for n = 4 it must be true for all n > 4.
What is a number divisible by 2 3 4 5 6 and has a remainder of 1?
If LCM(2, 3, 4, 5, 6) = L, then the required number is l*k + 1 where k is an integer.
Factor the expression 2 k squared minus k minus 4 equals 0?
2 k^2 - k - 4 = 0 2 (k^2 - (1/2)k - 2) = 0 2 ((k - 1/4)^2 - 1/16 - 2) = 0 2 ((k - 1/4)^2 - 33/16) = 0 2 (k - 1/4 - sqrt(33)/4)(k - 1/4 + sqrt(33)/4) = 0 32 (4k - 1 - sqrt(33))(4k - 1 + sqrt(33)) = 0
Are multiples of 4 always even why?
The multiples of 4 are numbers of the form 4*k where k is an integer.4*k = 2*(2*k) that is, 4*k is a multiple of 2 - in other word, it is even.
