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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 is a number divisible by 3 and 4 but not 9?
12 and 24 Those are 2 excellent examples but to give an equation for ALL numbers, here you go! Any number that is equal to m*3*4^k where k is any integer >0 and m is any integer except 0, 3 or -3. Example: 1*3*4^1 = 12 2*3*4^1 = 24 5*3*4^1 = 60 ... 1*3*4^2 = 48 2*3*4^2 = 96 5*3*4^2 = 240 ...
Do straights wrap around in poker, allowing a hand like Q-K-A-2-3 to be considered a straight?
No, straights do not wrap around in poker. In a standard game, the highest straight is A-K-Q-J-10, and the lowest straight is A-2-3-4-5. The hand Q-K-A-2-3 would not be considered a straight in poker.
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
Can you use an ace for a straight?
Yes, an ace can be used for a straight in poker. It can either be the highest card in a straight, such as in A-K-Q-J-10, or the lowest card, as in 5-4-3-2-A. However, it cannot be used to form a straight in both positions simultaneously.
Is the ace considered low or high in poker?
In poker, the ace can be considered both low and high. It can be the lowest card in a straight (A-2-3-4-5) or the highest card in a straight (10-J-Q-K-A).
How many phonemes are in quack?
There are 4 phonemes in "quack" - /k/ /w/ /ae/ /k/.
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
If y varies inversely as the square of x and y equals 4 when x equals 3 find y when x is 2?
y varies as 1/x^(2) y = k X 1/^(2) y = k/x^(2) k = y x^(2) When y = 4 & x = 3 Then k = 4(3^(2)) k = 4(9) k = 36 (The constant of proportionality Hence the general equation is y = 36/x^(2) When x = 2 y = 36/2^(2) y = 36 /4 y = 9 The answer!!!!!
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
