n=1 a(1) = 0.001 n=2 a(2) = 0.012 n=3 a(3) = 0.144 n=4 a(4) = 1.728 n=5 a(5) = 20.736 .... n=k a(k) = (12^(k-1))/1000 let n = k+1 a(k+1) = 12^(k)/1000 12(ak) = a(k+1) 12(12^k-1)/1000 = 12^k/1000 the 12 gets absorbed here. 12^(k-1+1)/1000 = 12^k/1000 Valid for k and k+1 therefore our equation A(n) = 12^(n-1)/1000, n(greater than or equal to) 1
The equation is xy = k where k is the constant of variation. It can also be expressed y = k over x where k is the constant of variation.
In the equation m = k + 3, m is the:
To solve the equation 5.6 + k = 10, you need to isolate the variable "k" on one side of the equation. Start by subtracting 5.6 from both sides to get k = 10 - 5.6. This simplifies to k = 4.4. Therefore, the value of k that satisfies the equation is 4.4.
12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132 and 144 are just a few (note the Least Common Multiple is 12, and any multipleof 12 will be a common multiple of 4 and 6)
Any number of the form 84*k, where k is an integer, is evenly divisible.
They are members of the infinite set of numbers of the form 84*k where k is any integer.
79 + k = 84 84 - 79 = k k = 5
They are members of the infinite set of numbers of the form 84*k where k is an integer. Since the set is infinite, it is not possible to list them.
A line in the X-Y plane follows an equation y = (slope X x) plus a constant k. In this instances, 5 = [(-12)(3)] + k, or k = 5 + 36 = 41. Therefore, the equation is: y = -12x + 41.
k2 + 3k =108k2 + 3k - 108 = 0(k - 9) (k + 12)k = -9 , k = 12ANSWER:k = -9 , k = 12I hope this helps!
If y and x are related inversely, then the equation for y can be said to be:y = k/xTo find the constant k, substitute 12 for y and 6 for x (a pair of values that are known to satisfy the equation).y = k/x12 = k/612 X 6 = k72 = kThe value of the variation constant k is 72.
It is any fraction of the form (21*k)/(84*k) where k is a non-zero integer, or a common factor of the two.
a sign is missing, but if 11k - 13 = 119 then 11k = 132 and k = 12.
n=1 a(1) = 0.001 n=2 a(2) = 0.012 n=3 a(3) = 0.144 n=4 a(4) = 1.728 n=5 a(5) = 20.736 .... n=k a(k) = (12^(k-1))/1000 let n = k+1 a(k+1) = 12^(k)/1000 12(ak) = a(k+1) 12(12^k-1)/1000 = 12^k/1000 the 12 gets absorbed here. 12^(k-1+1)/1000 = 12^k/1000 Valid for k and k+1 therefore our equation A(n) = 12^(n-1)/1000, n(greater than or equal to) 1
The equation is xy = k where k is the constant of variation. It can also be expressed y = k over x where k is the constant of variation.
center (3,0) radius 12 So the equation is (x-3)^2 + y^2 = 144 → y^2 + x^2 - 6x - 135 = 0 in general a circle of radius r centered at (h, k) has the equation: (x - h)^2 + (y - k)^2 = r^2