(a - t)/(b - t) = c => a - t = c(b - t) = cb - ct = bc - tc => tc - t = bc - a => t(c - 1) = bc - a => t = (bc - a)/(c - 1)
The geometric mean of n numbers (t{1}, t{2}, ..., t{n}) is given by (Π t{n})^(1/n) → geometric mean of 8.5 and 12.4 = (8.5 × 12.4)^(1/2) = 10.26645... ≈ 10.266
T = 2(pi)sqrt(I/(mgh)) f' = f0sqrt((1+(v/c))/(1-(v/c))) y(x,t) = Asin(kx -omega(t) + phi)
r=0,Tr-r = 0 = r(T-1), since T != 1, then T-1 is non zero so r must be zero.
T = Time (time to launch/ignition) .. used in countdown examples: T-30 = 30 seconds to launch/ignition T-1 = 1 second to launch/ignition T-0 = time of launch/ignition
what does the t and c mean on the pregnancy test
(a - t)/(b - t) = c => a - t = c(b - t) = cb - ct = bc - tc => tc - t = bc - a => t(c - 1) = bc - a => t = (bc - a)/(c - 1)
t s 1 = first up
16 T in a c means 16 Tablespoons in a Cup
int f1=1, f2=1, c=2; do { t=f1+f2; printf("%d\t",t); f1=f2; f2=t; c=c+1; }while(c
C- Control T- means test Line under the C - Negative Line under C and T- positive :)
Find I = ∫ tan³ x dx. The solution is: I = ½ tan² x - log cos x. * * * Here is how we can obtain this result: First, let t = tan x, s = sin x, and c = cos x; then, dI = t³ dx, ds = c dx, dc = -s dx, and dt = (1 + t²) dx; and, of course, t = s / c. By algebra, t³ = t(t² + 1) - t; thus, we have dI = t³ dx = t(t² + 1) dx - t dx = t dt - t dx. Now, d (t²) = 2t dt; thus, t dt = ½ d(t²). On the other hand, we have d log c = dc / c = -s dx / c = -t dx; thus, t dx = -d log c. Combining these results, we have dI = t dt - t dx = ½ d(t²) - d log c. This integrates readily, giving I = ½ t² - log c, which is the solution we sought. * * * We may check our result, by differentiating back: dt / dx = 1 + t²; and d(t²) / dt = 2t; thus, (d/dx)(t²) = 2t dt / dx = 2t (1 + t²). Also, we have d log c / dc = 1 / c; and dc / dx = -s; whence, (d/dx)(log c) = (dc / dx) / c = -s / c = -t. Then, dI / dx = ½ (d/dx)(t²) - (d/dx)(log c) = t (1 + t²) - t = t + t³ - t = t³, re-assuring us that we have integrated correctly.
2 x b x t x a x c to the power 2
The bases in DNA are: Adenine(A), Thymine(T), Guanine(G), Cytosine(C) when they pair up: A-T, C-T
choose the right
B-i-t-c-h
In DNA, the ratio of adenine (A) to thymine (T) and guanine (G) to cytosine (C) is always 1:1 due to complementary base pairing. So, the ratio of A to T' and G' to C's will also be 1:1.