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If t is real then [1 to infinity) ie all real numbers from 1 to infinity, including 1 but not infinity. If t is in the complex plane then the domain of t^2+1 is also the complex plane.
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How Laplace Transform is used solve transient functions in circuit analysis?
f(t)dt and when f(t)=1=1/s or f(t)=k=k/s. finaly can be solve:Laplace transform t domain and s domain L.
How do you find the domain?
The domain of a function, is the range of input values which will give you a real answer.For example the domain of x+1 would be all real numbers as any number plus 1 will be another real numberThe domain of x0.5 would be all positive numbers as the answer to square root of a negative number is not realNote:x0.5 means the square root of x* * * * *Not quite. A function is a one-to-one or many-to-one mapping from a set S to a set T (which need not be a different set). A function can be one whose domain is all the cars parked in a street and the range is the second character of their registration number.A mathematical function can have the complex field as its domain and range, so a real answer is not a necessary requirement for a function.
How do you make a the subject of t equals 0.5 x a plus b x h?
To make the following relationship: t = (1/2)a +bh a function of "a" (a = ...) start by subtracting both sides by "bh": t - bh = (1/2)a then divide both sides by (1/2), which is the same as multiplying by 2: 2(t - bh) = a from here you can distribute the 2 if desired: a = 2t - 2bh
Integration of tangent cubed of x?
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.
If T equals 3x plus 7y what is T when x equals 4 and y equals 6?
T = 3x + 7y; x = 4; y = 6T = 3x + 7y (substitute 4 for x, and 6 for y)T = 3(4) + 7(6) = 12 + 42 = 54
Can you factor the expression three t squared plus sixteen t plus five?
(3t + 1)(t + 5)
How do you factor 18x squared plus 9xt plus t squared?
(6x + t)(3x + t)
How do you solve-t5t squared plus t?
_t(5t squared t+)
Is t plus t equals 2t or t squared?
2t
What is 7 squared plus t?
72 = 49.72 + t = 49 + t
Does t plus t equals t squared or 2t?
t+t=2tand t*t= t squaredyou would only get t squared if you multiplied
What is the remainder when 15t2 squared minus t minus15t plus 3?
15t2 squared-t-15t+3=15t squared-14t+3
How do you factorise t squared?
t(t-1)
What is the answer to t squared plus t minus 42?
t2+ t - 42 it can't be simplified anymore
What does T squared and T squared mean?
T squared is T times T. T squared and T squared appears to be the addition of T squared with itself. That answer would be 2T squared or 2T^2
How could you represent the following in algebraic terms 7 squared plus t?
49+t
What is t plus v squared?
All that can be said, based on the available information is: t + v2
