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If q(10 - 10e-2t)mc find the current at t 1s?
i=dq/dt i=d(10)/dt - d(10e^-2t)/dt d(10)/dt=0 i= - d(10e^-2t)/dt i=-((0 X e^-2t) +(-2 X 10X e^-2t) ) {by uv methord } i=20e^-2t t=1s i=20 e^-2 {e^-2=0.135335 } i=2.707 Ans:2.707mA
What is 0.4 in standard index form?
4 x 10-1
What are the Prime factors of 75 in index form?
75 = 3 x 5 x 5 = 3 x 52
What is 470 in standard index form?
4.7x10^2
What is 60 in standard index form?
6 X 10 to the power of 1
How do you solve 5t plus 2 equals 3t-2t plus 5?
5t + 2 = 3t - 2t + 5 Collect like terms 5t - 3t + 2t = 5 - 2 4t = 3 t = ¾ Check: 5 x 3/4 = 3¾ + 2 = 5¾; 3x¾- 2x¾ + 5 = 5¾ QED
How do you express the prime factors of 80 in index form?
This is the prime factorization of 80 in index form: 80 = 2 x 2 x 2 x 2 x 5 = 24 x 5
What is is 5xy-x2t plus 2xy plus 3x2t?
1xy
What is the index form of 7 times 7 times 7?
7 x 7 x 7 is the expanded form of 343. The proper index form of this exponent would be73
What is 33344 in index form?
2^ x 521
How would this be re-arranged so that x is the subject y squared minus xy equals 9t squared minus 3xt?
x=y+3t Original formula: y2 -xy = 9t2 -3xt (assumes you meant "9 times t2 " and not (9t)2 ) y2 = xy +9t2 -3xt y2 - 9t2 = xy - 3xt (y-3t)(y+3t) = x(y-3t) (y-3t) = x
What is 1152 in index form?
27 x 32 = 1152
What is 63 as a product of prime factors in index form?
To express 63 as a product of prime factors in index form, we first need to find the prime factors of 63. 63 can be factored as 3 x 3 x 7. This can be written in index form as 3^2 x 7. Therefore, 63 as a product of prime factors in index form is 3^2 x 7.
how do i solve 3 x t x t =?
it became 3t^2
What is 0.4 in standard index form?
4 x 10-1
Prime factor of 171 in index form?
32 x 19
What are the prime factors of 30 in index form?
30 = 2 x 3 x 5
