2t x 3t = 6t2
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
4 x 10-1
75 = 3 x 5 x 5 = 3 x 52
To express 144 as a product of its prime factors in index form, we first find the prime factors of 144, which are 2 and 3. 144 can be written as 2^4 * 3^2 in index form, where 2 is raised to the power of 4 and 3 is raised to the power of 2. This shows that 144 can be expressed as the product of its prime factors in index form.
4.7x10^2
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
This is the prime factorization of 80 in index form: 80 = 2 x 2 x 2 x 2 x 5 = 24 x 5
1xy
7 x 7 x 7 is the expanded form of 343. The proper index form of this exponent would be73
2^ x 521
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
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.
The answer to the expression -3t x 7t is -21t². This is obtained by multiplying the coefficients (-3 and 7) to get -21, and then multiplying the variable terms (t x t) to get t².
27 x 32 = 1152
it became 3t^2
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
30 = 2 x 3 x 5