Answer:
See explaination and attachment.
Explanation:
Iteration method is a repetitive method applied until the desired result is achieved.
Let the given equation be f(x) = 0 and the value of x to be determined. By using the Iteration method you can find the roots of the equation. To find the root of the equation first we have to write equation like below
x = pi(x)
Let x=x0 be an initial approximation of the required root α then the first approximation x1 is given by x1 = pi(x0).
Similarly for second, thrid and so on. approximation
x2 = pi(x1)
x3 = pi(x2)
x4 = pi(x3)
xn = pi(xn-1).
please go to attachment for the step by step solution.
Answer:
Web Browser
Explanation:
Because you dont use a messaging app or presentation software to look up stuff its common knowledge
Ability to recognize words and understand vocabulary
Answer: Option 2.
<u>Explanation:</u>
Reading comprehension is the capacity to process content, comprehend its significance, and to incorporate with what the peruser definitely knows. Capacity to grasp content is affected by perusers' aptitudes and their capacity to process data.
For the students reading comprehension problems frequently include troubles in perceiving and suitably applying foundation information, poor disentangling and word acknowledgment abilities, restricted jargon information, immature understanding familiarity, a not exactly key way to deal with cognizance.
complete question
A certain amplifier has an open-circuit voltage gain of unity, an input resistance of 1 \mathrm{M} \Omega1MΩ and an output resistance of 100 \Omega100Ω The signal source has an internal voltage of 5 V rms and an internal resistance of 100 \mathrm{k} \Omega.100kΩ. The load resistance is 50 \Omega.50Ω. If the signal source is connected to the amplifier input terminals and the load is connected to the output terminals, find the voltage across the load and the power delivered to the load. Next, consider connecting the load directly across the signal source without the amplifier, and again find the load voltage and power. Compare the results. What do you conclude about the usefulness of a unity-gain amplifier in delivering signal power to a load?
Answer:
3.03 V 0.184 W
2.499 mV 125*10^-9 W
Explanation:
First, apply voltage-divider principle to the input circuit: 1
*5
= 4.545 V
The voltage produced by the voltage-controlled source is:
A_voc*V_i = 4.545 V
We can find voltage across the load, again by using voltage-divider principle:
V_o = A_voc*V_i*(R_o/R_l+R_o)
= 4.545*(100/100+50)
= 3.03 V
Now we can determine delivered power:
P_L = V_o^2/R_L
= 0.184 W
Apply voltage-divider principle to the circuit:
V_o = (R_o/R_o+R_s)*V_s
= 50/50+100*10^3*5
= 2.499 mV
Now we can determine delivered power:
P_l = V_o^2/R_l
= 125*10^-9 W
Delivered power to the load is significantly higher in case when we used amplifier, so a unity gain amplifier can be useful in situation when we want to deliver more power to the load. It is the same case with the voltage, no matter that we used amplifier with voltage open-circuit gain of unity.
Answer:
a) 2∪p/lb (l+b)dH
b) po exp( 4∪x/l)
Explanation:
please check the attachment for proper explanation and proper sign notations thanks.
