Square root of complex number -14?

romanna [79]

Maybe 4.80

Step-by-step explanation:

3 0

3 years ago

Consider a series circuit consisting of a resistor of R ohms, an inductor of L henries, and variable voltage source of V(t) volt

ser-zykov [4K]

Answer:

$I(t)=\frac{1}{3}(1-e^{-30t})$

Step-by-step explanation:

We are given that

$\frac{dI}{dt}+\frac{R}{L}I=\frac{V(t)}{L}$

R=150 ohm

L=5 H

V(t)=10 V

$P=\frac{R}{L}=\frac{150}{5}=30$

$I.F=e^{\int Pdt}=e^{\int 30 dt}=e^{30 t}$

$I(t)\times I.F=\int e^{30 t}\times 10 dt+C$

$I(t)\times e^{30 t}=\frac{10}{30}e^{30 t}+C$

$I(t)=\frac{1}{3}+Ce^{-30 t}$

I(0)=0

Substitute t=0

$0=\frac{1}{3}+C$

$C=-\frac{1}{3}$

Substitute the values

$I(t)=\frac{1}{3}-\frac{1}{3}e^{-30 t}$

$I(t)=\frac{1}{3}(1-e^{-30t})$

5 0

3 years ago

Please please help me I don't understand

Alona [7]

Answer:

4th option - over the interval (4,7) the local minimum is -7

Step-by-step explanation:

There's only one local minimum in this graph and it's the one between (4,7), so this is the only plausible answer.

4 0

2 years ago

Towers A and B are located 5 miles apart. A ranger spots a fire at a 42-degree angle from tower A. Another fire ranger spots the

lord [1]

Answer: The fire is 3.5 miles from tower B

Step-by-step explanation: Please refer to the attached diagram. The triangle in the attached diagram illustrates the clues given in the question. Both rangers are standing at points A and B respectively with a distance of 5 miles between them, which is line AB. Also, one ranger spots a fire from a tower at an angle of 42 degrees, which is point A. Another ranger spots the same fire from another tower, but from an angle of 64 degrees, which is point B. The fire is at point C on the triangle. Now we have a triangle with only one side known (5 miles) and three angles known (the third angle is computed as 180 - {64+42} which equals 74) which are 64 degrees, 42 degrees and 74 degrees.

The distance from the fire to tower B is calculated using the law of sines. (Note that this is not a right angled triangle, hence we cannot use trigonometric ratios). The law of sines is expressed as follows;

a/SinA = b/SinB or

a/SinA = c/SinC

Depending on the sides and angles we are given and the ones we are to calculate.

The distance from the fire to tower B is line BC, labeled as a in our diagram. Using the law of sines

a/SinA = c/SinC

(Note also that a is directly facing angle A, c is directly facing angle C, and so on)

a/SinA = c/SinC

a/Sin 42 = 5/Sin 74

By cross multiplication we now have

a (Sin 74) = 5 (Sin 42)

Divide both sides of the equation by Sin 74 and we now arrive at

a = 5 (Sin 42)/Sin 74

a = 5 (0.6691)/0.9613

a = 3.3455/0.9613

a = 3.4802

{rounded to the nearest tenth of a mile, a equals 3.5}

Therefore the distance from tower B to the fire is approximately 3.5 miles

3 0

2 years ago

Points A and B have coordinates (3,1,2) and (1,-5,4) respectively. Point C lies on line AB such that AC:BC=3:2. Find position ve

Sever21 [200]

Answer:

The position vector of point C is <-3 , -17 , 8> or -3i - 17j + 8k

Step-by-step explanation:

* Lets revise how to solve the problem

- If the endpoints of a segment are (x1 , y1 , z1) and (x2 , y2 , z2), and

point (x , y , z) divides the segment externally at ratio m1 :m2, then

$x=\frac{m_{1}x_{2}-m_{2}x_{1}}{m_{1} -m_{2}},y=\frac{m_{1}y_{2}-m_{2}y_{1}}{m_{1}-m_{2}},z=\frac{m_{1}z_{2}-m_{2}z_{1}}{m_{1}-m_{2}}$

* Lets solve the problem

∵ AB is a segment where A = (3 , 1 , 2) and B = (1 , - 5 , 4)

∵ Point C lies on line AB such that AC : BC=3 : 2

∵ From the ratio AC = 3/2 AB

∴ C divides AB externally

- Lets use the rule above to find the coordinates of C

- Let Point A is (x1 , y1 , z1) , point B is (x2 , y2 , z2) and point C is (x , y , z)

and AC : AB is m1 : m2

∴ x1 = 3 , x2 = 1

∴ y1 = 1 , y2 = -5

∴ z1 = 2 , z2 = 4

∴ m1 = 3 , m2 = 2

- By using the rule above

∴ $x=\frac{3(1)-2(3)}{3-2}=\frac{3-6}{1}=\frac{-3}{1}=-3$

∴ $y=\frac{3(-5)-2(1)}{3-2}=\frac{x=-15-2}{1}=\frac{-17}{1}=-17$

∴ $z=\frac{3(4)-2(2)}{3-2}=\frac{12-4}{1}=\frac{8}{1}=8$

∴ The coordinates fo point c are (-3 , -17 , 8)

* The position vector of point C is <-3 , -17 , 8> or -3i - 17j + 8k

3 0

3 years ago