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3 years ago

Match the terms with the correct definitions.

Engineering

1 answer:

arsen [322]3 years ago

7 0

I think answer should be the first one please give me brainlest let me know if it’s correct or not okay thanks bye

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What are the nine Historical periods?

Rufina [12.5K]

Answer:

https://quizlet.com/148993376/the-nine-distinct-periods-of-time-flash-cards/

Explanation:

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3 years ago

What is the function of a regulator?

elena55 [62]

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3 years ago

If x < 5 and x >c, give a value of c such that there

Arlecino [84]

we have

x<5

x>c

we know that

The solution is the intersection of both solution sets of the given inequalities.

The solutions of the compound inequality must be solutions of both inequalities.

The value of c could be 5 or any number greater than 5, such that there are no solutions to the compound inequality

Because

A number cannot be both less than 5 and greater than 5 at the same time

therefore

the answer is

for c_> there are no solutions to the compound inequality

7 0

3 years ago

a. Determine R for a series RC high-pass filter with a cutoff frequency (fc) of 8 kHz. Use a 100 nF capacitor. b. Draw the schem

Readme [11.4K]

Answer:

a) 199.04 ohms

b) attached in image

c) -0.696dB

Explanation:

We are given:

Fc = 8Khz = 8000hz

$C = 100nF = 100*10^-^9F$

a)Using the formula:

$F_c = \frac{1}{2pie*Rc}$

$8000= \frac{1}{2*3.14*R*100*10^-^9}$

$R =\frac{1}{2*3.14*100*10^-^9*8000}$

R = 199.04 ohms

b) diagram is attached

c) $H(w) = \frac{V_out(w)}{Vin(w)} = \frac{1}{1-j\frac{wc}{w}}$

$H(F) = \frac{1}{1-j\frac{fc}{f}}$

At F = 20KHz and Fc= 8KHz we have:

$H(F)= \frac{1}{1-j\frac{8}{20}} = \frac{1}{1-j(0.4)}$

$|H(F)|= \frac{1}{\sqrt{1^2+(0.4)^2}}$

=0.923

|H(F)| in dB = 20log |H(F)|

=20log0.923

= -0.696dB

5 0

3 years ago

The denominator of a fraction is 4 more than the numenator. If 4 is added to the numenator and 7 is added to the denominator, th

kati45 [8]

Answer:

$\frac{3}{7}$

Explanation:

Lets take the numerator of the fraction to be = x

So the denominator of the fraction is 4 more than the numerator = x+4

The fraction is ; $\frac{x}{4+x}$

Now add 4 to the numerator and add 7 to the denominator as;

$\frac{x+4}{4+x+7} =\frac{x+4}{x+11}$

This new fraction is equal to 1 half =1/2

write the equation as;

$\frac{x+4}{x+11} =\frac{1}{2}$

perform cross-product

2(x+4 )=1( x+11 )

2x+8 = x + 11

2x-x = 11-8

x=3

The original fraction is;

$\frac{x}{4+x} =\frac{3}{3+4} =\frac{3}{7}$

3 0

3 years ago