Please help me with this problem i really can't complete it

VX = V1·R2/(R2 +R1(1 +AVOL))VX ≈ -99.0001 μVVout = -V1·R2·AVOL/(R2 +R1(1 +AVOL))Vout ≈ 0.990001 V

Explanation:

It is useful to consider the voltage at X to be the superposition of two voltages divided by the R1/R2 voltage divider. It is also helpful to remember that OUT = -VX·AVOL.

__

a)

 

__

b)

Filling in the given numbers, we find VX to be ...

 

__

c)

As we said at the beginning, OUT = -VX·AVOL. Multiplying the expression for VX by -AVOL, we get ...

 

__

d)

As with the expression, the output voltage is found by multiplying VX by -AVOL:

 

You know where the glacier is now, and how far it moves in
one year.  The question is asking how close to the sea it will be
after many years.

Step-1 ... you have to find out how many years

Step-2 ... you have to figure out how far it moves in that many years

Step-3 ... you have to figure out where it is after it moves that far

The first time I worked this problem, I left out  the most important
step ... READ the problem carefully and make SURE you know
the real question.  The first time I worked the problem, I thought
I was done after Step-2. 



Step-1:  How many years is it from 2010 to 2030 ?

               (2030  -  2010)  =  20 years .

Step-2:  How far will the glacier move in 20 years ? 

               It moves 0.004 mile in 1 year.

              In 20 years, it moves 0.004 mile 20 times

              0.004 x 20  =  0.08 mile

Step-3: How far will it be from the sea after all those years ?

              In 2010, when we started watching it, it was 6.9 miles
              from the sea.

              The glacier moves toward the sea.
               In 20 years, it will be  0.08 mile closer to the sea.
               How close will it be ?

               6.9 miles  -  0.08 mile  =  6.82 miles  (if it doesn't melt)

You know where the glacier is now, and how far it moves in
one year.  The question is asking how close to the sea it will be
after many years.

Step-1 ... you have to find out how many years

Step-2 ... you have to figure out how far it moves in that many years

Step-3 ... you have to figure out where it is after it moves that far

The first time I worked this problem, I left out  the most important
step ... READ the problem carefully and make SURE you know
the real question.  The first time I worked the problem, I thought
I was done after Step-2. 



Step-1:  How many years is it from 2010 to 2030 ?

               (2030  -  2010)  =  20 years .

Step-2:  How far will the glacier move in 20 years ? 

               It moves 0.004 mile in 1 year.

              In 20 years, it moves 0.004 mile 20 times

              0.004 x 20  =  0.08 mile

Step-3: How far will it be from the sea after all those years ?

              In 2010, when we started watching it, it was 6.9 miles
              from the sea.

              The glacier moves toward the sea.
               In 20 years, it will be  0.08 mile closer to the sea.
               How close will it be ?

               6.9 miles  -  0.08 mile  =  6.82 miles  (if it doesn't melt)

13) 790°

14) 945°

15) -890°

16) -360°

17) 890°

18) -220°

Step-by-step explanation:

Please see the attached pictures.

Trace the lines and for every complete circle (4 quadrants), which I represented with a single colour, add 360°.

If it's just two quadrants, that's 180°.

One quadrant is 90°.

Some properties that might help:

- Adj ∠s on a straight line= 180°

- ∠s at a point= 360°

I've attached the maximum number of pictures so I can't draw on the questions anymore.

For question 18,

From 4th quadrant to 3rd quadrant, it is 180°.

The remaining value would be 90°-50°= 40°

Hence the measure of the angle is 180°+40°= 220°

If the angle moves clockwise from the starting point, it's a negative angle. That's why I added a minus sign for questions 15, 16 and 18.

Feel free to ask if you have any questions :)

Step-by-step explanation:

We know a maximum point on the height vs. time curve is at t=16 seconds. Then the height function can be written by filling in the known values in ...

  h(t) = (center height) + (wheel radius)·cos((frequency)·2π·(t -(time at max height)))

Since t is in seconds, we want the frequency in revolutions per second. That will be ...

  (3.2 rev/min)·(1 min)/(60 sec) = 3.2/60 rev/sec = 4/75 rev/sec

Then our height function is ...

  h(t) = 59 + 45·cos(8π/75·(t -16))

9 minutes is 9·60 sec = 540 sec, so we want to find the value of h(540).

  h(540) = 59 + 45·cos(8π/75·(540 -16))

  = 59 +45·cos(4192π/75)

  ≈ 59 + 45·0.944376 . . . . . calculator in radians mode

 ≈ 101.496937 . . . . feet

_____

The cosine function is a maximum when its argument is zero. We used the process of function translation to translate the maximum point to t=16 from t=0. That is, we replaced t in the usual cosine function with (t-16).

We can also evaluate the cosine function by subtracting multiples of 2π from the argument. When we do that, we find that Shirley's height at 9 minutes is the same as it is after 15 seconds. Some calculators evaluate smaller cosine arguments more accurately than they do larger argument values.

  1 cake and 6 pies

Step-by-step explanation:

Let c represent the number of cakes Alice and George can make. Likewise, let p represent the number of pies.

Then the amount of flour they need (in cups) is ...

  3c + 2p . . . . cups of flour needed

And the amount of sugar needed is ...

  2c + 1.5p . . . cups of sugar needed

__

They want to use all 15 cups of flour they have, and all 11 cups of sugar, so the amounts above need to match the amounts on hand:

  3c + 2p = 15

  2c + 1.5p = 11

We can solve these equations various ways. One of my favorite is to use a graphing calculator. (See attached) It tells us that Alice and George can make 1 cake and 6 pies with the flour and sugar they have.

_____

If you want to solve these equations algebraically, there are several methods for that, too. Here, we can multiply the first one by 2/3 and subtract the result from the second one:

  (2c +1.5p) -(2/3)(3c +2p) = (11) -(2/3)(15)

  2c +3/2p -2c -4/3p = 11 -10 . . . . . eliminate parentheses

  1/6p = 1 . . . . . . . . . . . . . . . . . . . . . .collect terms

  p = 6 . . . . . . . . . . . . . . . . . . . . . . . . multiply by 6

Substituting into the first equation gives ...

  3c +2·6 = 15 . . . substitute 6 for p

  3c = 3 . . . . . . . . subtract 12

  c = 1 . . . . . . . . . . divide by 3

These values tell us Alice and George can make 1 cake and 6 pies with the flour and sugar they have.

Step-by-step explanation:

We know a maximum point on the height vs. time curve is at t=16 seconds. Then the height function can be written by filling in the known values in ...

  h(t) = (center height) + (wheel radius)·cos((frequency)·2π·(t -(time at max height)))

Since t is in seconds, we want the frequency in revolutions per second. That will be ...

  (3.2 rev/min)·(1 min)/(60 sec) = 3.2/60 rev/sec = 4/75 rev/sec

Then our height function is ...

  h(t) = 59 + 45·cos(8π/75·(t -16))

9 minutes is 9·60 sec = 540 sec, so we want to find the value of h(540).

  h(540) = 59 + 45·cos(8π/75·(540 -16))

  = 59 +45·cos(4192π/75)

  ≈ 59 + 45·0.944376 . . . . . calculator in radians mode

 ≈ 101.496937 . . . . feet

_____

The cosine function is a maximum when its argument is zero. We used the process of function translation to translate the maximum point to t=16 from t=0. That is, we replaced t in the usual cosine function with (t-16).

We can also evaluate the cosine function by subtracting multiples of 2π from the argument. When we do that, we find that Shirley's height at 9 minutes is the same as it is after 15 seconds. Some calculators evaluate smaller cosine arguments more accurately than they do larger argument values.

  1 cake and 6 pies

Step-by-step explanation:

Let c represent the number of cakes Alice and George can make. Likewise, let p represent the number of pies.

Then the amount of flour they need (in cups) is ...

  3c + 2p . . . . cups of flour needed

And the amount of sugar needed is ...

  2c + 1.5p . . . cups of sugar needed

__

They want to use all 15 cups of flour they have, and all 11 cups of sugar, so the amounts above need to match the amounts on hand:

  3c + 2p = 15

  2c + 1.5p = 11

We can solve these equations various ways. One of my favorite is to use a graphing calculator. (See attached) It tells us that Alice and George can make 1 cake and 6 pies with the flour and sugar they have.

_____

If you want to solve these equations algebraically, there are several methods for that, too. Here, we can multiply the first one by 2/3 and subtract the result from the second one:

  (2c +1.5p) -(2/3)(3c +2p) = (11) -(2/3)(15)

  2c +3/2p -2c -4/3p = 11 -10 . . . . . eliminate parentheses

  1/6p = 1 . . . . . . . . . . . . . . . . . . . . . .collect terms

  p = 6 . . . . . . . . . . . . . . . . . . . . . . . . multiply by 6

Substituting into the first equation gives ...

  3c +2·6 = 15 . . . substitute 6 for p

  3c = 3 . . . . . . . . subtract 12

  c = 1 . . . . . . . . . . divide by 3

These values tell us Alice and George can make 1 cake and 6 pies with the flour and sugar they have.

I believe It's A) How easily does violence happen in society, If you put into thought we can't really live in peace without chaos. It's that balancing beam to keep one end off the ground to be able to live today.

I believe It's A) How easily does violence happen in society, If you put into thought we can't really live in peace without chaos. It's that balancing beam to keep one end off the ground to be able to live today.