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$\frac{x-1}{x-2}-\frac{2}{x}\,=\,\frac{1}{x-2}$

First let's get a common denominator.

$\frac{(x)(x-1)}{(x)(x-2)}-\frac{2(x-2)}{(x)(x-2)}\,=\,\frac{1x}{(x)(x-2)}$

We can multiply through by the denominator....however....you are right that x can't be 2 !

Also  x  can't be  0 . If we plug in  0  or  2  for  x  into the original equation, we will get an undefined result. Before we multiply through by the denominator, we must say that  x ≠ 0  and  x ≠ 2 .

$(x)(x-1)-2(x-2) \,=\,1x$         Distribute the  x  and the  -2 .

$x^2-x-2x+4 \,=\,x$           Subtract  x  from both sides and combine like terms.

$x^2-4x+4 \,=\,0$          Factor the left side and solve for  x  .

$(x-2)^2=0 \\~\\ x-2=0 \\~\\ x=2$

But, x cannot be 2 because it causes a zero in the denominator of the original problem.

So there is no solution.

Hey!! 

Let's get this equation of a circle into the form  (x - h)² + (y - k)²  =  r²   , where  (h, k)  is the center of the circle, and  r  is its radius.

x² + y² - 7  =  4y - 14x + 3

                                                  Add  7  to both sides.

x²  +  y²  =  4y - 14x + 10

                                                 Add  14x  to both sides, and subtract  4y  from both sides.

x² + 14x   +  y² - 4y    =  10

                                                 Add  49   and add  4  to both sides to complete the squares.

x² + 14x + 49  +  y² - 4y + 4   =   10 + 49 + 4

                                                                         Factor  x² + 14x + 49  as a perfect square trinomial.

(x + 7)²   +   y² - 4y + 4  =  63

                                                                         Factor  y² - 4y + 4  as a perfect square trinomial.

(x + 7)²   +   (y - 2)²  =  63

Now that the equation is in this form, we know that..

r²  =  63

And...

area of circle    =    pi * r²    =    pi * 63    =    63pi

Can you give an example where the zeros aren't counted?

A lot of times the zeros matter! For example

100    in scientific notation is   1 * 10²

The  "2" comes from the 2 zeros that are in 100......they are counted!

You are right that the distance between the points is  = √50

To simplify this, we can split 50 into its prime factors. So we get...

√50   =   √[ 2 * 5 * 5 ]

And...

√[ 2 * 5 * 5 ]    =    √2 * √[5 * 5]    =    √2 * √5²    =    √2 * 5    =    5√2    

When the height of the space above the liquid equals 0, the radius of the liquid's surface will be the same as the radius of the top of the tank.

So, plug in  0  for  h  and solve for  r  .

0  =  21 - $\frac72$ r      Add  $\frac72$r  to both sides of the equation.

$\frac72$r  =  21              Multiply both sides of the equation by  $\frac27$ .

r  =  6

This is the radius (in meters) of the liquid surface when there is no space above it, and so this is the radius of the top of the tank. So we have a circle with a radius of 6 meters.

circumference  =  2pi * radius

circumference  =  2pi * 6

circumference  =  12pi

This was the Golden Age. They invited artists and poets to live and work in the city. People were so impressed and this allowed for more merchants to trade.

This is indeed a messy problem. I'll solve for both variables, I guess.

1. Solve for a Variable

In this case, I will solve for x in the first equation, $\sqrt{x^2+y^2}=100$. 

2. Use Substitution to eliminate the x and solve for y

Plug in $\sqrt{10000-y^2}$ for x into the 2nd equation and solve for y. This is not going to look good...

Time to do the simplification process. 

You know what...

This is really boring and tedious...

There are 2 ordered pair solutions to this. They are the following:

$x≈68.53740254003346, y≈72.81912147963209$

$x≈86.60254037943613, y≈49.99999999828134$

I believe you are trying to access the inverse sine function. I have create pictures so that you can follow along!

Press the 2nd button that is encircled in red. Doing this will allow you, the user, to access secondary functions that are otherwise invisible.

Yes, $\text{asin}$ and $\sin^{-1}$ are the same operation. Make sure that the calculator is set to appropriate setting of either degrees or radians before performing the calculation. The process is the same for the other inverse trigonometric functions, too. 

Thanks!

The minute hand has moved 30° from 12 o'clock

The hour hand at 3 o'clock  is 90° from 12 o'clock

And It moves 30° every hour....so...in 5 minutes....it would have moved  (1/12) of 30° = [30/12]° =

(5/2)°  = 2.5°

So......the  [lesser]  angle  between the minute and hour hand at 3:05  is

[ 90 + 2.5] -  30  =  62.5°

- Aiyan

I hardly think your math teacher would give you 13 pages of mandatory math homework in one day that's due the next. Check the due date, it could be due in a few days or a week.

Yes! There can be such unique interest rate. However, it is relatively difficult to calculate such rate directly. Modern computers and calculators can find such rate if they are programmed correctly. In this case, for instance, you have to equate two different formulas. Namely, the first is to find the FV of $1 per period and the second is to find the PV of $1 per period as follows:
FV =500*(((1+R)^420 -1) / R) ....................................... (1)
PV =7500*(((1+R)^300-1) / (1+R)^300) / R)................(2) 
Then, you would equate the above 2 formulas as follows:
500*(((1+R)^420 -1) / R) =7500*(((1+R)^300-1) / (1+R)^300) / R, solve for R 
Through the process of iteration, the computer finds the interest rate, R, which comes to 0.00625 per month. Or, 0.00625 x 1200 =~7.50% compounded monthly.

Let the initial deposit that Michelle made = P

P x [1 + 0.10/4]^4

P x    1.025^4

=1.103812890625P  This is the balance of the deposit after the first year.

1.103812890625P + 1,375 = 1.103812890625^2P

1.103812890625P + 1,375 =1.21840289751P   Subtract 1.103812890625P from both sides.

1,375 =0.11459P      Divide both sides by 0.11459

P = 1,375 / 0.11459

P =11,999.30 - The initial principal that Michelle deposited in the Bank.

2x^2 + 4x - 7 = 0

Add 7 to both sides

2x^2 + 4x  =   7       divide through by 2

x^2 + 2x  = 7/2 

Take (1/2) of the coefficient on x  = (1/2)(2)  = 1

Square this  = 1   and add it to both sides

x^2 + 2x + 1  =  7/2 + 1        factor the left and simpify the right

(x + 1)^2  =   9/2                  take both roots

x + 1  =   ±√ [9/2]

x + 1  = ±3 /√2        rationalize the denominator on the righr

x + 1  =  ±3√2  / 2      subtract 1 from both sides

x =  ±3√2  / 2  -  1

x = ±3√2  / 2  - 2/2

x = [ -2  ± 3√2 ]   / 2

Thanks Alan :)

"A function has zeroes at -1, 3, and 5. We know that f(-2) and f(2) are negative, while f(4) and f(6) are positive. Sketch a graph of f(x)."

One possibility:

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Simpler than I expected it to be! Thanks!

Here are some of the intermediate steps Melody:

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How did you do that guest?

Look at the values near t = 0.

At t = 0, cos(0) is +1, and since the graph shows y₂(0) is positive, then a₂ must be positive.

For small values of b₁t, sin(b₁t) ≈ b₁t.  For b₁ positive, this is positive, and since the graph shows y₁(t) positive for small values of t, then a₁ must be positive.

a₁ and a₂ represent the amplitudes of y₁ and y₂ respectively.  Since it is clear from the graph that the amplitude of y₁ is greater than that of y₂, we must have a₁>a₂.

Summarising:    0 < a₂ < a₁

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_.

What is the cos of what?

ما هو ال Cos           من ماذا؟

Your question is not complete.

سؤالك ليس كاملا

Thanks Alan :)

Alright, then he must have accidentally said it wrong. I thought so at first because radiuseseseses sounded very incorrect but then he said the same thing twice so I got very confused. Thanks!