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The  radius  =   sqrt [ (0 +3)^2  + ( 0 - 4)^2 ] = sqrt [ 9 + 16 ] = sqrt (25)  = 5

So...the equation becomes

(x + 3)^2 + ( y - 4)^2  =  5^2

(x + 3)^2 + ( y - 4)^2  =  25

x * y   =  ( x + 2) (  y - 3)

So

t * ( t + 2)  =

(t  + 2) * [(t + 2) - 3 ]  = 

( t + 2) ( t - 1)

t^2 + t - 2    (1)

  And 

(t + 1) ( t + 1)  =

[ ( t + 1) + 2 ]  *  [ ( t + 1) - 3 ]  =

( t + 3) ( t - 2)  = 

t^2 + t -  6     (2)

So

(1)  - (2)  =

[ t^2 + t  - 2 ]  - [ t^2 + t - 6 ]  =

4

4 is a factor of the numerator: 

x+1 is a factor of the denominator

the numerator has 2 terms

We need to know what the original polynomial is  and what (possible) root we are testing

\(4^3 * 5^4 *6^2=2^6*5^4*2^2*3^2=2^8*3^2*5^4.\) There is actually a neat way to calculate the number of factors a number has. Suppose we have \(4^k\) , and the number of factors that number would have is \(4^{k+1}\) . Following the same example here, we have \(9*3*5=27*5=\boxed{135}\)   factors.

Just a wild guess....I'd say that  "D" isn't an asset for panel effectiveness

Thanks! Can you tell me what SD is? 

The sequel arrived to this question

HAHAHA!!!!......this website's info is reliable....well, mostly

"B"  is correct

lol

But yeah, B is the most resonable answer.

-15 + 14 - 3 =

-15 +  11  =

- 4    ........ correct !!!

-15+14-3

=-1-3

=-4

Yes it's -4, can't you just use a caculator?

To have two roots.....the discriminant must be a perfect square > 0.....so...

5a^2 - 26a - 8  +  4 (1)(a^2 - 4a + 9)   simplify

9a^2 - 42a + 28

Graphing this...it will be  > 0   and a perfect square   when the integer value for a  = 6

So  the polynomial  is

x^2   - 4x  - 21      factor ( x - 7) ( x + 3)  and the roots are  - 3  and 7

CORRECTED

The discriminant  is

(m - 2)^2  - 4m (m - 3)

To have multiple roots....this must be a perfect square > 0

(m - 2)^2  - 4m (m - 3)  simplify

m^2 - 4m + 4 - 4m^2 + 12m

-3m^2 + 8m + 4

Graphing this.....it is only > 0   when the integer values for m  =    0, 1, 2 or 3

There are two values for m that produce a perfect square

When m  = 1.....the perfect square is 9

When m = 3, the perfect square is 1

So....one possible quadratic is

x^2 + x - 2    factoring we have (x + 2) (x - 1)  and the roots are  -2 and 1

The other possible quadratic is

3x^2  - x     factor   x (3x - 1)   and the roots are  0 and 1/3

However....we require integer roots...so

x^2 + x - 2  is the quadratic     and m = 1

Remember that  x  is the additonal dollars that she  puts in her savings account aftr she earns $200.....she contributes $20 as long as she  only makes $200

After  $200, she puts  every  dollar into the account...

So  the final amount she contributes (saves) is 20 + x

And the total amount she makes is  200 +  x

So.....the  ratio of savings / total paycheck  = [ 20 + x ] / [ 200 + x ]

Example ...suppose she eans $240 dollars

She puts $40  into the account

So.....the ratio of savings / total paycheck  =  [ 20 + 40 ] / [200 + 40 ] = 60 / 240  = .25  = 25%

How many distinct, natural-number factors does 4^3 * 5^4 *6^2 have?
\(4^3 * 5^4 *6^2\\ =2^6 * 5^4 *2^2*3^2\\ =2^8 * 3^2 *5^4\\\)

2 is the only prime factor,  8

3 is the only prime factor,  2

5 is the only prime factor,  4

2 and 3 are the only prime factors  8*2=16

2 and 5 are the only prime factors  8*4=32

3 and 5 are the only prime factors  2*4=8

2 and 3 and 5 are the only prime factors  8*2*4=64

8+2+4+16+32+8+64 = 134

plus 1 I guess

So that is 135   

1440000 = 2^8×3^2×5^4 (14 prime factors, 3 distinct)

(8+1) x (2+1) x (4+1) =135 divisors.

There WERE 20 more people in A then B......then six people left......now there are only 14 more people in A than B....

THEN the six people went in to B.....Now there are only 8 more people in A than B.

I'd replace it with r for row. 

You can use any letter but I do not think that n is the best choice as it could lead to confusion,

(n for number of seats or n for number or rows ??? )

[Don't forget to thank Rom   ]

Are you saying that we should not use this website?

Don't worry, I always thought you meant it nicely :)

Omg im so sorry, things have been really busy with school and curricular activities but i know that's no excuse. Thank you so much! I really do appreciate it, you've helped me a lot. I am sorry if I did p**s you off in some way, I didn't mean it. 

Assume that a minute ago there were x people in Room B.

At that same time, there were: x + 20 in Room A.

x + 20 - 6 = x + 14 people in Room A now!.

Solve for x:
x/(x - 1) - 1/(x - 2) = (2 x - 5)/(x^2 - 3 x + 2)

Multiply both sides by x^2 - 3 x + 2:
1 - x + x (x - 2) = 2 x - 5

Expand out terms of the left hand side:
x^2 - 3 x + 1 = 2 x - 5

Subtract 2 x - 5 from both sides:
x^2 - 5 x + 6 = 0

The left hand side factors into a product with two terms:
(x - 3) (x - 2) = 0

Split into two equations:
x - 3 = 0 or x - 2 = 0

Add 3 to both sides:
x = 3 or x - 2 = 0

Add 2 to both sides:
x = 3 or x = 2

x/(x - 1) - 1/(x - 2) ⇒ 2/(2 - 1) - 1/(2 - 2) = ∞^~
(2 x - 5)/(x^2 - 3 x + 2) ⇒ (2 2 - 5)/(2 - 3 2 + 2^2) = ∞^~:
So this solution is incorrect

x/(x - 1) - 1/(x - 2) ⇒ 3/(3 - 1) - 1/(3 - 2) = 1/2
(2 x - 5)/(x^2 - 3 x + 2) ⇒ (2 3 - 5)/(2 - 3 3 + 3^2) = 1/2:
So this solution is correct

The solution is: x = 3      The answer is "B"

Factor the following:
2 x^3 + 5 x^2 - x - 6

The possible rational roots of 2 x^3 + 5 x^2 - x - 6 are x = ± 1/2, x = ± 3/2, x = ± 1, x = ± 2, x = ± 3, x = ± 6. Of these, x = -3/2, x = 1 and x = -2 are roots. This gives 2 x + 3, x - 1 and x + 2 as all factors:

 =(2 x + 3) (x - 1) (x + 2)

Thank You!