All Answers 358181 Answers

We have that

255/512  =  (1/4) [  1 - (1/2)^n  ]  /  [  1 - 1/2]    simplify

255/512 =  (1/2) [ 1 - (1/2)^n ]  

255 / 256 =  1 - (1/2)^n

(1/2)^n  =   1 -  255/256

(1/2)^n  =  1/256

(1/2)^n  =  (1/2)^8

So....n  = 8

Solve for x:

x^3 + 2 x^2 - 11 x - 12 = 0

Factor the left hand side.

The left hand side factors into a product with three terms:

(x - 3) (x + 1) (x + 4) = 0

Find the roots of each term in the product separately.

Split into three equations:

x - 3 = 0 or x + 1 = 0 or x + 4 = 0

Look at the first equation: Solve for x.

Add 3 to both sides:

x = 3 or x + 1 = 0 or x + 4 = 0

Look at the second equation: Solve for x.

Subtract 1 from both sides:

x = 3 or x = -1 or x + 4 = 0

Look at the third equation: Solve for x.

Subtract 4 from both sides:

x = 3      or      x = -1      or       x = -4

Thanks!! Glad to be of some help!!

num of cats  =  n

num of fleas per cat  =  2n

total num of fleas  =  (num of fleas per cat) * (num of cats)   =   2n * n   =   2n²

If we take the total number of fleas, subtract the fleas of two cats, then subtract  3  for every remaining cat, we get half the total number of fleas.

(total num of fleas) - 2(num of fleas per cat) - 3(n - 2)  =  (total num of fleas)/2

2n²  -  2(2n)  -  3(n - 2)   =   (2n²)/2

2n² - 4n - 3n + 6   =   n²

                                                 Subtract  n²  from both sides and combine like terms.

n² - 7n + 6  =  0

                                                 Factor the left side.

(n - 6)(n - 1)  =  0

n = 6      or      n = 1

n  must be  6  because we can't remove  2  cats if there is only  1  cat to begin with.

We have that there are n*2n = 2n^2 fleas. Then it says that 2 cats are removed, so 4n fleas are removed. then it says that out of the remaining n-2 cats, you remove 3 fleas from them, so you remove 3 * ( n-2 ) = 3n-6 fleas after removing 4n fleas. So it is saying that 2n^2 = 2 ( 7n - 6 ) = 14n - 12. We move all of the equation to one side to get 2n^2 - 14n + 12 = 0. This is a example of using a quadratic equation. Where the equation is an^2 + bn + c = 0. The solution for n is (-b+-*sqrt(b^2-4ac))/2a. In this equation, a = 2, b = -14, and c = 12. We fill it in to get (14+-sqrt(196-96))/4 = (14+-sqrt(100))/4 = (14+- 10)/4 = 1 or 6. Now, you can't have 1 cat because in the question is says that you remove 2 cats, and you can't have -1 cats, which you would have if you removed 2 cat's from 1 cat. So The answer is 6 cats.

*+- is plus or minus

P.S. Sorry, I can't do LaTeX that well.

Boom! You rock, hectictar! It's correct! I understand it better now...

temperature  =  -t² + 12t + 50

And we want to know what  t  is when the temperature is  77  .

77   =   -t² + 12t + 50        Now we want to solve this equation for  t .

                                         Get one side of the equation equal to zero.

t² - 12t + 27   =   0

                                         Now we can factor the left side like this..

(t - 9)(t - 3)   =   0

                                         Set each factor equal to zero.

t - 9  =  0          or          t - 3  =  0

t  =  9               or          t  =  3

The largest value of  t  that causes the expression    -t² + 12t + 50   to be  77  is   9   .

\(\frac{r+9}{r-3}\,=\,\frac{r-2}{r+5}\)

Note that  r  cannot be  3  or  -5  since these values cause a zero in the denominator of the original equation. Now multiply both sides of the equation by  r - 3 , and multiply both sides by  r + 5 .

(r + 9)(r + 5)  =  (r - 2)(r - 3)

                                                  Multiply out the factors.

r² + 14r + 45   =   r² - 5r + 6

                                                  Subtract  r²  from both sides of the equation.

14r + 45  =  -5r + 6

                                                  Add  5r  to both sides.

19r + 45  =  6

                                                  Subtract  45  from both sides.

19r  =  -39

                                                  Divide both sides by  19 .

r  =  -39/19

x / (x + 4)  =  -9 / (x + 3)                                            Numbers that cannot be answers:  -4, -3

Cross multiply:

x(x + 3)  =  -9(x + 4)

x² + 3x  =  -9x - 36

x² + 3x + 9x + 36  =  0

x² + 12x + 36  =  0

(x + 6)(x + 6)  =  0

x  =  -6

Hello, tertre...

\(6x^2+3x=0, x(6x+3)=0, \) so one root of the equation is 0, and the other is \(6x+3=0, 6x=-3, x=\frac{-3}{6}, \frac{-1}{2}\)

Solution: 

Since there are the same amounts of all the balls in Alice's bag and Bob's bag, and there is an equal chance of each ball being selected, the color of the ball that Alice puts in Bob's bag doesn't matter. Without loss of generality, let the ball Alice puts in Bob's bag be red.

For both bags to have the same contents, Bob must select one of the 2 red balls out of the 6 balls in his bag.

So the desired probability is \(\frac{2}{6}=\frac{1}{3}\)

56x² + 27  =  89x - 8

56x² - 89x + 35  =  0

Since  x = 5/7  is a solution; one factor will be  7x - 5: the other factor will be  8x - 7:

(7x - 5)(8x - 7)  =  0

Either 7x - 5  =  0     or     8x - 7  =  0

                7x  =  5                   8x  =  7

                  x  =  5/7                  x  =  7/8

We can count the total number of handshakes by counting the lawyers shaking hands with each of the politicians and then the politicians shaking hands among themselves.

If each lawyer shakes hands with each of the politicians, then each lawyer shakes hands with four people. Since there are three lawyers, 12 handshakes occur.

If each politician shakes hands with each of the other politicians, the first shakes hands with three others, the second shakes hands with two others (not counting the handshake with the first that has already transpired), and the last two each shake the hand of the other. Thus, there are 3+2+1=6 handshakes that occur.

So,  12+6=18 total handshakes take place.

First let's factor all the numerators and denominators.

\(\frac{r^2-5r+4}{r^2-8r+7}\,=\,\frac{r^2-2r-15}{r^2-r-20} \\~\\ \frac{(r-4)(r-1)}{(r-7)(r-1)}\,=\,\frac{(r-5)(r+3)}{(r-5)(r+4)} \)

Now we can reduce both fractions.

\(\frac{(r-4)}{(r-7)}\,=\,\frac{(r+3)}{(r+4)}\)                 and      r ≠ 1 ,  r ≠ 5

Multiply both sides of the equation by  (r - 7) ,  and multiply both sides by  (r + 4) .

(r - 4)(r + 4)   =   (r + 3)(r - 7)         and      r ≠ 7 ,  r ≠ -4

r² - 16   =   r² - 4r - 21

                                         Subtract  r²  from both sides.

-16  =  -4r - 21

                                         Add  21  to both sides.

5   =   -4r

                                         Divide both sides by  -4 .

\(-\frac54\)   =   r                         This is not a restrited value, so this is the solution. 

The expression  \(\frac{x+1}{8x^2-65x+8}\)   is not defined when the denominator equals zero.

So to find all the  x  values that cause the expression to be undefined, set the denominator = 0 .

8x² - 65x + 8  =  0            Now we need to solve this equation for  x .

                                         We can factor the left side like this....

8x² - 64x - x + 8  =  0

                                         Factor  8x  out of the first two terms, factor  -1  out of the last two terms.

8x(x - 8) - 1(x - 8)  =  0

                                         Factor  x-8  out of both terms.

(x - 8)(8x - 1)  =  0

                                         Set each factor equal to zero and solve for  x .

x - 8  =  0          or          8x - 1  =  0

x  =  8               or          8x  =  1

                                        x  =  1/8

So the expression is not defined when  x = 8  and when  x = 1/8  .

The largest of these values is  8 .

cross multiply

We expand both sides to find

\(\begin{align*} (w+13)(w+13)&=(3w+7)(2w+4)\\ w^2+26w+169&=3w(2w+4)+7(2w+4)\\ w^2+26w+169&=6w^2+12w+14w+28\\ w^2+26w+169&=6w^2+26w+28\\ w^2+169&=6w^2+28\\ 141&=5w^2\\ \frac{141}{5}&=w^2.\\ \end{align*}\)



So, expressed as a decimal, our answer is \(\frac{141}{5}=\boxed{28.2}\) .

First we divide both sides by 2 to get \(x(x-10)=-25\). Expanding the left side and bringing the constant over, we get \(x^2-10x+25=0\) . We can factor this into \((x-5)(x-5)\), so the only possible value for \(x\)  is \(\boxed{5}\) , which is also our answer.

First we need to find the solutions of the equation...

x² - 7x - 4  =  40

                                    Subtract  40  from both sides so that the right side is zero.

x² - 7x - 44  =  0

                                    Now we can factor the left side like this..

(x + 4)(x - 11)  =  0

                                    Set each factor equal to zero and solve each for  x .

x + 4  =  0         or          x - 11  =  0

x  =  -4              or          x  =  11

The solutions of the equation are  -4  and  11 .

Twice the nonnegative difference of  -4  and  11  is...

2 * | -4 - 11 |   =   2 * | -15 |   =   2 * 15   =   30

2x(x - 10)   =   -50

                                             Distribute  2x  to both terms in parenthesees.

(2x)(x) + (2x)(-10)   =   -50

2x² - 20x   =   -50

                                             Add  50  to both sides.

2x² - 20x + 50   =   0

                                             Divide through by  2 .

x² - 10x + 25   =   0

                                             Factor the left side of the equation.

(x - 5)(x - 5)  =  0

(x - 5)²  =  0

                                             Take the square root of both sides.

x - 5  =  0

x  =  5

The only possible value of  x  is  5 , so the sum of all the possible values is  5  .

(w + 13)²  =  (3w + 7)(2w + 4)

                                                                   First multiply out the parenthesees on both sides.

(w + 13)(w + 13)   =   (3w + 7)(2w + 4)

w² + 13w + 13w + 169   =   6w² + 12w + 14w + 28

w² + 26w + 169   =   6w² + 26w + 28

                                                                   Subtract  26w  from both sides.

w² + 169  =  6w² + 28

                                                                   Subtract  w²  from both sides.

169  =  5w² + 28

                                                                   Subtract  28  from both sides.

141  =  5w²

                                                                   Divide both sides by  5 .

141/5  =  w²

w²  =  28.2

after  1  year, the population   =   298 +  0.08 * 298   =   298(1 + 0.08)   =   298(1.08)

after  2  years, the population   =   [ 298(1.08) ](1.08)   =   298(1.08)²

after  x  years, the population   =   298(1.08)^x

f(x)   =   298(1.08)^x

f(3)   =   298(1.08)³   ≈   375

Solve for w:
(w + 13)^2 = (2 w + 4) (3 w + 7)

Write the quadratic polynomial on the right-hand side in standard form.

Expand out terms of the right-hand side:
(w + 13)^2 = 6 w^2 + 26 w + 28

Move everything to the left-hand side.
Subtract 6 w^2 + 26 w + 28 from both sides:
-28 - 26 w - 6 w^2 + (w + 13)^2 = 0

Write the quadratic polynomial on the left-hand side in standard form.
Expand out terms of the left-hand side:
141 - 5 w^2 = 0

Isolate terms with w to the left-hand side.
Subtract 141 from both sides:
-5 w^2 = -141

Divide both sides by a constant to simplify the equation.
Divide both sides by -5:
w^2 =± 141/5

hmm, i'm getting something different

\(\frac{x}4 + \frac{y}{12} = 1 \)

If x=0   y=12   so  the y intercept b = 12

If y=0   x=4     so the x intercept  a = 4

Thinking of the graph in my head I can see this has a negative gradient.

m= - rise / run = - 12/4 = -3

a+b+m = 4+12+ -3 = 13