All Answers 356329 Answers

yes thanks

Find the sum of all real numbers that are not in the domain of f(g(x)).

Enter (A,B,C,D) in order below if A, B, C, and D are the coefficients of the partial fractions expansion of

\(\displaystyle12\cdot\frac{x^3+4}{(x^2-1)(x^2+3x+2)} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{x+1} + \frac{D}{(x+1)^2}\)

12\cdot\frac{x^3+4}{(x^2-1)(x^2+3x+2)} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{x+1} + \frac{D}{(x+1)^2}

\(\small{ \begin{array}{|rcll|} \hline 12\cdot\dfrac{x^3+4}{(x^2-1)(x^2+3x+2)} &=& 12\cdot\dfrac{x^3+4}{(x-1)(x+1)(x+2)(x+1)} \\ \\ \hline \\ 12\cdot\frac{x^3+4}{(x-1)(x+1)(x+2)(x+1)} &=& \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{x+1} + \frac{D}{(x+1)^2} \qquad | \qquad \cdot (x-1)(x+2)(x+1)^2 \\ \hline \end{array} }\)

\(\begin{array}{|rcll|} \hline 12(x^3+4) &=& A\cdot (x+2)(x+1)^2 + B\cdot (x-1)(x+2)^2 \\ &+& C\cdot (x-1)(x+2)(x+1) + D\cdot (x-1)(x+2) \\ \hline \end{array}\)

\(\small{ \begin{array}{lrcll} \hline \mathbf{x = 1:} & 12(1 + 4) &=& A\cdot (1+2)(1+1)^2 + B\cdot 0 + C\cdot 0 + D\cdot 0 \\ & 60 &=& A\cdot 3 \cdot 4 \\ & 60 &=& 12A \\ & \mathbf{A} &\mathbf{=}& \mathbf{5} \\ \\ \hline \mathbf{x = -1:} & 12(-1 + 4) &=& A\cdot 0 + B\cdot 0 + C\cdot 0 + D\cdot (-1-1)(-1+2) \\ & 36 &=& -2D \\ & \mathbf{D} &\mathbf{=}& \mathbf{-18} \\ \\ \hline \mathbf{x = -2:} & 12(-8 + 4) &=& A\cdot 0 + B\cdot (-2-1)(-2+1)^2 + C\cdot 0 + D\cdot 0 \\ & -48 &=& -3B \\ & 48 &=& 3B \\ & \mathbf{B} &\mathbf{=}& \mathbf{16} \\ \\ \hline \mathbf{x =0:} & 12(0 + 4) &=& A\cdot 2 + B\cdot (0-1)(0+1)^2 + C\cdot (0-1)(0+2)(0+1)\\ & &+& D\cdot (0-1)(0+2) \\ & 48 &=& 2A-B-2C-2D \\ & 48 &=& 2\cdot 5-16-2C-2\cdot(-18) \\ & 48 &=& 10-16-2C+36 \\ & 48 &=& 30-2C \\ & 2C &=& 30-48 \\ & 2C &=& -18 \\ & \mathbf{C} &\mathbf{=}& \mathbf{-9} \\ \hline \end{array} }\)

\(\displaystyle 12\cdot\frac{x^3+4}{(x^2-1)(x^2+3x+2)} = \frac{5}{x-1} + \frac{16}{x+2} - \frac{9}{x+1} - \frac{18}{(x+1)^2}\)

\(\mathbf{(A,B,C,D) = (5,16,-9,-18)}\)

 Enter (A,B,C,D,E) in order below if A, B, C, D, and E are the coefficients of the partial fractions expansion of
\(f(x) = 3\cdot\frac{x^4+x^3+x^2+1}{x^2+x-2} = Ax^2 + Bx + C + \frac{D}{x+2} + \frac{E}{x-1}.\)

 Enter (A,B,C,D,E) in order below if A, B, C, D, and E are the coefficients of the partial fractions expansion of

\(\displaystyle f(x) = 3\cdot\frac{x^4+x^3+x^2+1}{x^2+x-2} = Ax^2 + Bx + C + \frac{D}{x+2} + \frac{E}{x-1}. \)

f(x) = 3\cdot\frac{x^4+x^3+x^2+1}{x^2+x-2} = Ax^2 + Bx + C + \frac{D}{x+2} + \frac{E}{x-1}.

\(\small{ \begin{array}{|rcll|} \hline 3\cdot\dfrac{x^4+x^3+x^2+1}{x^2+x-2} &=& 3\cdot\dfrac{x^4+x^3+x^2+1}{(x+2)(x-1)} \\ \\ \hline \\ 3\cdot\dfrac{x^4+x^3+x^2+1}{(x+2)(x-1)} &=& Ax^2 + Bx + C + \dfrac{D}{x+2} + \dfrac{E}{x-1} \qquad | \qquad \cdot (x+2)(x-1) \\ \hline \end{array} }\)

\(\begin{array}{|rcll|} \hline 3(x^4+x^3+x^2+1) &=& A\cdot x^2(x+2)(x-1) + B\cdot x(x+2)(x-1) \\ &+& C\cdot (x+2)(x-1) + D\cdot (x-1) + E\cdot(x+2) \\ \hline \end{array}\)

\(\small{ \begin{array}{lrcll} \hline \mathbf{x = 1:} & 3(1 +1+1+1) &=& A\cdot 0 + B\cdot 0 + C\cdot 0 + D\cdot 0 + E\cdot (1+2) \\ & 12 &=& 3E \\ & \mathbf{E} &\mathbf{=}& \mathbf{4} \\ \\ \hline \mathbf{x = -2:} & 3(16-8+4+1) &=& A\cdot 0 + B\cdot 0 + C\cdot 0 + D\cdot (-2-1) + E\cdot 0 \\ & 3\cdot 13 &=& -3D \\ & 13 &=& - D \\ & \mathbf{D} &\mathbf{=}& \mathbf{-13} \\ \\ \hline \mathbf{x = 0:} & 3(0+0+0+1) &=& A\cdot 0 + B\cdot 0 + C\cdot(0+2)(0-1) + D\cdot (0-1) + E\cdot (0+2) \\ & 3 &=& -2C-D+2E \\ & 3 &=& -2C-(-13)+2\cdot 4 \\ & 3 &=& -2C+13+8 \\ & 3 &=& -2C+ 21 \\ & 2C &=& 21 -3 \\ & 2C &=& 18 \\ & \mathbf{C} &\mathbf{=}& \mathbf{9} \\ \\ \hline \mathbf{x = -1:} & 3(1 -1+1+1) &=& A\cdot 1(-1+2)(-1-1) + B\cdot (-1)(-1+2)(-1-1) \\ & &+& C\cdot (-1+2)(-1-1) + D\cdot (-1-1) + E\cdot (-1+2) \\ & 6 &=& -2A+2B-2C-2D+E \\ & 6 &=& -2A+2B-2\cdot 9 +2\cdot 13+ 4 \qquad | \qquad : 2 \\ & 3 &=& -A+B-9 +13+ 2 \\ & 3 &=& -A+B+6 \\ & \mathbf{-A+B} &\mathbf{=}& \mathbf{-3} \qquad (1) \\ \\ \hline \mathbf{x = 2:} & 3(16+8+4+1) &=& A\cdot 4(2+2)(2-1) + B\cdot 2(2+2)(2-1) \\ & &+& C\cdot (2+2)(2-1) + D\cdot (2-1) + E\cdot (2+2) \\ & 87 &=& 16A+8B+4C+D+4E \\ & 87 &=& 16A+8B+4\cdot 9 -13+4\cdot 4 \\ & 87 &=& 16A+8B+36 -13+16 \\ & 87 &=& 16A+8B+39 \\ & 16A+8B&=& 87 - 39 \\ & 16A+8B&=& 48 \qquad | \qquad : 8 \\ & \mathbf{2A+B} &\mathbf{=}& \mathbf{6} \qquad (2) \\ \hline \end{array} }\)

\(\begin{array}{|lrclcrclr|} \hline (2) & 2A+B &=& 6 &\qquad & -A+B &=& -3 & (1) \\ & & & &\qquad & B &=& -3+A \\ & 2A-3+A &=& 6 \\ & 3A &=& 9 \\ & \mathbf{A } &\mathbf{=}&\mathbf{ 3} \\ & & & &\qquad & B &=& -3+3 \\ & & & &\qquad & \mathbf{B} &\mathbf{=}&\mathbf{ 0} \\ \hline \end{array} \)

\(\displaystyle f(x) = 3\cdot\frac{x^4+x^3+x^2+1}{x^2+x-2} = 3x^2 + 0\cdot x + 9 - \frac{13}{x+2} + \frac{4}{x-1}\)

\(\mathbf{(A,B,C,D,E) = (3,0,9,-13,4)}\)

Find the sum of the values of x where the vertical asymptotes of f(g(x)) are located.

Both questions are to do with medians of a triangle, and the solutions come easiest from the fact that the medians trisect each other. You should have been made aware of this prior to looking at these questions.

The second question is the easier to see, so look at your  diagram for that.

The medians AM and CN cross at the point P.

By trisect, I mean that the point P splits the line AM into two parts in the ratio 2 : 1, and similarly P splits the line CN into two parts in the ratio 2 : 1.

The 2 part is the bit from the vertex, the 1 part is the bit from the side.

So, AP = 2PM (sounds like some time in the afternoon !), and CP = 2PN.

In question 2 then, rather than the earlier ' round the houses ' approach, say that

\(\displaystyle BN=AB/2=3,\text{ so }CN^{2}=3^{2}+(3\sqrt{3})^{2}=36\text{ (by Pythagoras)}\),

\(\displaystyle \text{so }CN=6,\text{ so }CP=(2/3)6=4.\)

Question 2 also can be used to give an insight into question 1.

Look at the diagram and in particular the triangles APC and PMC.

Considering the line AM to be a base line for both triangles, both APC and PMC have the same height.

Therefore, since AP is twice the length of PM it follows that the area of the triangle APC will be twice the area of the triangle PMC.

That's a clue as to how to tackle question 1.

Try it, and post again if it's still a problem.

What is the horizontal asymptote as x approaches positive infinity ?

y=1

The following exercise is designed to be solved using technology such as calculators or computer spreadsheets.
If a fair coin is tossed 100 times, the probability that one gets at least 61 heads is given by the following.

This is the p-value if you toss 100 coins and get 61 heads. Use technology to calculate this number. Round your answer to three decimal places.

With calculator texas instrument TI-89:

\(\begin{array}{|rcll|} \hline P(X\ge61) &=& 1 -\text{ binomcdf }(100,0.5,60) \\ &=& 1 - 0.98239989989 \\ &=& 0.01760010011 \\ &\approx& 0.018 \\ \hline \end{array} \)

Find the largest value of x where the plots of 

\(f(x) = - \frac{2x+5}{x+3} \text{ and } g(x) = 4\cdot \frac{x+1}{x-4}\)
intersect.

Find the largest value of x where the plots of

\(\displaystyle f(x) = - \frac{2x+5}{x+3} \qquad \text{ and } \qquad g(x) = 4\cdot \frac{x+1}{x-4}\)

f(x) = - \frac{2x+5}{x+3} \text{ and } g(x) = 4\cdot \frac{x+1}{x-4}
intersect.

\(\begin{array}{|rcll|} \hline - \dfrac{2x+5}{x+3} &=& 4\cdot \frac{x+1}{x-4} \\\\ -(2x+5)(x-4) &=& 4\cdot (x+1)(x+3) \\ 4\cdot (x+1)(x+3) &=& -(2x+5)(x-4) \\ 4\cdot (x^2+3x+x+3) &=& -(2x^2 -8x +5x -20) \\ 4\cdot (x^2+4x+3) &=& -(2x^2 -3x -20) \\ 4x^2+16x+12 &=& -2x^2 +3x +20 \\ 4x^2+2x^2+16x -3x+12-20 &=& 0 \\ 6x^2+13x-8 &=& 0 \\\\ x_{1,2} &=& \dfrac{ -13\pm \sqrt{13^2-4\cdot 6 \cdot (-8) } } {2\cdot 6} \\\\ &=& \dfrac{ -13\pm \sqrt{169+192 } } {12} \\\\ &=& \dfrac{ -13\pm \sqrt{361} } {12} \\\\ &=& \dfrac{ -13\pm 19 } {12} \\\\ x_{\text{max}} &=& \dfrac{ -13+ 19 } {12} \\\\ &=& \dfrac{ 6 } {12} \\\\ &=& \dfrac{ 1 } {2} \\\\ \mathbf{x_{\text{max}}} &\mathbf{=}&\mathbf{ 0.5 } \\ \hline \end{array}\)

Enter (A,B,C) in order below if A, B, and C are the coefficients of the partial fractions expansion of 

\(\frac{2(x^2+x-1)}{x(x^2-1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}\)

Enter (A,B,C) in order below if A, B, and C are the coefficients of the partial fractions expansion of 

\(\displaystyle \frac{2(x^2+x-1)}{x(x^2-1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}. \)
 \frac{2(x^2+x-1)}{x(x^2-1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}.

\(\small{ \begin{array}{|rcll|} \hline \dfrac{2(x^2+x-1)}{x(x^2-1)} &=& \dfrac{2(x^2+x-1)}{x(x-1)(x+1)} \\ \\ \hline \\ \dfrac{2(x^2+x-1)}{x(x-1)(x+1)} &=& \dfrac{A}{x} + \dfrac{B}{(x-1)} + \dfrac{C}{(x+1)} \qquad | \qquad \cdot x(x-1)(x+1) \\\\ 2(x^2+x-1) &=& \dfrac{A\cdot x(x-1)(x+1)}{x} + \dfrac{B\cdot x(x-1)(x+1)}{(x-1)} + \dfrac{C\cdot x(x-1)(x+1)}{(x+1)} \\\\ && \boxed{ 2(x^2+x-1) = A (x-1)(x+1) + B x (x+1) + C x(x-1) } \\\\ \hline \end{array} }\)

\(\small{ \begin{array}{lcll} \hline \mathbf{x = 0:} & 2(0+0-1) &=& A (0-1)(0+1) + B \cdot 0 (0+1) + C \cdot 0 (0-1) \\ & -2 &=& -A \\ & \mathbf{A} &\mathbf{=}& \mathbf{2} \\ \\ \hline \mathbf{x = 1:} & 2(1+1-1) &=& A (1-1)(1+1) + B\cdot 1 (1+1) + C \cdot 1(1-1) \\ & 2 &=& A \cdot 0(1+1) + B\cdot 1 (1+1) + C \cdot 1\cdot 0 \\ & 2 &=& 2B \\ & \mathbf{B} &\mathbf{=}& \mathbf{1} \\ \\ \hline \mathbf{x = -1:} & 2(1-1-1) &=& A (-1-1)(-1+1) + B\cdot (-1) (-1+1) + C \cdot (-1)(-1-1) \\ & -2 &=& A (-1-1)\cdot 0 + B\cdot (-1) \cdot 0 + C \cdot (-1)(-1-1) \\ & - 2 &=& 2C \\ & \mathbf{C} &\mathbf{=}& \mathbf{-1} \\ \hline \end{array} }\)

\(\mathbf{(A,B,C) = (2,1,-1)}\)

9)In cyclic quadrilateral ABCD, AB = 2, BC = 3, CD = 10, and DA = 6. Let P be the intersection of lines AB and CD. Find the length BP.

Draw AC

AC^2  =  2^2  + 3^2   -  2(3)(2)cos (ABC)  

AC^2  = 13  - 12cos(ABC)   (1)

And since ABC is obtuse and ADC  is supplemental to ABC, cos(ADC)  = -cos(ABC)

So

AC^2   = 6^2  + 10^2 - 2(6)(10)cos(ADC)

AC^2  = 6^2 + 10^2  - 120 [ -cos(ABC) ]

AC^2  = 6^2 + 10^2 + 120cos(ABC)   

AC^2  = 136 + 120cos(ABC)   (2)

Subtract (1) from (2)

0  = 123  + (120 + 12) cos(ABC)

0  =  123  + 132cos(ABC)

-123 / 132  =  cos (ABC)

-41/44 = cos(ABC)   

So  sin (ABC)  =  √ [ 1 -  (41/44)^2 ]  =   √ (255) / 44   =  sin PBC

Likewise.....draw  DB

DB^2  = 6^2  + 2^2  - 2(6)(2)cos(DAB)

DB^2  = 40 - 24cos(DAB)

And sincw DAB is obtuse and DCB is supplemental to DCB, cos(DCB) = -cos(DAB)

So

DB^2  = 10^2 + 3^2  - 2(10)(3)cos(DCB)

DB^2 =  10^2 + 3^2  - 60 [-cos(DAB) ]  

DB^2   =  109 + 60cos(DAB)

Subtract (3) from (4)

0  = 69 + (60 + 24) cos(DAB)

-69/84  = cos(DAB)

-23/28  = cos(DAB)

So sin (DAB)  = √  [ 1 - (23/28)^2 ]  = √ (255)/28  =  sin PCB

sin PBC / sin PCB   = PC/ PB

(√ (255) / 44)  / ( √ (255)/28)  = PC / PB

28 / 44 = PC /PB

7/11 = PC /PB

PC   =  (7/11)PB

And we have that

(PB + AB) * PB   = PC(PC + CD)

(PB + 2) * PB  = (7/11)PB [ (7/11)PB + 10)

PB^2 + 2PB  =  (49/121)PB^2 + (70/11)PB

Let PB  = x

x^2 + 2x =  (49/121)x^2 + (70/11)x

(121 - 49) x^2 / 121   +  (2 - 70/11)x  = 0

(72/121)x^2  -  (48/11)x  = 0

(1/11)x [ (72/11)x - 48 ] = 0

So..... either x  =  0   {reject}    or

(72/11)x  - 48  = 0

(72/11)x  =  48

  x =  48(11)/72  =  (2/3) * 11  =   22/3   =  PB  = BP

Proof   {PC  = (7/11)PB  =  (7/11)(22/3)  = 14/3 }

(PB + AB) * PB   = PC(PC + CD)

(22/3 + 2) * (22/3)  =  (14/3) * ( 14/3 + 10)

616 / 9    =   616 / 9

Again thanks so much

Thanks!

Prob  that the spinner lands on B  0 times  =  3/4 * 3/4   =  9/16

Prob that the spinner lands on  1 time  =  (1/4) (3/4) + (3/4)1/4)  =  6/16 = 3/8

Prob that the spinner lands on B 2 times  = (1/4) (1/4) =  1/16

The third answer

2.  M =  ( 1.5 sqrt (3), 3)      N  =  (0, 3)

The slope of AM  =   3 / (1.5sqrt (3) )   =  2/sqrt(3)

And the equation of  AM  =  

y  = [2/sqrt(3) ] x

The slope of CN  =   [ 3 / -3sqrt (3)]  = -1/sqrt(3)

And the equation of AN  =

y  =  [ -1/sqrt(3)] x  +  3

To find the x coordinate of P, we have

[2/sqrt(3) ] x  =   [ -1/sqrt(3)] x  +  3

3/sqrt (3)x  =  3

x =  3 *sqrt (3) / 3   =  sqrt (3)

And the y coordinate  of P

y  = [2/sqrt(3)] (sqrt (3)  =  2

So  CP  =  √ [ ( 3sqrt (3) - sqrt (3) )^2  + (2 - 0)^2 ]  = √ (2sqrt(3))^2 + 4 )  =

√  [ (sqrt (12) )^2  + 4 ]  =  √ [ 12 + 4]  =  √16   =   

4

Let Daniella gross pay = P, then we have:

P - [15%*P) = $51,000

P - 0.15P     =$51,000

0.85P           =$51,000      divide both sides by 0.85

P = $51,000 / 0.85

P = $60,000 - The least salary that Daniella must earn.

$10,000 - $1,250 - $3,500 - (0.15*M) =0

$5,250 - (0.15M) =0           add (0.15M) to both sides

$5,250 =0.15M                   divide both sides by 0.15

M =$5,250 / 0.15 = $35,000 - his last tax bracket. So, his total Income was:

$35,000 + $35,000 + $25,000 + $20,000(non-taxable amount)

=$115,000 - Dr. Jones Salary.

Ckeck:

$115,000 - $20,000 =$95,000 Taxable Income

On the first $25,000 he pays 5% tax =$1,250

On the next $35,000 he pays 10% tax =$3,500

$95,000 - $25,000 - $35,000 =$35,000 - his last tax bracket.

On this last $35,000 he pays 15% tax =$5,250

So, his total taxes =$1,250+$3,500+$5,250 =$10,000

Thank you sir for your help! On my homework the answer was 0.018 because they wanted in decimal form rounded three places.

OK, young person! Here is your result:

0.5^100*∑[100! / ((61+n)!(39-n)!)], n=0 to 38 =0.0176001 x 100 =1.76001%

By the way, that is the accurate probability of getting at least 61 Heads in 100 coin flips.

I believe just one.....there are two instances of   43  in   (wait for it)....Row 43 !!!

Note that 43 does not appear in any of the first 9 rows

And  in Row 10  ....C(10,2)  =   45.... and this is the smallest number in the triangle from this point forward   [ except for the elements  C(n,0), C(n,1), C(n-1, n), C (n, n) ]  that appear on each row from row  11  to row 42.....all these elements  will be < 43......and 43 will not appear in any row after row 43

Did my response aide you in your quest for mathematical success?