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As I said in the answer that the Math Package that I have used "iteration method" to get that answer. "Iteration method" is a fancy phrase for "trial and error method"!!. The computer "guesses" and then tries to see if the equation balances, or how close it is to the real answer and repeats the process through iteration and interpolation, until it finds the "right" answer that balances the equation.

It does this very rapidly, since it can do hundreds or even thousands of these iterations in one second.

implicit differentiation

\(\begin{array}{|rcll|} \hline e^y &=& x \quad & | \quad \text{chain rule} \\ e^y\cdot y' &=& 1 \\ y' &=& \dfrac{1}{e^y} \quad & | \quad \dfrac{1}{e^y}= x \\ y' &=& \dfrac{1}{x} \\ \hline \end{array}\)

System of equations: Solve the system of equation

{xy=8}

{x²+y²=20}

Formula:

\(\begin{array}{|rcll|} \hline (x+y)^2 &=& x^2+2xy+y^2 \\ x^2+y^2 &=& (x+y)^2-2xy \quad & | \quad xy = 8 \\ x^2+y^2 &=& (x+y)^2 - 8 \\ \hline \end{array}\)

1.

\(\begin{array}{|rcll|} \hline x^2+y^2 &=& 20 \quad & | \quad x^2+y^2 = (x+y)^2-16 \\ (x+y)^2-16 &=& 20 \\ (x+y)^2 &=& 20+16 \\ (x+y)^2 &=& 36 \\ \mathbf{x+y} &\mathbf{=}& \mathbf{\pm 6} \\ \hline \end{array}\)

Two Systems of equation now:

\(\begin{array}{|lrcl|lrcl|} \hline (1)& xy &=& 8 &(1) & xy &=& 8 \\ &\mathbf{y} &\mathbf{=}& \mathbf{\frac{8}{x}} & & \mathbf{y} &\mathbf{=}& \mathbf{\frac{8}{x}} \\\\ (2)& x+y &=& 6 &(2)& x+y &=& -6 \\ & x+\frac{8}{x} &=& 6 & & x+\frac{8}{x} &=& -6 \\ & x^2+8 &=& 6x & & x^2+8 &=& -6x \\ & x^2-6x+8 &=& 0 & & x^2+6x+8 &=& 0 \\ & x_{1,2}&=& \frac{6\pm \sqrt{36-4\cdot 8} }{2} & & x_{3,4}&=& \frac{-6\pm \sqrt{36-4\cdot 8} }{2} \\ & x_{1,2}&=& \frac{6\pm 4 }{2} & & x_{3,4}&=& \frac{-6\pm 4 }{2} \\ & x_{1,2}&=& 3\pm 1 & & x_{3,4}&=& -3\pm 1 \\\\ & \mathbf{x_1 =4} && \mathbf{x_2 = 2} & & \mathbf{x_3 = -2} && \mathbf{x_4 =-4} \\ & \mathbf{y_1} =\frac{8}{4}\mathbf{=2} && \mathbf{y_2} = \frac{8}{2}\mathbf{=4} & & \mathbf{y_3} = \frac{8}{-2}\mathbf{=-4} && \mathbf{y_4} = \frac{8}{-4}\mathbf{=-2} \\ \hline \end{array} \)

the first equation is right, the answer for x is 1.97568, but how can u get this?

2.5           

P(x)   =   a(x + b)²(x - c)

P(x)   =   a(x + b)(x + b)(x - c)      Multiplied out.....

P(x)   =   ax³ - acx² + 2abx² - 2abcx + ab²x - ab²c     So  P(x)  does have a degree of  3 .

P(x)  has multiplicity of  2  at  (-3, 0)  and passes through  (2, 0)  so...

P(x)  =  a(x + 3)²(x - 2)

P(x)  passes through  (0, 36)  so  P(0)  =  36

P(0)  =  a(0 + 3)²(0 - 2)

36   =   a(0 + 3)²(0 - 2)

36   =   a(9)(-2)

36   =   -18a

a   =   -2

0  =  -x - 3

                      Add  x  to both sides of the equation.

x  =  -3

-3  is  ≤  1  so  -3  is a solution to  f(x) = 0

0  =  x/2 + 1

                      Subtract  1  from both sides.

-1  =  x/2

                      Multiply both sides by  2 .

-2  =  x

-2  is not  >  1  so  -2  is not a solution to  f(x)  =  0

The only value of  x  that makes  f(x) = 0  is  -3

So the sum of all values of  x  such that  f(x) = 0  is just  -3 .

Part A:

We want the equation of the line with a slope of  4  that passes through the point  (1, 2)

We know a point  (1, 2)  and we know the slope  4

So we can write the equation in point-slope form.

y - 2   =   4(x - 1)            To get this into slope-intercept form, distribute the  4 .

y - 2   =   4x - 4              Add  2  to both sides.

y  =  4x - 2

Part B:

Is the line   x + 4y - 20  =  0   perpendicular to the line  y  =  4x - 2   ?
 

The slope of the line   y  =  4x - 2   is  4.

To find the slope of the line  x + 4y - 20  =  0  , let's get the equation into slope-intercept form.

x + 4y - 20  =  0       Add  20  to both sides.

x + 4y  =  20            Subtract  x  from both sides.

4y   =   -x + 20         Divide both sides by  4 .

y   =   -\(\frac14\)x + 5          Now we can see that the slope of this line is  -\(\frac14\) .

Since   -\(\frac14\)   is the negative reciprocal of  4  , these lines are perpendicular.

Yes, the two ships are sailing perpendicular to each other.

Point  A  is the current location of the plane.

Point  B  is the point on the ground directly under the plane.

Point  C  is the location of the airport.

m∠BAC   =   90° - 6°   =   84°

a)

plane's ground distance to the airport =  BC

tan 84°  =   BC / AB

tan 84°  =   BC / 3900

3900 tan 84°  =  BC

BC  ≈  37106.021    feet

b)

distance between plane and runway   =   AC

cos 84°   =   AB / AC

cos 84°   =   3900 / AC

AC cos 84°  =  3900

AC  =  3900 / cos 84°

AC  ≈  37310.412    feet

I agree with Melody...I think it's false.

The zeros, or roots, of a polynomial are the values of the independent variable that cause a polynomial to be equal to zero, not the factors.

The answer is C. 

I'd say 'false' but it is a very weirdly worded question.

\( \frac{1}{M} +\frac{1}{N} = \frac{1}{P} \;\;find\;\;P \\ \frac{N}{MN} +\frac{M}{NM} = \frac{1}{P}\\ \frac{N+M}{MN} = \frac{1}{P}\\ \frac{MN}{N+M} = \frac{P}{1}\\ P=\frac{MN}{N+M} \)

These questions all use the same fact.

Opposite angles in a cyclic quadriateral are Supplementary, That means they add up to 180 degrees.

Now you can work all of them out by yourself.

Yes, it is a good start :)

Assuming no friction or terminal velocity:

h = 0 when it hits the ground     find t

0 = 16t^2+vt + s       v = 0  as you drop it     and s= 16 at the starting point

0= -16t^2 + 16

-16 = - 16t^2

1= t^2                t=1 sec

Thanks Guys

The question asks about the type of roots--not what the roots are. 

Because of this technicality, we can combine knowledge from Descartes' Rule of Signs and the Fundamental Theorem of Algebra and the Complex Conjugate Root Theorem in order to answer this question. 

The Descartes' Rule of Signs states that the number of sign changes in \(f(x)\), when the terms are ordered by degree, indicates the number of positive solutions or less than this by an even number. Let's consider the original function \(f(x)=5x^2-7x-6\):

\(f(x)=\underbrace{5x^2-7}\underbrace{x-6}\\ \hspace{16mm}+1\hspace{7mm}+0\hspace{10mm}=\text{1 sign change}\)

Notice that +5x² and -7x have a different sign, so I marked that change with a +1. -7x and -6 do not have a signage change, so I marked it with a +0.

The Fundamental Theorem of Algebra is relevant because it indicates how many total solutions that a function can have. In this case, there are 2 roots since the degree of the polynomial is 2. 

The Complex Conjugate Root Theorem states that if polynomials have real coefficients with complex roots, then they will come in pairs; in other words, there will always be an even number of non-real roots.  

I would say that it would be wise to consolidate this information into a table.

Based on the rules mentioned previously, only one possibility remains: Both roots are real.

\(5x^2-7x-6=0\)

First, do grouping.

\(5x^2-10x+3x-6=0\)

\(5x(x-2)+3(x-2)=0\)

\((5x+3)(x-2)=0\)

Then "tee-up" and solve for \(x\).

\(5x+3=0, x-2=0\)

\(x=-\frac{3}{5}, x=2\)

The roots are very much so real.

\(\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+...}}}}}}}}}}}}\)

With LaTeX