Completing the square is method of solving a quadratic equation that involves finding a value to add to both the left and right side of the equation. This value has the extra benefit of making one side of the equation a perfect trinomial which makes the function easier to identify and/or graph.

Let's begin by factoring

Take the coefficient of the **divide** it by **square** it

We added these values but the equation remains balanced because they are added to both sides of the equation.

3 years ago

**The sequence is a geometric sequence.**

- In an Arithmetic sequence there is a common difference
#d# between any two consecutive terms - In a Geometric sequence there is a common ratio
#r# for any two consecutive terms

The sequence given is :

1) Checking if the sequence is an arithmetic sequence:

As observed

2) Checking if the sequence is a geometric sequence:

Since

**So the sequence is a geometric sequence.**

3 years ago

A piecewise continuous function is a function that is continuous except at a finite number of points in its domain.

Note that the points of discontinuity of a piecewise continuous function do not have to be removable discontinuities. That is we do not require that the function can be made continuous by redefining it at those points. It is sufficient that if we exclude those points from the domain, then the function is continuous on the restricted domain.

For example, consider the function:

graph{(y - x/abs(x))(x^2+y^2-0.001) = 0 [-5, 5, -2.5, 2.5]}

This is continuous for all

The discontinuity at

At

So the left limit and right limit disagree with one another and with the value of the function at

If we exclude the finite set of discontinuities from the domain, then the function restricted to this new domain will be continuous.

In our example, the definition of

If we graph

Slightly confusingly, the function

graph{tan(x) [-10.06, 9.94, -4.46, 5.54]}

Meanwhile, the sawtooth function

graph{3/5(abs(sin(x * pi/2))-abs(cos(x * pi/2))-abs(sin(x * pi/2)^3)/6+abs(cos(x * pi/2)^3)/6) * tan(x * pi/2)/abs(tan(x * pi/2))+1/2 [-2.56, 2.44, -0.71, 1.79]}

Via changing velocity I pressume you mean an object that accelerates or decelerates.

**If acceleration is constant**

If you have initial and final speed:

Usually

If the above method does not work because you are missing some values, you can use the equation below. The distance traveled

where

Therefore, if you know the distance, initial speed and acceleration you can find the time by solving the quadratic equation that is formed. However, if acceleration if not given, you will need the final speed of the object

and substitute to the distance equation, making it:

Factor

So you got 2 equations. Pick one of them, which will help you solve with the data you are given:

Below are two other cases where acceleration is not constant. **FEEL FREE TO IGNORE THEM** if acceleration in your case is constant, since you placed it in the Precalculus category and the below contain calculus.

**If acceleration is a function of time #a=f(t)#**

The definition of acceleration:

If you still don't have enough to solve, that means you have to go to distance. Just use the definition of speed and move on, as if I analyze it further it will only confuse you:

The second part of this equation means integrading acceleration with respect to time. Doing that gives an equation with only

**If acceleration is a function of speed #a=f(u)#**

The definition of acceleration:

2 years ago

The partial fraction decomposition suggests that the function can be broken down into the sum of two other functions, or;

Where we need to solve for

We can now cancel the denominator on each side, leaving;

Now we can solve for

The

Solving for

We can substitute

This time, the

Now that we have our values for

2 years ago

Look at changes of signs to find this has

Then do some sums...

#f(x) = -3x^4-5x^3-x^2-8x+4#

Since there is one change of sign,

#f(-x) = -3x^4+5x^3-x^2+8x+4#

Since there are three changes of sign

Since

Newton's method can be used to find approximate solutions.

Pick an initial approximation

Iterate using the formula:

#a_(i+1) = a_i - f(a_i)/(f'(a_i))#

Putting this into a spreadsheet and starting with

#x ~~ 0.41998457522194#

#x ~~ -2.19460208831628#

We can then divide

Notice the remainder

Check the discriminant of the approximate quotient polynomial:

#-3x^2+0.325x-4.343#

#Delta = b^2-4ac = 0.325^2-(4*-3*-4.343) = 0.105625 - 52.116 = -52.010375#

Since this is negative, this quadratic has no Real zeros and we can be confident that our original quartic has exactly

2 years ago

The general standard form for the equation of a circle is

for a circle with center

Given

We can transform this into the standard form by the following steps:

Move the

Complete the square for each of the

Re-write the

Often we would leave it in this form as "good enough",

but technically this wouldn't make the

So more accurately:

with center at

An exponential function is in the general form

#y=a(b)^x#

We know the points

#8=a(b^-1)=a/b#

#2=a(b^1)=ab#

Multiply both sides of the first equation by

#8b=a#

Plug this into the second equation and solve for

#2=(8b)b#

#2=8b^2#

#b^2=1/4#

#b=+-1/2#

Two equations seem to be possible here. Plug both values of

**If** **:**

#2=a(1/2)#

#a=4#

Giving us the equation:

**If** **:**

#2=a(-1/2)#

#a=-4#

Giving us the equation:

**However!** In an exponential function,

The only valid function is

#color(green)(y=4(1/2)^x#

Given that a point

Then

This is not a polar graph!!!

We are give the coordinates of (-4,3)

Suppose we viewed this in the context of Cartesian form and use

Then

So we would have

Suppose the graph was only plotted over the range

Then the above graph would not be continuous but be a line from

All we need now is the angle that that line makes to the x-axis and the length of that line.

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The Polar angle

Let the angle from the line to the negative x-axis be

Then

But

so

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Given that a point

Then

2 years ago

The first thing that has to be done is to factorise the denominator.

#a^3 - b^3 = (a - b )(a^2 + ab + b^2 ) #

here then

So

Now

#(8x - 1) /(x^3 - 1 )= A/(x - 1 ) +( Bx + C)/(x^2 + x + 1 )#

Multiplying through by

#8x - 1 = A(x^2 + x + 1 ) + (Bx + C )(x - 1 )" " " " color(red)(("*"))#

We now have to find the values of

Note that if we use

Substitute x = 1 in equation

#7 = 3A + 0 rArr A = 7/3#

To find B and C it will be necessary to compare the coefficients on both sides of the equation

#rArr 8x -1 = Ax^2 + Ax + A + Bx^2 + Cx - Bx - C #

This can be 'tidied up' by collecting like terms and letting

#8x - 1 = 7/3 x^2 + 7/3 x + 7/3 + Bx^2 + Cx - Bx - C #

Compare

#0 = 7/3 + B rArr B = -7/3 #

Now compare constant terms.

# - 1 = A - C = -7/3 - C rArr C= -10/3 #

Finally

#(8x - 1 )/(x^3 - 1 ) =( 7/3)/(x - 1 ) + ( -7/3x - 10/3)/(x^2 +x + 1 ) #