Pre-Calculus › Pre-Calculus
What is the sum of the alternating series below?
The alternating series follows a geometric pattern.
We can evaluate the geometric series from the formula.
Convert to polar form.
Write the Cartesian to polar conversion formulas.
Substitute the coordinate point to the equations to find .
Since is not located in between the first quadrant, this is not the correct angle. The correct location of this coordinate is in the third quadrant. Add
radians to get the correct angle.
Therefore, the answer is .
Find the derivative of the following function
To find the derivative of this function, we simply need to use the Power Rule. The Power rule states that for each term, we simply multiply the coefficient by the power to find the new coefficient. We then decrease the power by one to obtain the degree of the new term.
For example, with our first term, , we would multiply the coefficient
by the power
to obtain the new coefficient of
. We then decrease the power by one from 4 to 3 for the new degree. Therefore, our new term is
. We then simply repeat the process with the remaining terms.
Note that with the second to last term, our degree is 1. Therefore, multiplying the coefficient by the power gives us the same coefficient of 8. When the degree decreases by one, we have a degree of 0, which simply becomes 1, making the entire term simply 8.
With our final term, we technically have
Therefore, multiplying our coefficient by our power of 0 makes the whole term 0 and thus negligible.
Our final derivative then is
Solve this polynomial by factoring if it is factorable:
Not factorable
To factor this polynomial it is prudent to recognize that there will only be two factors since the highest power is .
Then ask what numbers multiply to equal postive 35.
Next, what numbers can multiply to equal positive 12.
Let's try 7 and 5 for the last term and 3 and 4 for the first term.
Be sure to put an "x" in the first term of each factor.
Choose the signs based on what the polynomial calls for. In our case we choose negative signs to get positive 35.
Foil these two factors and we get .
Write the parametric equation for the line y = 5x - 3.
x = 5t - 3
y = t
x = t
y = 5t - 3
x = 5t - 3
y = 5t - 3
x = t
y = t
In the equation y = 5x - 3, x is the independent variable and y is the dependent variable. In a parametric equation, t is the independent variable, and x and y are both dependent variables.
Start by setting the independent variables x and t equal to one another, and then you can write two parametric equations in terms of t:
x = t
y = 5t - 3
Write a vector equation describing the line passing through P1 (1, 4) and parallel to the vector = (3, 4).
First, draw the vector = (3, 4); this is represented in red below. Then, plot the point P1 (1, 4), and draw a line
(represented in blue) through it that is parallel to the vector
.
We must find the equation of line . For any point P2 (x, y) on
,
. Since
is on line
and is parallel to
,
for some value of t. By substitution, we have
. Therefore, the equation
is a vector equation describing all of the points (x, y) on line
parallel to
through P1 (1, 4).
Determine the value of of the function
In order to determine the value of of the function we set
The value becomes
As such
Find , then find its magnitude.
and
are both vectors.
In vector addition, you simply add each component of the vectors to each other.
x component: .
y component: .
z component: .
The new vector is
.
To find the magnitude we use the formula,
Thus its magnitude is 5.
Find if
and
.
To find the direction vector going from to
, subtract the x and y-coordinates of
from
.
Find the zeros and asymptotes for
.
Zero: ; Asymptote:
Zeros: ; Asymptote:
Zero: ; Asymptotes:
Zeros: ; Asymptotes:
Zero: ; Asymptotes:
To find the information we're looking for, we should factor this equation:
This means that it simplifies to .
When the equation is in the form of a fraction, to find the zero of the function we need to set the numerator equal to zero and solve for the variable.
To find the asymptote of an equation with a fraction we need to set the denominator of the fraction equal to zero and solve for the variable.
Therefore our equation has a zero at -3 and an asymptote at -2.