So focusing on x^4 + 5x^2 - 36, we will be completing the square. Firstly, what two terms have a product of -36x^4 and a sum of 5x^2? That would be 9x^2 and -4x^2. Replace 5x^2 with 9x^2 - 4x^2: 
Next, factor x^4 + 9x^2 and -4x^2 - 36 separately. Make sure that they have the same quantity inside of the parentheses: 
Now you can rewrite this as 
, however this is not completely factored. With (x^2 - 4), we are using the difference of squares, which is 
. Applying that here, we have 
. x^4 + 5x^2 - 36 is completely factored.
Next, focusing now on 2x^2 + 9x - 5, we will also be completing the square. What two terms have a product of -10x^2 and a sum of 9x? That would be 10x and -x. Replace 9x with 10x - x: 
Next, factor 2x^2 + 10x and -x - 5 separately. Make sure that they have the same quantity on the inside: 
Now you can rewrite the equation as 
. 2x^2 + 9x - 5 is completely factored.
<h3><u>Putting it all together, your factored expression is
</u></h3>
You can turn 5 into 10/2 and add to x/2
then you need to use simple algebra to left the x alone in the equation so you need to multiply both sides by 2 so you get
2y=10+x
then you need to subtract both sides a 10 so
2y-10=x
now change the x to f'(x) and y to x so:
f'(x)=2x-10
Answer:
The correct answer is 12.
To get this you will divide 60minutes into 5.
You will get the answer "12"
Thus, there are 12, 5 minute segments, in one hour.
Hope this helps, good luck!
Answer:
Option B 13 units is correct.
Step-by-step explanation:
The formula used to find distance between 2 points is:
x₁ = 12 x₂ = 0 y₁= 9 and y₂=4
Putting values in the formula:
So, Option B 13 units is correct.
Answer:
Below, you can see the graph of the function:
f(x) = x + cos(k*x)
for different values of k, as follows:
red: k = 1
green: k = 2
orange: k = 0.
Now let's find the values of k such that our function does not have local maxima nor local minima.
First, remember that for a given function f(x), the local maxima or minima points are related to the zeros of the first derivate of f(x).
This means that if:
f'(x0) = 0.
Then x0 is a maxima, minima or an inflection point.
Then if a function is such that the f'(x) ≠ 0 , ∀x, then this function will not have local maxima nor minima.
Now we have:
f(x) = x + cos(k*x)
then:
f'(x) = 1 - k*sin(k*x)
This function will be zero when:
1 = k*sin(k*x)
1/k = sin(k*x)
now, remember that -1 ≤ sin(θ) ≤ 1
then if 1/k is smaller than -1, or larger than 1, we will not have zeros.
And this will happen if -1 < k < 1.
