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seropon [69]
3 years ago
5

If you have a radius of 6 what is the area of a circle?

Mathematics
1 answer:
tekilochka [14]3 years ago
3 0

Answer:

36pi, or 113.097

Step-by-step explanation:

area = pi*(r)^2

area = pi(6)^2

area = 36pi, or 113.097

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For what values of x is the rational expression below undefined? x-7/x^2-7x-8
alekssr [168]
We have

f(x)=(x-7)/(x²<span>-7x-8)

resolving the quadratic equation
using a graph tool-------- > see the attached figure
x=-1  x=8
therefore
</span>(x-7)/(x²-7x-8)=(x-7)/[(x+1)*(x-8)

the answer is  -1  and  8



5 0
3 years ago
A graphing calculator is recommended. A function is given. g(x) = x4 − 5x3 − 14x2 (a) Find all the local maximum and minimum val
Taya2010 [7]

Answer:

The local maximum and minimum values are:

Local maximum

g(0) = 0

Local minima

g(5.118) = -350.90

g(-1.368) = -9.90

Step-by-step explanation:

Let be g(x) = x^{4}-5\cdot x^{3}-14\cdot x^{2}. The determination of maxima and minima is done by using the First and Second Derivatives of the Function (First and Second Derivative Tests). First, the function can be rewritten algebraically as follows:

g(x) = x^{2}\cdot (x^{2}-5\cdot x -14)

Then, first and second derivatives of the function are, respectively:

First derivative

g'(x) = 2\cdot x \cdot (x^{2}-5\cdot x -14) + x^{2}\cdot (2\cdot x -5)

g'(x) = 2\cdot x^{3}-10\cdot x^{2}-28\cdot x +2\cdot x^{3}-5\cdot x^{2}

g'(x) = 4\cdot x^{3}-15\cdot x^{2}-28\cdot x

g'(x) = x\cdot (4\cdot x^{2}-15\cdot x -28)

Second derivative

g''(x) = 12\cdot x^{2}-30\cdot x -28

Now, let equalize the first derivative to solve and solve the resulting equation:

x\cdot (4\cdot x^{2}-15\cdot x -28) = 0

The second-order polynomial is now transform into a product of binomials with the help of factorization methods or by General Quadratic Formula. That is:

x\cdot (x-5.118)\cdot (x+1.368) = 0

The critical points are 0, 5.118 and -1.368.

Each critical point is evaluated at the second derivative expression:

x = 0

g''(0) = 12\cdot (0)^{2}-30\cdot (0) -28

g''(0) = -28

This value leads to a local maximum.

x = 5.118

g''(5.118) = 12\cdot (5.118)^{2}-30\cdot (5.118) -28

g''(5.118) = 132.787

This value leads to a local minimum.

x = -1.368

g''(-1.368) = 12\cdot (-1.368)^{2}-30\cdot (-1.368) -28

g''(-1.368) = 35.497

This value leads to a local minimum.

Therefore, the local maximum and minimum values are:

Local maximum

g(0) = (0)^{4}-5\cdot (0)^{3}-14\cdot (0)^{2}

g(0) = 0

Local minima

g(5.118) = (5.118)^{4}-5\cdot (5.118)^{3}-14\cdot (5.118)^{2}

g(5.118) = -350.90

g(-1.368) = (-1.368)^{4}-5\cdot (-1.368)^{3}-14\cdot (-1.368)^{2}

g(-1.368) = -9.90

7 0
3 years ago
Answer the following questions.
SOVA2 [1]

Answer:

43

Step-by-step explanation:


5 0
2 years ago
Function that has <br> Domain: x≠-1<br> Range: y≠2
Degger [83]
I don't understand your question.
4 0
3 years ago
PLEASE!!! WILL MARK BRAINLIEST IF YOU'RE RIGHTIn this triangle, what is the value of x? enter your answer, rounded to the neares
OLga [1]
\bold{ANSWER:}
x = 61.1

\bold{SOLUTION:}

6 0
2 years ago
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