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Elanso [62]
3 years ago
10

Find the relative extrema, if any, of the function. Use the Second Derivative Test if applicable. (If an answer does not exist,

enter DNE.) g(x) = x3 − 15x
Mathematics
1 answer:
belka [17]3 years ago
3 0

Answer:

We have the function g(x) = x^3 -15*x

First, to find extrema, we can find the zeros of the first derivative.

g'(x) = 3*x^2 -15

g'(x) = 0 = 3*x^2 - 15

x^2 = 15/3 = 5

x = √5

x = -√5

Now, watching at the second derivative we have:

g''(x) = 6*x

so when we have

g''(√5) = 6*√5 > 0 then x = √5 is a local minimum

g''(-√5) = -6*√5 < 0, then x = -√5 is a local maximum.

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ozzi

Answer:

7. 22 2/5   8. 10 8/10

Step-by-step explanation:

7. 7x3=21 + 7x1/5= 22 2/5

8.  2x5=10 + 2x4/10= 10 8/10

5 0
2 years ago
Which of the following polygons has nine lines of symmetry?
anastassius [24]
So we know d. is out and that's all I got help wise sorry its not much..:)
3 0
3 years ago
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A book is 0.33 meters long. How many millimeters long is it?
Phoenix [80]

we know that

1\ meter=1,000\ millimeters

so

by proportion

Find how many millimeters are in 0.33 meters

\frac{1,000}{1} \frac{mm}{m} =\frac{x}{0.33} \frac{mm}{m} \\ \\x=1,000*0.33 \\ \\ x=330\ mm

therefore

<u>the answer is</u>

330\ mm

7 0
2 years ago
What are the endpoint coordinates for the midsegment of △PQR that is parallel to PQ¯¯¯¯¯?
andriy [413]

Answer:

M(x₄ ,y₄) = (-3.5 , 0.5)  and

N (x₅ ,y₅) = ( -1 , -0.5 )

Step-by-step explanation:

Let the endpoint coordinates for the mid segment of △PQR that is parallel to PQ be

M(x₄ ,y₄) and N(x₅ ,y₅) such that MN || PQ

point P( x₁ , y₁) ≡ ( -3 ,3 )

point Q( x₂ , y₂) ≡ (2 , 1 )

point R( x₂ , y₂) ≡ (-4 , -2)  

To Find:

M(x₄ ,y₄) = ?  and

N (x₅ ,y₅) = ?

Solution:

We have Mid Point Formula as

Mid\ point(x,y)=(\frac{x_{1}+x_{2} }{2}, \frac{y_{1}+y_{2} }{2})

As M is the mid point of PR and N is the mid point of RQ so we will have

Mid\ pointM(x_{4} ,y_{4})=(\frac{x_{1}+x_{3} }{2}, \frac{y_{1}+y_{3} }{2})

Mid\ pointN(x_{5} ,y_{5})=(\frac{x_{2}+x_{3} }{2}, \frac{y_{2}+y_{3} }{2})

Substituting the given value in above equation we get

Mid\ pointM(x_{4} ,y_{4})=(\frac{-3+-4 }{2}, \frac{3+-2} }{2})

∴ Mid\ pointM(x_{4} ,y_{4})=(\frac{-7} }{2}, \frac{1}{2})

∴ Mid\ pointM(x_{4} ,y_{4})=(-3.5, 0.5)

Similarly,

Mid\ pointN(x_{5} ,y_{5})=(\frac{2+-4 }{2}, \frac{1+-2 }{2})

∴ Mid\ pointN(x_{5} ,y_{5})=(\frac{-2 }{2}, \frac{-1}{2})

∴ Mid\ pointN(x_{5} ,y_{5})=(-1, -0.5)

∴ M(x₄ ,y₄) = (-3.5 , 0.5)  and

  N (x₅ ,y₅) = ( -1 , -0.5 )

3 0
3 years ago
Please Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Travka [436]

Answer:

I cant see the picture. could you put it in the comments

Step-by-step explanation:

5 0
3 years ago
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