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Mariana [72]
3 years ago
10

8 points and brainliest answer if you get it right

Mathematics
1 answer:
MissTica3 years ago
6 0
Y= 17 would be the answer.

For something to be a function, it needs an input (x) and an output (y).

There is no x is this answer.

I hope this helps!
~cupcake
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What is the answer to 4(10 + 9)
riadik2000 [5.3K]

Answer:

76

Step-by-step explanation:

10+9= 19

19*4=76

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Which one of the following equations represents a circle with its center of origin?
babymother [125]

Answer:

  • A) x² + y² = 5

Step-by-step explanation:

<u>The equation of the circle:</u>

  • (x - h)² + (y - k)² = r²

The center is (h, k)

<u>Wen the center is the origin we have, h = 0, k = 0:</u>

  • x² + y² = r²

The only equation in same format is the first one

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Step-by-step explanation:

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2 years ago
Complete parts a through c for the given function.
victus00 [196]

Answer:

a. The critcal points are at

x=0,-5,3

b. Then, x = -5   is a maximum and x=3 is a minimum

c. The absolute minimum is at   x = 3  and the absolute maximum is at  x = -5.

Step-by-step explanation:

(a)

Remember that you need to find the points where

f'(x)=0

Therefore you have to solve this equation.

20x^4  + 40x^3 - 300x^2 = 0

From that equation you can factor out    20x^2  and you would get

20x^2 (  x^2  +2x - 15)  = 0

And from that you would have   20x^2 = 0  , so x = 0.

And you would also have  x^2 +2x-15 = 0.

You can factor that equation as    x^2 +2x -15 = (x+5)(x-3) = 0

Therefore   x=-5 ,   x=3.

So the critcal points are at

x=0,-5,3

b.  

Remember that a function has a maximum at a critical point if the second derivative at that point is negative. Since

f''(x) = 80x^3 + 120x^2 -600x\\f''(-5) = 80(-5)^3 + 120(-5)^2 -600(-5) = -4000 < 0\\\\f''(3) = 80(3)^3 + 120(3)^2 -600(3) = 1440 > 0 \\

Then, x = -5   is a maximum and x=3 is a minimum

c.

The absolute minimum is at   x = 3  and the absolute maximum is at  x = -5.

6 0
2 years ago
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Evaluate g (c) = 4 - 3x when c = -3, 0, and 5
Lana71 [14]

Answer:

6.09

Step-by-step explanation:

multiplayer ok so that is how I did it ok

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