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choli [55]
3 years ago
13

Find the equation and solve it.

Mathematics
1 answer:
DiKsa [7]3 years ago
5 0
4(3x + 1)= 76
*There are 4 sides to a square and they are all equal in length, so you would mulitply 3x +1 by 4
*The total perimeter is 76
*Distribute the 4 into the equation

12x + 4=76
*subtract 4 on both sides

12x=72
*divide 12 on both sides

x=6
*Plug 6 in for x in the equation

3(6) +1
*this means that each side of the square would be 19 meters
*If you want to check to make sure you solved for x correctly, just plug it into the equation[4(3x + 1)]

Hope this helps!
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Which relation is a function? A. X -14,-9-1,4,6 Y 6,2,-3,2,4 B. X -2,1,9,9,15 Y -3,-3,0,-8,12 C. X -4,6,6,7,9 Y -1,-3,-1,15,1 D.
Butoxors [25]
❅☃ To find a function you need to look at the x values, if one of the same x value is repeated more then once, it is considered a non function.
❅☃ Option A:
-14
-9
-1
4
6
As you can see no repeated values.
Therefore A is already our function but lets look at the others just in case. 
❅☃ Option B
-2
1
9
9
15
9 was repeated therefore it is a non function.
❅☃ Option C
-4
6
6
7
9
6 was again repeated. Non function.
❅☃ And lastly, D.
-16
-11
10
10
16
10 was reapted. Non function.
❅☃ Your answer is A
8 0
3 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.

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3 years ago
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