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Here's (7)

Let x be the width of the garden

The, the amount of fencing available for the lengths is 300-2x.....and since there are 3 equal "lengths," the length of one is just (300 - 2x)/3

So....the area is

A = (x)((300 - 2x) / 3)) = 100x - (2/3)x^2

A' = 100 - (4/3)x

Set this to 0

100 - (4/3)x = 0

(4/3)x = 100

x = 75

Taking the second derivative, we have

y = -(4/3)....so this curve is concave down everywhere....so x = 75 is a max

And the area is (75)(300-2(75))/3  =  (75)(150/3) = (75)(50) = 3750 ft²

Here's (6)

y = 2 + 3x - x^3

y ' = 3 - 3x^2

y'' = -6x

Notice that this will be concave down on ( 0, ∞) because at all x's in this interval, the second derivative will be negative

15.4x-13.7=14x-17.2

Step1: Add 13.7 to both sides

Step2: Subtract 14x from both sides

step 3: think about what you neen to get rid of next and decide how this can be  achieved.

$$\\K^{0.3}L^{0.7}/L\\\\
=K^{0.3}L^{0.7-1}\\\\
=K^{0.3}L^{-0.3}\\\\
=\frac{K^{0.3}}{L^{0.3}}\\\\
=\left(\frac{K}{L}\right)^{0.3}$$

If you have to do it by hand this is where you start

 $$235\frac{714286}{1000000}$$

I did it on the site calc

$${\mathtt{0.714\: \!286}} = {\frac{{\mathtt{5}}}{{\mathtt{7}}}}$$

235 and 5/7

Here's (5)

y = 2x^3 - 3x^2 - 12x

y' = 6x^2 - 6x - 12

y'' = 12x - 6

Set this to 0

12x - 6 = 0    factor

6 (2x -1) = 0       divide by 6 on both sides

2x - 1 = 0

x = 1/2        and this is a possible inflection point

Test a point on either side of x = 1/2 in the second derivative (I'll pick 0 and 1)

2(0) - 1 = -1

2(1) -1 = 1

So, since the second derivative has different values on either side of x = 1/2......this is an inflection point.

well 100% = 1

so your question is very simple  :)

$$\\(x-2)(x-0.5)(x+4)\\\\
(x^2-0.5x-2x+1)(x+4)\\
(x^2-2.5x+1)(x+4)\\
(x+4)(x^2-2.5x+1)\\
x(x^2-2.5x+1) + 4(x^2-2.5x+1)\\$$

You should be able to finish it :))

15 * 60  square feet 

Huh?

What is you question again ???

??

https://www.google.com.au/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=weather%20in%20Lierge%20Luxembourge

Here's (4)

y = x^4 - 4x^3

y' = 4x^3 - 12x^2

y'' = 12x^2 - 24x

Set this to 0

12x^2 - 24x = 0     factor

12x ( x - 2) = 0

And x = 0  and x = 2 are possible inflection points.

Test a point on either side of 0 in the second derivative (I'll use -1 and 1)......putting -1 into the second derivative we have a positive and putting 1 into the second derivative gives a negative.....so x = 0 is an inflection point.

And putiing 3 into the second derivative gives a positive (and since 1 gives a negative), then x = 2 is another inflection point.

A camel might never make its way out of the rift valley.  

I just googled it - there are not any advertised camel safaris from South Africa to Saudi Arabia.

Australia does export camels to Saudi Arabia though.  Maybe I could go with them and detour to Johanesburg and then go overland from there?

These days.....the camel might be faster than flying.......

Here's (3)

y = x^5 - 20x^2

y' = 5x^4 - 40x

Set this to 0

5x^4 - 40x =  0      factor 

5x(x^3 - 8) = 0    divide by 5 on both sides

x(x^3 - 8) = 0

So x = 0 is one critical point

x^3 = 8      take the cube root of both sides

x = 2     and this is another critical point

And taking the second derivative we have

y'' = 20x^3 - 40

And putting 0 into this we have a negative.....so we have a max when x = 0

And putting 2 into this we have a positive......so we have a min at x = 2

Are we traveling by camel?😊

Here's (2)

y = 5x^2 + 2x

y' = 10x + 2

Set this to 0

10x + 2 = 0

10x = -2

x = -(1/5)

Note that the second derivative is just

y'' = 2

This curve is continuous and concave up at all points so we have a minimum at  x = -(1/5)

Here's (1)

y = x^2 + x - 2

y' = 2x + 1

Set this to 0

2x + 1 = 0

2x = -1

x = -1/2

And this is the critical point.

Using the second derivative, we have

y'' = 2

This indicates that the curve is concave up at every point.  Thus, the curve is at an absolute minimum at x = -1/2. And since this polynomial is continuous everywhere, the curve is decreasing on (-∞, -1/2) and  increasing on (-1/2, ∞)

Nice , thank you chris 

So we have

20x + 8y ≤ 500

And the values that "work" are (10, 22) and (16, 22)

Thanks!

180 = C * A

20 = C *A^3

Note that if we divide the first by A on both sides, we have  180 / A  = C ......and substituting this into the second thing, we have

20 = (180 / A ) A^3 

20 = (180) A ^2     divide both sides by 180

20/180 = 1/9 = A^2    taking the square root of both sides, we have

A = ±(1 / 3)

Therefore, C = 180 / ±(1 / 3) =  ±540

Thus, the solutions for (A, C) are (1/3 , 540) and (-1/3, -540)

I'm howling because i am a wolf of course - That is what I am supposed to do for cry'n out loud.