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The max are is a square  that is  (20/4)  =  5 ft on  a side

Max Area =  side^2 =  5^2  = 25  ft^2

\(\sqrt{160} = \sqrt{16} \times \sqrt {10} = 4 \sqrt{10}\)

\(\sqrt{252} = \sqrt{36} \times \sqrt 7 = 6 \sqrt 7\)

\(\sqrt{245} = \sqrt{49} \times \sqrt{5} = 7 \sqrt 5 \)

\(\sqrt{180} = \sqrt{36} \times \sqrt{5} = 6 \sqrt 5\)

\({4 \sqrt {10} \over 6 \sqrt 7} \times {7 \sqrt 5 \over 6 \sqrt5} = {4 \sqrt {10} \over 6 \sqrt 7} \times {7 \over 6} = {28 \sqrt {10} \over 36 \sqrt 7} = {28 \sqrt {10} \over 36 \sqrt 7} \times {\sqrt 7 \over\sqrt 7} = {28 \sqrt {70} \over 252} = \color{brown}\boxed{\sqrt{70} \over 9} \)

By power of a point, CM = 5*sqrt(3).

how many positive perfect cubes does (3^5)(5^12) have?

Floor[5/3 +1] * [12 / 3 + 1] ==2  x  5 ==10 perfect cubes.

6, a, b ==6, 36, 216 [Common ratio==6]

a, b, 1296==36, 216, 1296

Let \(l \) = local stamps and let \(f\) = foreign stamps. 

We have the system: 

\(l = {4 \over 5}(l + f)\)

\(l - 20 = {8 \over 15} (f + l) \)

Now, just solve for l and f...

There are 10 perfect cubes.

BuilderBoi: You missed the count by 1. Should read: 10 / 21

When y =  1

x ranges from  1  to  28

When y = 2

x ranges from  1 to  27

When y = 3

x  ranges from 1 to  26

...

When y = 22

x ranges from 1 to 2

When y = 23

x  only has the value 1

So the  number  of  positive integer  values  = the sum of the first  28 positive integers  =

(28) (29)  / 2   = 

406

Let b = (-1/2)

So

x^4 - 3x^2 + 2  + (-1/2) [ 2x^4 - 6x^2 + 2x - 1 ]  =

x^4  - 3x^2 + 2   - x^4 + 3x^2 - x + 1   =

-x + 3

The smallest possible degree ( as found by the guest)  =   1

The smallest possible degree is 1.

Since f(x) is divisible by x^3, f(x) is of the form ax^5 + bx^4 + cx^3.

You then want ax^5 + bx^4 + cx^3 + 2 to be divisible by (x + 1)^3.  Using long division, you get the equations

-10a  + 6b - 3c = 0

4a - 3b + 2c = 0

-a + b - c + 2 = 0

==> a = 6, b = 16, c = 12

So f(x) = 6x^5 + 16x^4 + 12x^3.

\( $\lfloor \sqrt[3]{n} \rfloor = 5$\)

Note that     (125)^(1/3)  =  5

And that      (216)^(1/3)  = 6

So....the number of positive integers that satisfy this  =     215 - 125 + 1 =  91

        B

A       8      D      3     C

Note that  triangles  ABD  and DBC are under the same height

And....triangles under the same height are to each other as their bases

So

[ABD ]   =  [ 8 / ( 8 + 3)  ] [ ABC]   =  (8/11) (120)  =    960 / 11 

There are: 1000 / 3==333  + 1000 / 4==250  -  1000 / 12 ==83  -  1000 / 15 ==67  -  1000 / 30==33

333  +  250  -  83  -  67  -  33 ==400 such integers

Oh, wait that's a cube root...

Yea, you're right, I thought it meant \(n\sqrt{3}\) not \(\sqrt[3]{n}\)

Hey BuilderBoi, 

I have a question...

How is it 3?

Using the floor rule \(\mathrm{If}\:\lfloor \:n\:\rfloor =x\:\mathrm{then}\:x\le \:n \)

We get \(5\le \sqrt[3]{n}<6\)

That is \(125\le \:n<216\)

-Vinculum

Hello, 

Take a look at the graph below:

The points where they intersect are (1, -4);(2, -5)

-Vinculum

Hello, 

This question was answered by CPhill,

Please take a look:

 https://web2.0calc.com/questions/algebra_13724

If you have a question, reply back.

-Vinculum

Hello, 

So to convert from percents to decimals you move the decimal right 2 spaces. 

Example: 45.0% = 0.45

Or you can divide the percent by 100.

Example: We have 67.0%, 

Divide by 100

67.0/100 = 0.67

-Vinculum

Hi sherwyo!

Let \(F(x,y)=x^2+xy+y^2+x+y+1\)

Then,

\(F_x=2x+y+1\)

\(F_y=x+2y+1\)

Where \(F_x \text{ and, } F_y \) are the partial derivatives.

Now, since we want a minimum point, we set these partial derivatives to zero:

\(2x+y+1=0\)                    (1)

\(x+2y+1=0\)                    (2)

Multiply equation (2) by -2 and then add to equation (1), to get:

\(-3y-1=0 \implies y=-\dfrac{1}{3}\)

Thus,

\(x=-2y-1=\dfrac{2}{3}-1=-\dfrac{1}{3}\)

So the minimum is: \(F(-\frac{1}{3},-\frac{1}{3})=\dfrac{2}{3}\)

WolframAlpha:

There are:

[- 107 - 433] / 4 + 1 ==136 terms

There is no integer solution for n < 100!! Make sure your question is correct.