Write an expression to match the product of 8 and three times 11

In this 3x3 square, you can use only numbers from 1-9, to make all the rows and columns equal to 15. Good luck to the person sol

bulgar [2K]

Answer:

Hello,

Step-by-step explanation:

The well-known magic square. (first apparition in China).

Here it the methode du Marquis de Liouville: (odd square:3,5,7,9,11,13,...)

We are going to put successively the number from 1 to n² (here n=3)

We imagine that the square is put on a sphere;

We begin in the middle of the last line where we put 1

ICI:

We move in direction SE of one case and put le next number

until we reach of multiple of n

After have reached a multiple of n, we move verticaly of one case

and we go to ICI until we reach n²

8 0

3 years ago

2(6x+3) hurry please thanks

cricket20 [7]

Answer:

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░░░▌░▄▄▄▐▌▀▀▀░░ This is Bob

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▀█▌░░░▄░▀█▀░▀ ░░ Copy And Paste Him onto all of ur brainly answers

░░░░░░░▄▄▐▌▄▄░░░ So, He Can Take

░░░░░░░▀███▀█░▄░░ Over brainly

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Step-by-step explanation:

btw the answer is

(12x+6)

5 0

3 years ago

A rectangular field with one side along a river is to be fenced. Suppose that no fence is needed along the river, the fence on t

Andreas93 [3]

Answer:

a) Side Parallel to the river: 200 ft

b) Each of the other sides: 400 ft

Step-by-step explanation:

Let L represent side parallel to the river and W represent width of fence.

The required fencing (F) would be $F=L+2W$ .

We have been given that field must contain 80,000 square feet. This means area of field must be equal to 80,000.

$LW=80,000...(1)$

We are told that the fence on the side opposite the river costs $20 per foot, and the fence on the other sides costs $5 per foot, so total cost (C) of fencing would be $C=20L+5(2W)\Rightarrow 20L+10W$ .

From equation (1), we will get:

$L=\frac{80,000}{W}$

Upon substituting this value in cost equation, we will get:

$C=20(\frac{80,000}{W})+10W$

$C=\frac{1600,000}{W}+10W$

$C=1600,000W^{-1}+10W$

To minimize the cost, we need to find critical points of the the derivative of cost function as:

$C'=-1600,000W^{-2}+10$

$-1600,000W^{-2}+10=0$

$-1600,000W^{-2}=-10$

$-\frac{1600,000}{W^2}=-10$

$-10W^2=-1,600,000$

$W^2=160,000$

$\sqrt{W^2}=\pm\sqrt{160,000}$

$W=\pm 400$

Since width cannot be negative, therefore, the width of the fencing would be 400 feet.

Now, we will find the 2nd derivative as:

$C''=-2(-1600,000)W^{-3}$

$C''=3200,000W^{-3}$

$C''=\frac{3200,000}{W^3}$

Now, we will substitute $W=400$ in 2nd derivative as:

$C''(400)=\frac{3200,000}{400^3}=\frac{3200,000}{64000000}=0.05$

Since 2nd derivative is positive at $W=400$ , therefore, width of 400 ft of the fencing will minimize the cost.

Upon substituting $W=400$ in $L=\frac{80,000}{W}$ , we will get:

$L=\frac{80,000}{400}\\\\L=200$

Therefore, the side parallel to the river will be 200 feet.

4 0

3 years ago

What is the greatest common factor of 20x6y+40x4y2−10x5y5 ?

mestny [16]

The greatest common factor: 20x^6y 40x^4y^2 10x^5y^5 is 10x^4y.
20x^6y = 10x^4y*2x^2
40x^4y^2 = 10x^4y*4y
10x^5y^5 = 10x^4y*y^4

8 0

3 years ago

Select the correct answer.

zhannawk [14.2K]

You have a good day and good luck on that question

8 0

3 years ago