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Bess [88]
3 years ago
11

A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a​ single-strand electric

fence. With 700 m of wire at your​ disposal, what is the largest area you can​ enclose, and what are its​ dimensions?
Mathematics
1 answer:
otez555 [7]3 years ago
3 0

Answer:

dimension are 350 m and 350 m  and

area is 122500 m²

Step-by-step explanation:

Given data

perimeter is 700 m

to find out

the largest area and its​ dimensions

solution

we consider x and y are the sides

so area is xy

perimeter is 2x+ y = 700

so we can say y = 700 -2x

so area will be = x (700 - 2x)

area = 700x - 2x²

we take derivative here and equal to zero

dA/dt = 700x - 2x²

700 - 4x = 0

x = 700/4

x = 175

so will be

y = 700 -2x

y = 700 -2(175)

y = 350

so dimension are 350 m and 350m

area is 122500 m²

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Hitman42 [59]

Answer:

5 power 2 is the answer

Step-by-step explanation:

7 0
3 years ago
Pls help!Please help
mamaluj [8]

Answer:

4,6,8,10

$10,$15,$20,$25

Step-by-step explanation:

Part A :

4,6,8,10

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because the top row is adding by two and the bottom row is adding by 5

5 0
3 years ago
When a number is tripled,the results is -33
defon

Answer:

x = -11

Step-by-step explanation:

Considering tripled means multiplied by 3 because of tri, we know that we can use the equation 3x = -33.

To solve this we can divide both sides by 3:

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4 0
3 years ago
A right triangle has legs that measure 7 cm and 6 cm. What is the length of the hypotenuse in meters?
dalvyx [7]
<h2>Greetings!</h2>

Answer:

\sqrt{85}  or 9.22

Step-by-step explanation:

You need to remember the following equation:

a^{2} + b^{2} = c^{2}

Where a and b is the lengths of the two sides and c is the hypotenuse.

So simply plug these values into this:

6^{2} + 7^{2} = h^{2}

36 + 49 = h^2

36 + 49 = 85

So the hypotenuse is the square root of 85:

\sqrt{85}  = 9.22


<h2>Hope this helps!</h2>
7 0
3 years ago
An oxygen tank contained 212 12/3 liters of oxygen before 27 1/3 liters were used. If the tank can hold 240 3/8 liters, how much
lukranit [14]
So, initially   240\frac{3}{8} -212 \frac{12}{3} were empty:

this is :

240  \frac{3}{8} - 212  \frac{12}{3}  = 240  \frac{3}{8} - 212  +4 = 240  \frac{3}{8} - 216= 24  \frac{3}{8}

12/4 is 4 so that's why i substituted it with 4.

now, later 27 1/3 were used so we add this tho the original empty space

 24  \frac{3}{8}+ 27  \frac{1}{3} =  24  \frac{9}{24}+ 27  \frac{8}{24} = 51  \frac{17}{24}


which is the result!





5 0
3 years ago
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