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Mathematics

1 answer:

S_A_V [24]3 years ago

4 0

Answer:

The weight of square is 8 grams.

Step-by-step explanation:

Given that,

The weight of the circle = 2 grams

The weight of the triangle = 4 grams

We need to find the weight of the square. Let the weight of each square is w.

On LHS, 2 circles and 4 squares are there.

On RHS, 2 squares, 4 triangles and 2 circles are there.

For a balanced position,

Weight on LHS = Weight on RHS

$2\times \text{weight of circle}+4\times \text{weight of squares}=2\times \text{weight of squares}+4\times \text{weight of triangles }+2\times \text{weight of circle}$ 2(2)+4(w) = 2(w) + 4(4) +2(2)

4 + 4w = 2w +16 +4

4w-2w = 20-4

2w = 16

w = 8

So, weight of square is 8 grams.

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The owner of a cattle ranch would like to fence in a rectangular area of 3000000 m^2 and then divide it in half with a fence dow

Reika [66]

If you notice the picture below, the amount of fencing, or perimeter, that will be used will be 3w + 2l

now $\bf \begin{cases}
A=l\cdot w\\\\\
A=3000000\\
----------\\
3000000=l\cdot w\\\\
\frac{3000000}{w}=l
\end{cases}\qquad thus
\\\\\\
P=3w+2l\implies P=3w+2\left( \cfrac{3000000}{w} \right)\implies P=3w+\cfrac{6000000}{w}
\\\\\\
P=3w+6000000w^{-1}\\\\
-----------------------------\\\\
now\qquad \cfrac{dP}{dw}=3-\cfrac{6000000}{w^2}\implies \cfrac{dP}{dw}=\cfrac{3w^2-6000000}{w^2}
\\\\\\
\textit{so the critical points are at }
\begin{cases}
0=w^2\\\\
0=\frac{3w^2-6000000}{w^2}
\end{cases}$

solve for "w", to see what critical points you get, and then run a first-derivative test on them, for the minimum

notice the $0=w^2\implies 0=w$ so. you can pretty much skip that one, though is a valid critical point, the width can't clearly be 0

so.. check the critical points on the other