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3 years ago

Enter the number that belongs in the green box (please enter both numbers for the empty boxes)

Mathematics

1 answer:

VLD [36.1K]3 years ago

3 0

Given:

The equation is:

$5x-2=4+2x$

To find:

The number that belongs in the green box and another box.

Solution:

We have,

$5x-2=4+2x$

Subtracting 2x from both sides, we get

$5x-2-2x=4+2x-2x$

$3x-2=4$

Adding 2 on both sides, we get

$3x-2+2=4+2$

$3x=6$

We need to divide both sides by 3 to isolate the variable x.

On dividing both side by 3, we get

$\dfrac{3x}{3}=\dfrac{6}{3}$

Therefore, the missing values are 3 and 3, and the required equation is $\dfrac{3x}{3}=\dfrac{6}{3}$ .

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A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.

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1. Divide wire b in parts x and b-x.

2. Bend the b-x piece to form a triangle with side (b-x)/3

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$2 \pi r=x$

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4. Let f(x)= $\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}$

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)= $\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x$ =0

$\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0$

$(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b$

$x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

6. So one part is $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$ and the other part is b- $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

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