$Giventhat,\left[\right(3-x\left)\right(x+2)+(-x+4\left)\right(7x+2)-(x-y\left)\right(2x-y\left)\right]-3x^2-7x+5$

$=\left[\right(3x+6-x-2x)+(-x^2-2x+28x+8)-(2x^2-xy-2xy+y^2\left)\right]-3x²-7x+5$

$=\left[\right(x-x^2+6)+(26x+8-7x^2)-(2x^2-3xy+y^2\left)\right]-3x^2-7x+5$

$=x-x^2+6+26x+8-7x^2-2x^2+3xy-y^2-3x^2-7x+5$

$=(-13x^2+3xy-y^2+20x+19)answer$

$At 10\% profit,$

$The \cos t price of first shirt =\frac{100}{100+10}\times 594=\frac{100}{110}\times 594 = 540 Tk.$

$Again, at 10\% loss$

$The \cos t price of second shirt=\frac{100}{100-10}\times 594=\frac{100}{90}\times 594 = 660 Tk.$

$Now,thetotalcostpriceoftwoshirts=540+660=1,200 Tk.$

$And the total sell price of two shirts = 594 x 2 = 1,188 Tk.$

$\therefore Loss 1.200-1,188 = 12 Tk.$

$\therefore Percentage loss = (\frac{12}{1200}\times 100)\% = 1\% \left(answer\right)$

$Let, x liters of 40\% alcohol needs to be added to make a mixture that is 25\% alcohol.$

$According to question,$

$40\% of x + 10\% of 4 = 25\% of (x + 4)$

$=\frac{40x}{100}+\frac{10\times 4}{100}= \frac{25(x+4)}{100}$

= $40x+40=25x+100 [Multiplying both sides by 100]$

= $40x - 25x = 100 - 40$

= $15x = 60$

$\therefore x = \frac{60}{15} = 4 \left(answer\right)$

$Here, 30 minutes = \frac{1}{2} hour$

$24 machines can make 5 devices in = \frac{1}{2} hour$

$\therefore I machine can make 1 devices n =\frac{1\times 24}{2\times 5} hour$

$\therefore 4 machines can make 15 devices n = \frac{1\times 24\times 15}{2\times 5\times 4} = 9 hours \left(answer\right)$

$As shown in the figure,$

$△ABC is an equilateral triangle inscribed in a circle centered at O.$

$Here, the radius of the circle is 10cm.$

$As the triangle is an equilateral, so the value of each angle is 60° and half of it is 30°.$

$Here, OD \perp BC and OD line divides BC equally.$

$Now, from △BOD,$

$Cos30°= \frac{Base}{Hypotenuse}= \frac{BD}{OB} =\frac{BD}{10}$

$=\frac{\sqrt{3}}{2}=\frac{BD}{10}$

$\therefore BD = 5\sqrt{3}$

$\therefore One Arm of the triangle is = BD + CD = 5\sqrt{3} + 5\sqrt{3} = 10\sqrt{3} cm$

$Now, we know that,$

$Area of a equilateral triangle is \frac{\sqrt{3}}{4}{\left(Arm\right)}^{2}c{m}^2$

$=\frac{\sqrt{3}}{4}{\left(10\sqrt{3}\right)}^2 c{m}^2 =\frac{\sqrt{3}}{4}(100\times 3)c{m}^2=75\sqrt{3} c{m}^2 \left(answer\right).$