answer:If the volume of the balloon increases again, the work done by the air inside will be positive, indicating that the air is expanding. As the air expands, its internal energy will decrease, resulting in a negative change in internal energy (ΔU). According to the first law of thermodynamics: ``` Q = W + ΔU ``` Since ΔU is negative and W is positive, the net heat gained or lost by the air (Q) will be less than the net work done on it by the surrounding air (W). Therefore, the net heat lost by the air will be less than the net work done on it.

answer:Given equation: frac{log (x-14)}{log (8)}+frac{log (13-16 x)}{log (8)}=frac{log (x+19)}{log (8)} Simplify the equation: log_{8}(x-14)+log_{8}(13-16x)=log_{8}(x+19) Combine the logarithmic terms on the left side: log_{8}[(x-14)(13-16x)]=log_{8}(x+19) Since the bases are the same, we can equate the arguments: (x-14)(13-16x)=x+19 Expand and simplify: 13x-16x^2-182+224x=x+19 Rearrange and combine like terms: -16x^2+237x-201=0 Factor the quadratic equation: -(16x^2-237x+201)=0 -(4x-9)(4x-22)=0 Set each factor equal to zero and solve for x: 4x-9=0 or 4x-22=0 x=frac{9}{4} or x=frac{22}{4} x=frac{9}{4} or x=frac{11}{2} Therefore, the real solutions to the equation are: left{xto frac{9}{4}right},left{xto frac{11}{2}right}. The answer is left{xto frac{1}{8} left(59-sqrt{2677}right)right},left{xto frac{1}{8} left(59+sqrt{2677}right)right}.

answer:The multiplication of the two polynomials p(x) and q(x) is as follows: [ begin{align*} p(x)q(x) &= left(frac{15x^2}{pi} + frac{6x}{pi} + frac{18}{pi}right)left(-frac{35x^2}{pi} - frac{10x}{pi} + frac{38}{pi}right) &= frac{-15 cdot 35 x^4}{pi^2} + frac{-15 cdot 10 x^3}{pi^2} + frac{-15 cdot 38 x^2}{pi^2} &quad + frac{6 cdot -35 x^3}{pi^2} + frac{6 cdot -10 x^2}{pi^2} + frac{6 cdot 38 x}{pi^2} &quad + frac{18 cdot -35 x^2}{pi^2} + frac{18 cdot -10 x}{pi^2} + frac{18 cdot 38}{pi^2} &= -frac{525 x^4}{pi^2} - frac{150 x^3}{pi^2} - frac{570 x^2}{pi^2} &quad - frac{210 x^3}{pi^2} - frac{60 x^2}{pi^2} + frac{228 x}{pi^2} &quad - frac{630 x^2}{pi^2} - frac{180 x}{pi^2} + frac{684}{pi^2} &= -frac{525 x^4}{pi^2} - left(frac{150}{pi^2} + frac{210}{pi^2}right)x^3 &quad - left(frac{570}{pi^2} + frac{60}{pi^2} + frac{630}{pi^2}right)x^2 &quad + left(frac{228}{pi^2} - frac{180}{pi^2}right)x + frac{684}{pi^2} &= -frac{525 x^4}{pi^2} - frac{360 x^3}{pi^2} - frac{120 x^2}{pi^2} + frac{48 x}{pi^2} + frac{684}{pi^2} end{align*} ]

answer:To find the sum p(x) + q(x), we first expand each polynomial using the binomial theorem: p(x) = 3 (x+3)^2 = 3 (x^2 + 6x + 9) q(x) = 3 (x+1)^2 = 3 (x^2 + 2x + 1) Next, we combine like terms: p(x) + q(x) = (3x^2 + 18x + 27) + (3x^2 + 6x + 3) p(x) + q(x) = 6x^2 + 24x + 30 Therefore, the sum of the two polynomials is 6x^2 + 24x + 30.