Question

решите, умоляю!!!! дам много баллов

Answer 1

Ответ:

$5x = \frac{1}{3} {x}^{2} + 12 \\ 5x - \frac{1}{3} {x}^{2} - 12 = 0 \\ \frac{1}{3} {x}^{2} - 5x + 12= 0 \\ d = {b}^{2} - 4ac = {( - 5)}^{2} - 4 \times \frac{1}{3} \times 12= 25 - 16 = 9 \\ x1 = \frac{ - b - \sqrt{d} }{2a} = \frac{ - ( - 5) - \sqrt{9} }{2 \times \frac{1}{3} } = \frac{5 - 3}{ \frac{2}{3} } = 3 \\ x2 = \frac{ - b + \sqrt{d} }{2a} = \frac{ - ( - 5) + \sqrt{9} }{2 \times \frac{1}{3} } = \frac{5 + 3}{ \frac{2}{3} } = 12 \\ \\ 5 \times 3 = 15 \\ \frac{1}{3} \times {12}^{2} + 12 = \frac{1}{3} \times 144 + 12 = 48 + 12 = 60$

То есть точки пересечения: (3;15), (12;60)

Answer 2

$\displaystyle\bf\\y=5x \ \ , \ \ y=\frac{1}{3} x^{2} +12\\\\\\\frac{1}{3} x^{2} +12=5x \ |\cdot 3\\\\\\x^{2} -15x+36=0\\\\D=(-15)^{2} -4\cdot 36=225-144=81=9^{2} \\\\\\x_{1} =\frac{15-9}{2} =3\\\\\\x_{2} =\frac{15+9}{2} =12\\\\y_{1} =5\cdot x_{1} =5\cdot 3=15\\\\y_{2} =5\cdot x_{2} =5\cdot 12=60\\\\Otvet:(3 \ ; \ 15) \ , \ (12 \ ; \ 60)$