Question

12t+18)(1,6-0,2t)=0 СРОЧНО!!!!!​

Answer 1

Решение.

$\bf (12t+18)(1,6-0,2t)=0$  

 Ни в коем случае не надо раскрывать скобки, потому что, если произведение равно 0 , то хотя бы один из множителей равен 0 .

 Приравниваем каждый множитель к 0 .

$\bf a)\ 12t+18=0\ \ \ \to \ \ \ 12t=-18\ \ ,\ \ t=-\dfrac{18}{12}\ \ ,\ \ t=-\dfrac{3}{2}\ \ ,\ \ \boxed{\bf t=-1,5}\\\\\\b)\ 1,6-0,2t=0\ \ \ \to \ \ \ 1,6=0,2t\ \ ,\ \ t=\dfrac{1,6}{0,2}\ \ ,\ \ t=\dfrac{16}{2}\ \ ,\ \ \boxed{\bf t=8}$

Ответ:  $\bf t_1=-1,5\ \ ,\ \ t_2=8\ .$

Answer 2

To solve the equation (12t+18)(1.6-0.2t) = 0, we can use the zero product property, which states that if a product of factors is equal to zero, then at least one of the factors must be equal to zero.

So we set each factor equal to zero and solve for t:

1. 12t + 18 = 0 Subtracting 18 from both sides: 12t = -18 Dividing both sides by 12: t = -18/12 Simplifying the fraction: t = -3/2 or t = -1.5

2. 1.6 - 0.2t = 0 Subtracting 1.6 from both sides: -0.2t = -1.6 Dividing both sides by -0.2 (which is equivalent to multiplying by -5): t = (-1.6) / (-0.2) t = 8

Therefore, the solutions to the equation (12t+18)(1.6-0.2t) = 0 are t = -3/2, t = -1.5, and t = 8.