Question

$5$

$\frac{1}{6-9x}$ dt is equal to $0.2$ lambda

Answer 1

25

Answer 2

The problem statement provides two pieces of information about a mathematical expression:

1. The value of the expression is given as $5$.
2. The value of the expression is also given as $0.2 \lambda$.

The expression itself is $\frac{1}{6-9x}$ dt. In the context of calculus, this typically refers to an integrand, and 'dt' indicates integration with respect to 't'. However, the problem does not explicitly state an integral sign. Given the format, it is most likely that the value of the integral (or some quantity derived from the expression) is directly equated to 5 and to $0.2 \lambda$.

Let's denote the value of the expression as $V$.

From the first line, we have:

$V = 5$

From the second line, we have:

$V = 0.2 \lambda$

By equating the two expressions for $V$, we get:

$5 = 0.2 \lambda$

To solve for $\lambda$, we divide both sides by $0.2$:

$\lambda = \frac{5}{0.2}$

To simplify the division, we can write $0.2$ as a fraction or multiply the numerator and denominator by 10:

$\lambda = \frac{5}{1/5}$

$\lambda = 5 \times 5$

$\lambda = 25$

The integral part $\int \frac{1}{6-9x} dt$ is evaluated as $\frac{t}{6-9x} + C$. If the problem intended for this integral to be directly equal to 5 and $0.2 \lambda$, then it would imply $\frac{t}{6-9x} + C = 5$ and $\frac{t}{6-9x} + C = 0.2 \lambda$. This would still lead to $5 = 0.2 \lambda$, confirming the interpretation that the numerical value is given directly.