What is the scale factor from ABC to DEF?

Mathematics

1 answer:

sdas [7]3 years ago

3 0

Within similar triangles, the angles are the same. So, we need to look at the side lengths then.

72 to 12

66 to 11

42 to 7

In each pair of corresponding side lengths, the original side length from triangle ABC is divided by 6 (or multiplied by 1/6) to get the smaller side length on triangle DEF.

The scale factor from ABC to DEF is 1/6.

Hope this helps!

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seropon [69]

Answer:

Following are the solution to the given point:

Step-by-step explanation:

Please find the comp[lete question in the attached file.

Given:

$\bold{ \lim_{n \to \ \infty} (1+ \frac{r}{n})^{nt} =e^{rt}}$

In point 1:

$\to y = (1+ \frac{r}{n})^{nt}$

In point 2:

$\to \ln (y)= nt \ln(1+ \frac{r}{n})$

In point 3:

Its key thing to understand, which would be that you consider the limit n to $\infty$ , in which r and t were constants!

$=lim_{n \to \ \infty} \ln (y) = lim_{n \to \ \infty} nt \ln(1+\frac{r}{n})\\\\= lim_{n \to \ \infty} \frac{\ln(1+\frac{r}{n})}{\frac{1}{nt}}\\\\= lim_{n \to \ \infty} \frac{\frac{-r}{\frac{n^2}{(1+\frac{r}{n})}}}{- \frac{1}{n^2t}}\\\\= lim_{n \to \ \infty} \frac{\frac{rn^2t}{n^2}}{(1+\frac{r}{n})}\\\\= lim_{n \to \ \infty} \frac{rt}{(1+\frac{r}{n})}\\\\= \frac{rt}{(1+\frac{r}{0})}\\\\=rt$