$\qquad \textit{Amount for Exponential Growth} \\\\ A=P(1 + r)^t\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &a\\ r=rate\to 10\%\to \frac{10}{100}\dotfill &0.10\\ t=\textit{elapsed time}\dotfill &x\\ \end{cases} \\\\\\ A=a(1 + 0.10)^{x}\implies A=a(1.1)^x~\hfill \boxed{y=a(1.1)^x}$

5 0

2 years ago

Use the differential to approximate the expression. Then use a calculator to approximate the quantity, and give the absolute val

scoray [572]

9514 1404 393

Answer:

0.0056

Step-by-step explanation:

f(x) = √(49 +x)

f'(x) = 1/(2√(49 +x))

A linear approximation of f(x) expanded about x=0 is ...

f(x) ≈ f(0) + f'(0)x = 7 +x/(2·7)

Then for √53, we have x=4

f(4) ≈ 7 +4/14 = 7 2/7 . . . . . approximate √53 using differentials

The calculator value of √53 is about 7.280110, so the difference in results is ...

approx - actual ≈ 7.285714 -7.280110 = 0.005604 ≈ 0.0056

7 0

3 years ago