Provide an example of a situation that displays exponential growth

How to do question 22?

Aleks04 [339]

Answer:

A = 20sinθ(6 + 5 cosθ) cm²

Step-by-step explanation:

Drop perpendiculars DE and CF to AB.

Then, we have congruent triangles ADE and BCF, plus the rectangle CDEF.

The formula for the area of the trapezium is

A = ½(a + b)h

DE = 10sinθ

AE = 10cosθ

BF = 10cosθ

EF = CD = 12 cm

AB = AE + EF + BF = 10cosθ + 12 + 10 cosθ = 12 + 20cosθ

A = ½(a + b)h

= ½(12 +12 + 20 cosθ) × 10 sinθ

=(24 + 20 cosθ) × 5 sinθ

= 4(6 + 5cosθ) × 5sinθ

= 20sinθ(6 + 5 cosθ) cm²

6 0

3 years ago

How do I solve 13.65=h+4.88 it’s stupid I know but I need help!!

murzikaleks [220]

Answer:

8.77

Step-by-step explanation:

13.65 - 4.88 = 8.77

8.77 = h

4 0

3 years ago

Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.

Aleks [24]

Answer:

It means $\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$ also converges.

Step-by-step explanation:

The actual Series is::

$\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$

The method we are going to use is comparison method:

According to comparison method, we have:

$\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n$

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

$=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}$

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}$

Now:

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}$