Pls help with this question asap!

B(1)= -12

Leto [7]

Given:

$b(1)=-12$

$b(n)=b(n-1)\cdot 4$

To find:

The third term of the sequence.

Solution:

We have,

$b(n)=b(n-1)\cdot 4$

For n=2, we get

$b(2)=b(2-1)\cdot 4$

$b(2)=b(1)\cdot 4$

$b(2)=-12\cdot 4$ $[\because b(1)=-12]$

$b(2)=-48$

For n=3, we get

$b(3)=b(3-1)\cdot 4$

$b(3)=b(2)\cdot 4$

$b(3)=-48\cdot 4$ $[\because b(2)=-48]$

$b(3)=-192$

Therefore, the 3rd term of the sequence is -192.

7 0

2 years ago

Suppose you toss a 6-sided die 40 times, with 14 tosses landing on the five. What percentage of the tosses would be fives?

Alex777 [14]

$\dfrac{14}{40}\cdot100\%=35\%$

3 0

3 years ago

Read 2 more answers

The Gotemba walking trail up Mount Fuji is about 9km long. Walkers need to return from the 18km walk by 8pm. Toshi estimates tha

stiv31 [10]

Answer:

Toshi must begin his walk at 11:00 AM in order that he can return by 8:00 PM.

Step-by-step explanation:

Since the Gotemba walking trail up Mount Fuji is about 9km long, and walkers need to return from the 18km walk by 8pm, if Toshi estimates that he can walk up the mountain at 1.5km / h on average, and down at twice that speed , these speeds taking into account meal breaks and rest times, to determine what is the latest time he can begin his walk so that he can return by 8pm the following calculation must be performed:

Climb: 1.5 km / h

Descent: 2 x 1.5 km / h = 3 km / h

Climb: 9 km / 1.5 km / h = 6 hours

Descent: 9km / 3 km / h = 3 hours

Total: 9 hours

8 PM = 20:00

20:00 - 09:00 = 11:00

Thus, Toshi must begin his walk at 11:00 AM in order that he can return by 8:00 PM.

3 0

3 years ago

Value of log2 8 + log 3(1/3)

Dvinal [7]

There’s no answer choices but i got log(28•3^1/3)

5 0

2 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:

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