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3 years ago

Find the 6th term of the geometric sequence whose common ratio is 3/2 and whose first term is 5

Mathematics

1 answer:

Fynjy0 [20]3 years ago

5 0

Answer:

a(6) = 37.97

Step-by-step explanation:

If the first term a(1) is 5 and the common ratio r is 3/2, then the general formula for this geometric sequence is

a(n) = 5*r^(n - 1).

Thus, the 6th term is a(6) = 5*(3/2)^(6 - 1), or a(6) = 5(3/2)^5 = 37.97

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MArishka [77]

Answer:

c − 135

Explanation:

Let's say that the amount of money in Bailey's checking account is

c dollars.

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From this variable and number, we can state the algebraic expression

c − 135

<h2>Hope this helps!</h2>

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3 years ago

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3 years ago

Identify the coefficient of -13x

Vika [28.1K]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Solve the equation for x. Sqrt X-6+3 = 10 x=1 x=13 x = 43 x= 55

Jlenok [28]

Answer is 55
Two ways you can approach this first you can subtract the three from the 10 and get 7. Square that to get 49. Now you know that x-6 has to equal 49 and the only option greater than 49 is 55
Second you can just take the problem step by step.
Subtract three from both sides.
Sqrt x-6 =7
Square both sides
X-6=7^2
x-6=49
Add six to each side
X=55

8 0

3 years ago

Calculate the limit of the function with L'Hospital rule

mr_godi [17]

Answer:

L=24

Step-by-step explanation:

L'Hopital's Rule for Evaluating Limits:

Rule is that if $\lim_{x \to a} \frac{f(x)}{g(x)}$ takes $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form, then,

$\lim_{x \to a} \frac{f(x)}{g(x)}= \lim_{x \to a} \frac{f'(x)}{g'(x)}$

where $f'(x)=\frac{df(x)}{dx}$ and $g'(x)=\frac{dg(x)}{dx}$

Now coming to the problem,

$L= \lim_{x \to \frac{\pi}{6} } \frac{cot^{3}x-3cotx}{cos(x+\frac{\pi}{3} )}$

Here $f(x)=cot^{3}x-3cotx$ and $g(x)=cos(x+\frac{\pi}{3} )$

Substituting $x=\frac{\pi}{6}$ in f(x) and g(x),

$f(\frac{\pi}{6})=cot^{3}\frac{\pi}{6}-3cot\frac{\pi}{6}\\=3\sqrt{3}-3\sqrt{3}\\ =0$

$g(\frac{\pi}{6})=cos(\frac{\pi}{6}+\frac{\pi}{3})\\=cos\frac{\pi}{2}\\=0$

Since L takes the form $\frac{0}{0}$ , using l'hopital's rule

$L= \lim_{x \to \frac{\pi}{6}} \frac{cot^{3}x-3cotx}{cos(x+\frac{\pi}{3})}= \lim_{x \to \frac{\pi}{6}} \frac{3cot^{2}x(-cosec^{2}x)-3(-cosec^{2}x)}{-sin(x+\frac{\pi}{3})}$

now substituting $x=\frac{\pi}{6}$ ,

$L= \lim_{x \to \frac{\pi}{6}} \frac{3cot^{2}\frac{\pi}{6}(-cosec^{2}\frac{\pi}{6})-3(-cosec^{2}\frac{\pi}{6})}{-sin(\frac{\pi}{6}+\frac{\pi}{3})}\\=\frac{3*3^{2}(-2^{2})+3(2^{2})}{-1}\\=24$

6 0

3 years ago