Mathematics, 16.12.2019 22:31 cheergirl21

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6. given the series: 4 + 12 + 36 +

find the 11th term of the series and find the sum of the first 11 terms.

7. a ball is dropped from a height of 40 feet and bounces back up to 90% of its previous height on each successive bounce. how far will the ball have traveled by the time it comes to a stop?

8. for each series below, decide whether it converges or diverges. find the sum, if it exists.

a) 80 + 20 + 5 + 5/4 +
b) 2/9 + 4/3 + 8 +

Answers: 2

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Answers

Answer from: jjdj5218

Q6. a₁₁ = 236 196; S₁₁ = 354 292

Q7. 354 292

Q8. a) Converges, S = 320/3; b) diverges

Step-by-step explanation:

6. Geometric sequence

The first three terms of your sequence are 4, 12, 36.

Each term differs from the previous one by a factor of 3, so it is a geometric sequence.

Each term has the form

aₙ = a₁rⁿ⁻¹

In your sequence, a₁ = 4 and r = 3.

Thus, the formula for the nth term is

aₙ = 4(3)ⁿ⁻¹

The 11th term is

a₁₁ = 4(3)¹¹⁻¹ = 4(3)¹⁰ = 4 × 59 049 = 236 196

The formula for the sum of the first n terms of a geometric series is

Sum = a₁[(1 - rⁿ)/(1 - r)]

For the sum over the first 11 terms,

Sum = 4[(1 - 3¹¹)/(1 - 3)

= 4(1 - 177 147)/(-2)

= -2(-177 146)

= 354 292

7. Bouncing ball

Mathematically, the ball never stops bouncing. The height of each bounce just gets infinitesimally small.

So , you have an infinite geometric series in which the first term is 40 ft and each successive term is 90 % of the previous term.

The general formula for the nth term is

aₙ = a₁rⁿ⁻¹

with a₁ = 40 ft and r = 0.90.

Since |r| <1, we have a convergent series, and the formula for the sum is

S = a₁/(1-r)

∴ S = 40/(1 - 0.90) = 40/0.10 = 400 ft

The bouncing ball will have travelled 400 ft.

8. Test for convergence

a) 80 + 20 + 5 + 5/4 + ...

r = a₂/a₁ = 20/80 = ¼

r < 1, so the series converges.

S = 80/( 1 - ¼) = 80/¾ = 80 × ⁴/₃ = 320/3

The sum of the series is 320/3.

b) 2/9 + 4/3 + 8 + ...

r = a₂/a₁ = (⁴/₃)/(²/₉) = ⁴/₃ × ⁹/₂ = 2 × 3 = 6.

r > 1, so the series diverges.

Answer from: jakeyywashere

6 236,196 and 354,294

Step-by-step explanation:

$\text{Find the 11th term:}\\a_1=4\qquad r=\dfrac{12}{4}=3\qquad n=11\\\\a_n=a_1\cdot r^{n-1}\\a_{11}=4\cdot (3)^{11-1}\\.\quad =4\cdot3^{10}\\.\quad =4\cdot 59,049\\.\quad =\large\boxed{236,196}\\\\\\\text{Find the sum:}\\S_n=\dfrac{a_1(1-r^n)}{1-r}\\\\\\S_{11}=\dfrac{4(1-3^{11})}{1-3}\\\\\\.\quad =\dfrac{4(1-177,147)}{-2}\\\\.\quad =-2(-177,146)\\\\.\quad =\large\boxed{354,294}$

7 400

Step-by-step explanation:

$\sum\limits^{n=1}_\infty 40(0.9)^{n-1}\\\\a_1=40\qquad r=0.9\\\\S_\infty =\dfrac{a_1}{1-r}\\\\\\.\quad =\dfrac{40}{1-0.9}\\\\\\.\quad =\dfrac{40}{0.1}\\\\\\.\quad =\large\boxed{400}$

8 a) converges

b) diverges

Step-by-step explanation:

A series converges if: -1 < r < 1

a) 80 + 20 + 5 + $\frac{5}{4}$

$a)\ r=\dfrac{a_2}{a_1}=\dfrac{20}{80}=\dfrac{1}{4}\implies \text{converges}\\\\\\b)\ r={a_2}\div{a_1}=\dfrac{4}{3}\div \dfrac{2}{9}=\dfrac{4}{3}\times \dfrac{9}{2}=2\times 3=6\implies \text{diverges}$

Answer from: Quest

That is so true, i can’t believe how true this is

Answer from: Quest

center (1,3) with radius 6

step-by-step explanation:

we must complete the square to find the center and radius of the circle.

first make sure the x and y squared terms have 1 as their coefficients. we also make sure x and y terms together.

$x^2-2x+y^2-6y=26$

we now create space between the x and y terms with parenthesis.

$(x^2-2x)+(y^2-6y)=26$

we complete the square by taking the middle terms -2x and the -6y - divide each and square them.

$\frac{-2}{2} =(-1)^{2} =1$

$\frac{-6}{2} =(-3)^{2} =9$

we add the squares to both sides.

$(x^2-2x+1)+(y^2-6y+9)=26+1+9$

simplify.

$(x^2-2x+1)+(y^2-6y+9)=36$

and write the quadratics in factored form.

$(x-1)^{2} +(y-3)^{2} =36$

the center is (h,k) or (1,3). the radius is the square root of 36 which is 6.

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