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3 years ago

Two terms in a geometric sequence are a3=729 and a4=243.

Mathematics

1 answer:

lesantik [10]3 years ago

8 0

<h2>The first term of the given sequence (a) = 6561</h2>

Step-by-step explanation:

Let the first term = a and common difference = d

Given,

$a_{3}$ = 729 and $a_{4}$ = 243

To find, the first term of the given sequence (a) = ?

We know that,

The nth term of a G.P.

$a_{n} =ar^{n-1}$

The 3rd term of a G.P.

$a_{3} =ar^{3-1}$

⇒ $ar^{2}$ = 729 ..............(1)

The 4th term of a G.P.

$a_{4} =ar^{4-1}$

⇒ $ar^{3}$ = 243 ..............(2)

Dividing equation (2) by (1), we get

$\dfrac{ar^{3}}{ar^{2}}$ = $\dfrac{243}{729}$

⇒ $r=\dfrac{1}{3}$

Put $r=\dfrac{1}{3}$ in equation (1), we get

$a(\dfrac{1}{3})^{2}$ = 729

⇒ $a(\dfrac{1}{9})$ = 729

⇒ a = 9 × 729 = 6561

∴ The first term of the given sequence (a) = 6561

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Let C be the curve of intersection of the parabolic cylinder x^2 = 2y, and the surface 3z = xy. Find the exact length of C from

Maslowich

I've attached a plot of the intersection (highlighted in red) between the parabolic cylinder (orange) and the hyperbolic paraboloid (blue).

The arc length can be computed with a line integral, but first we'll need a parameterization for $C$ . This is easy enough to do. First fix any one variable. For convenience, choose $x$ .

Now, $x^2=2y\implies y=\dfrac{x^2}2$ , and $3z=xy\implies z=\dfrac{x^3}6$ . The intersection is thus parameterized by the vector-valued function

$\mathbf r(x)=\left\langle x,\dfrac{x^2}2,\dfrac{x^3}6\right\rangle$

where $0\le x\le 4$ . The arc length is computed with the integral

$\displaystyle\int_C\mathrm dS=\int_0^4\|\mathbf r'(x)\|\,\mathrm dx=\int_0^4\sqrt{x^2+\dfrac{x^4}4+\dfrac{x^6}{36}}\,\mathrm dx$

Some rewriting:

$\sqrt{x^2+\dfrac{x^4}4+\dfrac{x^6}{36}}=\sqrt{\dfrac{x^2}{36}}\sqrt{x^4+9x^2+36}=\dfrac x6\sqrt{x^4+9x^2+36}$

Complete the square to get

$x^4+9x^2+36=\left(x^2+\dfrac92\right)^2+\dfrac{63}4$

So in the integral, you can substitute $y=x^2+\dfrac92$ to get

$\displaystyle\frac16\int_0^4x\sqrt{\left(x^2+\frac92\right)^2+\frac{63}4}\,\mathrm dx=\frac1{12}\int_{9/2}^{41/2}\sqrt{y^2+\frac{63}4}\,\mathrm dy$

Next substitute $y=\dfrac{\sqrt{63}}2\tan z$ , so that the integral becomes

$\displaystyle\frac1{12}\int_{9/2}^{41/2}\sqrt{y^2+\frac{63}4}\,\mathrm dy=\frac{21}{16}\int_{\arctan(3/\sqrt7)}^{\arctan(41/(3\sqrt7))}\sec^3z\,\mathrm dz$

This is a fairly standard integral (it even has its own Wiki page, if you're not familiar with the derivation):

$\displaystyle\int\sec^3z\,\mathrm dz=\frac12\sec z\tan z+\frac12\ln|\sec x+\tan x|+C$

So the arc length is

$\displaystyle\frac{21}{32}\left(\sec z\tan z+\ln|\sec x+\tan x|\right)\bigg|_{z=\arctan(3/\sqrt7)}^{z=\arctan(41/(3\sqrt7))}=\frac{21}{32}\ln\left(\frac{41+4\sqrt{109}}{21}\right)+\frac{41\sqrt{109}}{24}-\frac98$

4 0

4 years ago

Can someone help i put a random answer

Gre4nikov [31]

P= 36 units............,

4 0

3 years ago

Read 2 more answers

Translate and solve: 46 less than y is at least -184

natima [27]

Translate: y - 46 > -184

Possible answer:<u> y > - 138</u>

Solution:

y - 46 > -184

= y > - 184 + 46

= y > - 138

#CarryOnLearning

8 0

3 years ago

PLEASE HELP AND SHOW WORK THANKYOU IN ADVANCE BY A MILLION:))))))

Mazyrski [523]

Answer:

C (12+1/2) hours

Step-by-step explanation:

The picture represents the number of hours she spend watching tv for 2 weeks.

One day she watched for 0 hours.

One day she watched for 1/4 hours.

One day she watched for 1/2 hours.

Three days she watched for 3/4 hours.

Three days she watched for 1 hours.

Four days she watched for (1+1/4) hours.

One day she watched for (1+1/2) hours.

-----------------------------------------------------------

So for those two weeks (those 14 days) she watched

0+1/4+1/2+3(3/4)+3(1)+4(1+1/4)+1(1+1/2)

1/4+1/2+9/4+3+[4+1]+[1+1/2]

(1/2+1/2)+(3+4+1+1)+(1/4+9/4)

(1)+(9)+(10/4)

(1+9)+(5/2)

(10)+(5/2)

(20/2)+(5/2)

25/2

(12+1/2)

8 0

3 years ago

What is 3/4 multiplied by 6/8?

Misha Larkins [42]

Answer:

9/16

Step-by-step explanation:

3/4 * 6/8

Simplify the fractions by dividing the top and bottom of the second by 2

3/4 * 3/4

Multiply the numerators

3*3 = 9

Multiply the denominators

4*4 = 16

Numerator over denominator

9/16

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3 years ago