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3 years ago

Determine 3rd term in the geometric sequence whose first term is -8 and whose common ratio is 6

Mathematics

2 answers:

k0ka [10]3 years ago

8 0

it would be -288. Correct answer

Brrunno [24]3 years ago

3 0

-8, -48, -288
or
-8, -4/3, -2/9

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((81x^3 y^(-1/3))^1/4)/(x^2 y)^1/4 simplify

malfutka [58]

$\bf a^{\frac{{ n}}{{ m}}} \implies \sqrt[{ m}]{a^{ n}} \qquad \qquad
\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}
\\\\\\
\left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}\\\\
-------------------------------\\\\$

$\bf \cfrac{\left( 81x^3y^{-\frac{1}{3}} \right)^{\frac{1}{4}}}{(x^2y)^{\frac{1}{4}}}\implies \cfrac{81^{\frac{1}{4}}x^{3\cdot {\frac{1}{4}}}y^{-\frac{1}{3}\cdot {\frac{1}{4}}}}{x^{2\cdot {\frac{1}{4}}}y^{\frac{1}{4}}}\implies
\cfrac{81^{\frac{1}{4}}x^{\frac{3}{4}}y^{-\frac{1}{12}}}{x^{\frac{2}{4}} y^{\frac{1}{4}}}$

$\bf \cfrac{81^{\frac{1}{4}}x^{\frac{3}{4}}x^{-\frac{2}{4}}}{ y^{\frac{1}{4}}y^{\frac{1}{12}}}\implies \cfrac{(3^4)^{\frac{1}{4}}x^{\frac{3}{4}-\frac{2}{4}}}{y^{\frac{1}{4}+\frac{1}{12}}}\implies \cfrac{3^{4\cdot \frac{1}{4}}x^{\frac{1}{4}}}{y^{\frac{4}{12}}}\implies \cfrac{3^{\frac{4}{4}}x^{\frac{1}{4}}}{y^{\frac{1}{3}}}
\\\\\\
\cfrac{3\sqrt[4]{x}}{\sqrt[3]{y}}$

7 0

3 years ago

S is between B and C. True or false

Brut [27]

False because I looked it up online

8 0

3 years ago

Read 2 more answers

Let z=3+i, then find a. Z² b. |Z| c.<img src="https://tex.z-dn.net/?f=%5Csqrt%7BZ%7D" id="TexFormula1" title="\sq

zysi [14]

Given z = 3 + i, right away we can find

(a) square

z ² = (3 + i )² = 3² + 6i + i ² = 9 + 6i - 1 = 8 + 6i

(b) modulus

|z| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(z) = arctan(1/3)

Then

z = |z| exp(i arg(z))

z = √10 exp(i arctan(1/3))

z = √10 (cos(arctan(1/3)) + i sin(arctan(1/3))

Any complex number has 2 square roots. Using the polar form from part (d), we have

√z = √(√10) exp(i arctan(1/3) / 2)

and

√z = √(√10) exp(i (arctan(1/3) + 2π) / 2)

Then in standard rectangular form, we have

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right)\right)$

and

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)$

We can simplify this further. We know that z lies in the first quadrant, so

0 < arg(z) = arctan(1/3) < π/2

which means

0 < 1/2 arctan(1/3) < π/4

Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

and since cos(x + π) = -cos(x) and sin(x + π) = -sin(x),

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

Now, arctan(1/3) is an angle y such that tan(y) = 1/3. In a right triangle satisfying this relation, we would see that cos(y) = 3/√10 and sin(y) = 1/√10. Then

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

So the two square roots of z are

$\boxed{\sqrt z = \sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

and

$\boxed{\sqrt z = -\sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

3 0

3 years ago

Read 2 more answers

John leaves school to go home.his bus drives 6 kilometers north and then goes 7 kilometers west.how far is John's house from the

otez555 [7]

Answer:

John is 9.21 km form the school.

Step-by-step explanation:

John leaves school to go home. His bus drives 6 kilometres north and then goes 7 kilometres west. It is required to find John's distance from the school. It is equal to the shortest path covered or its displacement. So,

$d=\sqrt{6^2+7^2} \\\\d=9.21\ km$

So, John is 9.21 km form the school.

5 0

3 years ago

Anyone know the answer to this?

Kaylis [27]

I think its a

because the line segments connect each other.

3 0

3 years ago