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3 years ago

What is the next term of the geometric sequence? 4/9,-4/3,4,...

Mathematics

2 answers:

Wittaler [7]3 years ago

6 0

Answer: 9 is the answer

Step-by-step explanation:

Gnesinka [82]3 years ago

3 0

Answer:

- 12

Step-by-step explanation:

This is a geometric sequence since there is a common ratio between each term. In this case, multiplying the previous term in the sequence by

−

gives the next term. In other words,

⋅

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Geometric Sequence:

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This is the form of a geometric sequence.

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Substitute in the values of

and

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(

)

⋅

(

−

)

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Combine

and

(

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)

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(

−

)

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Substitute in the value of

to find the

th term.

(

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)

(

)

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Simplify the numerator.

Tap for more steps...

⋅

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Simplify the expression.

Tap for more steps...

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HA<br> 1) Simplify the expression.<br> 9 - 5*4 + 3

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Answer:

Step-by-step explanation:

9-5*4+3

9-20+3

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= -8

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Answer:

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A bakery sold a total of 400 cupcakes in a day, and 184 of them were vanilla flavored. What percentage of cupcakes sold that day

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Answer:

Step-by-step explanation:

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3 years ago

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which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago

Which of the lines in the picture is parallel to line l? Explain how you know.

Nadya [2.5K]

Answer:

line k is parallel

Step-by-step explanation:

because they will never intersect

5 0

3 years ago

What is the next term of the geometric sequence? 4/9,-4/3,4,... ​

What is the next term of the geometric sequence? 4/9,-4/3,4,...