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3 years ago

Apply root test to the series.

Mathematics

1 answer:

Angelina_Jolie [31]3 years ago

8 0

$\displaystyle\lim_{n\to\infty}\sqrt[n]{\left|\left(\frac{5n+15}{2n-1}\right)^n\right|}=\lim_{n\to\infty}\frac{5n+15}{2n-1}=\dfrac52$

Since this limit exceeds 1, the series diverges.

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Dafna1 [17]

Answer number 3

Step-by-step explanation:

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3 years ago

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Billie solved the equation below by completing the square, but she got the incorrect solution. In which step did Billie first ma

Tanzania [10]

Answer:

step 2

Step-by-step explanation:

we have

$x^{2} +10x-49=0$ ---> given problem

step 1

Move the constant to the right side

$x^{2} +10x=49$

the step 1 is correct

step 2

Complete the square

$x^{2} +10x+5^2=49+5^2$

$x^{2} +10x+5^2=74$

The step 2 is not correct

step 3

Rewrite as perfect squares

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step 4

take square root both sides

$x+5=\pm\sqrt{74}$

step 5

Find the values of x

$x=-5\pm\sqrt{74}$

$x=-5+\sqrt{74}$

$x=-5-\sqrt{74}$

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4 years ago

[3x^(2/3)] • [7x^(1/4)] photo of equation above

aalyn [17]

$\bf a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n}
\qquad \qquad
\sqrt[ m]{a^ n}\implies a^{\frac{ n}{ m}}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
\left( 3x^{\frac{2}{3}} \right)\left( 7x^{\frac{1}{4}} \right)\implies 3\cdot 7x^{\frac{2}{3}}\cdot x^{\frac{1}{4}}\implies 21x^{\frac{2}{3}+\frac{1}{4}}\implies 21x^{\frac{(4)2+(3)1}{12}}
\\\\\\
21x^{\frac{11}{12}}\implies 21\sqrt[12]{x^{11}}$

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4 years ago

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Aliun [14]

I think your answer will be A. Aleta can buy 10 tops if she only buys 2 pair of pants.

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3 years ago

23. Starting with x=0.81818181 , express x as a fraction in simplest form.

White raven [17]

You could do x=81 with the little line on the top (represents the number going on forever, and example is 0.818181818181818181818181818818181818 etc.)

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3 years ago

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