<img src="https://tex.z-dn.net/?f=%7Cx-1%7C%3D%5Csqrt%7Bx%7D" id="TexFormula1" title="|x-1|=\sqrt{x}" alt="|x-1|=\sqrt{x}" align

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kenny6666 [7]

3 years ago

="absmiddle" class="latex-formula">

Mathematics

1 answer:

vodka [1.7K]3 years ago

7 0

Isolate the radical, then raise each side of the equation to the power of its index.
Exact Form:
x
=
3
+
√
5
2
,
3
−
√
5
2
x
=
3
+
5
2
,
3
-
5
2

Decimal Form:
x
=
2.61803398
…
,
0.38196601
…

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Karen will divide her garden into equal parts. She will plant corn in __8 12 of the garden. What is the fewest number of parts s

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Answer:

Step-by-step explanation:

Given that Karen will plant corn on 8 / 12 of his garden ;

To obtain the fewest number of parts garden can be divided into ;

Reduce the fraction of corn area to its lowest form ; the denominator gives the whole division of the garden and numerator gives the size of area used for corn with respect to the entire division.

8 /1 2 = 2 /3 (lowest term)

Number of parts garden is divided into = 3

Number of parts for corn ; 2 of the 3 divisions

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Joan is buying airline tickets for her family. Each airline ticket costs $232 with a baggage fee of $13. The airline charges a b

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Use the Ratio Test to determine the convergence or divergence of the series. If the Ratio Test is inconclusive, determine the co

jeka57 [31]

Answer:

<h2>A. The series CONVERGES</h2>

Step-by-step explanation:

If $\sum a_n$ is a series, for the series to converge/diverge according to ratio test, the following conditions must be met.

$\lim_{n \to \infty} |\frac{a_n_+_1}{a_n}| = \rho$

If $\rho$ < 1, the series converges absolutely

If $\rho > 1$ , the series diverges

If $\rho = 1$ , the test fails.

Given the series $\sum\left\ {\infty} \atop {1} \right \frac{n^2}{5^n}$

To test for convergence or divergence using ratio test, we will use the condition above.

$a_n = \frac{n^2}{5^n} \\a_n_+_1 = \frac{(n+1)^2}{5^{n+1}}$

$\frac{a_n_+_1}{a_n} = \frac{{\frac{(n+1)^2}{5^{n+1}}}}{\frac{n^2}{5^n} }\\\\ \frac{a_n_+_1}{a_n} = {{\frac{(n+1)^2}{5^{n+1}} * \frac{5^n}{n^2}\$

$\frac{a_n_+_1}{a_n} = {{\frac{(n^2+2n+1)}{5^n*5^1}} * \frac{5^n}{n^2}\\$

aₙ₊₁/aₙ =

$\lim_{n \to \infty} |\frac{ n^2+2n+1}{5n^2}| \\\\Dividing\ through\ by \ n^2\\\\\lim_{n \to \infty} |\frac{ n^2/n^2+2n/n^2+1/n^2}{5n^2/n^2}|\\\\\lim_{n \to \infty} |\frac{1+2/n+1/n^2}{5}|\\\\$

note that any constant dividing infinity is equal to zero

$|\frac{1+2/\infty+1/\infty^2}{5}|\\\\$

$\frac{1+0+0}{5}\\ = 1/5$

$\rho = 1/5$

Since The limit of the sequence given is less than 1, hence the series converges.

5 0

3 years ago