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quester [9]
3 years ago
9

I have to round an answer to the nearest tenth. My original answer is 15.98. Do I go to 16, or 15.9? Please answer soon.

Mathematics
1 answer:
puteri [66]3 years ago
3 0

Answer:

16

Step-by-step explanation: becuase you have to round it think to urself is it closer to 15.9 or 16

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PLZ HELP Word Problems on the system of equations:
ahrayia [7]

Answer:

let the two number be x and y then,

x+y=12

x-y=-4

-----------

2x=8

or, x = 4 put the value of x in first equation

4+y= 12

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or, y= 8 hence the two number is 4 and 8

8 0
3 years ago
Type the correct answer in the box use/ for the fraction bar
Zarrin [17]

Answer:

see explanation

Step-by-step explanation:

given the 2 equations

y = x² - 2x - 19 → (1)

y + 4x = 5 → (2)

substitute y = x² - 2x - 19 into (2)

x² - 2x - 19 + 4x = 5 ( subtract 5 from both sides )

x² + 2x - 24 = 0 ← in standard form

(x + 6)(x - 4) = 0 ← in factored form

equate each factor to zero and solve for x

x + 6 = 0 ⇒ x = - 6

x - 4 = 0 ⇒ x = 4

substitute each value of x into (1) for corresponding y- coordinate

x = - 6 : y = (- 6)² - 2(- 6) - 19 = 36 + 12 - 19 = 29 ⇒ (- 6, 29)

x = 4 : y = 4² - 2(4) - 19 = 16 - 8 - 19 = - 11 ⇒ (4, - 11)

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6 0
3 years ago
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.

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I think it would be 7.5!!

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