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viva [34]
3 years ago
6

The sum of two numbers is -14 one number is 20 less than the other find the numbers

Mathematics
1 answer:
Lisa [10]3 years ago
3 0

Answer:

x=-17,y=3

Step-by-step explanation:

x+y=-14

x=x

y=x+20

x+(x+20)=-14

2x+20=-14

2x=-14-20

2x=-34

x=-34/2

x=-17

y=-17+20

y=3

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What is this simplified? Please help
Naya [18.7K]

Answer:

The end result is -1/(x + 1)

Step-by-step explanation:

In order to find the answer to this, we first need to factor the denominator. Since it is a quadratic, we try to find number that multiply to the last term (8) and add to the middle term (9). In this case, the numbers 8 and 1 would work. This allows us to use those numbers in parenthesis along with x as a fully factored form.

x^2 + 9x + 8 = (x + 1)(x + 8)

Now that we have this factored we can take the original equation and factor a -1 out of the top.

(-1)(x + 8)/(x + 1)(x + 8)

Since there is an (x + 8) on the top and bottom, we can cancel those.

-1/(x + 1)

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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.

5 0
2 years ago
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Which u need help on ?
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Can someone please answer this for me i cant figure it out.
Jlenok [28]
<h3>Answer:</h3>

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<h3>Step-by-step explanation:</h3>

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Now, you have ...

... (x^2)^(1/3)

and the rule of exponents tells you to multiply the exponents.

... = x^(2·1/3) = x^(2/3)

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