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2 years ago

One positive number is five times another number is the difference between the two numbers is 1676 find the numbers

Mathematics

2 answers:

jonny [76]2 years ago

7 0

The two numbers are X=-419 X=-2095

AleksAgata [21]2 years ago

4 0

Answer:

The two numbers are -419 and -2095.

Step-by-step explanation:

5x=y

x-y=1676

----------------

x-5x=1676

-4x=1676

x=1676/-4

x=-419

5(-419)=y

y=-2095

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2 years ago

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1 year ago

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2 years ago

Which represents a function?

VLD [36.1K]

Answer:

$\boxed{ \text{Option \: D}}$

Step-by-step explanation:

All the relations are not functions. We can determine ( identify ) whether a relation is function or not by drawing a vertical line intersecting the graph of the relation. This is called vertical line test.

If the vertical line intersects the graph of a relation at one point , the relation is a function .
If it cuts at more than one point , it is not a function. It means that if there are more points of the graph of a relation of a vertical line , same first component ( pre - image ) has more images ( second component ) which is not the function by definition.

--------------------------------------------------

Let's check all of the options :

☐ Option A :

The vertical line cuts the graph at two points. So , the graph does not represent a function.

☐ Option B

No! This is also not a function as the vertical line cuts the graph at two points.

☐ Option C

Nah! This too can't be called a function as the vertical line cuts the graph at two points.

☑ Option D

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Yayy!! We found our answer. It's ' Option D '.

Hope I helped ! ツ

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2 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