5 ordered pairs of y=2x+5 please help its due tomorrow

Can you solve it, please

weeeeeb [17]

Answer:20

Step-by-step explanation:

3.1 *4.2=13.02 12.4*21=260.4 260.4/13.02=20

8 0

2 years ago

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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

2 years ago

Which equation is represented by the graph?

Trava [24]

Answer:

y (x+1) this is your answer

3 0

3 years ago

What is an equation of the line that passes through the point (-7,-6)(−7,−6) and is parallel to the line x-y=7x−y=7?

kotykmax [81]

Answer:

y = x + 1

Step-by-step explanation:

The gradient of a line can be defined by the equation:

m (gradient) = (y1 – y2 ) ÷ (x1 – x2) ----> "1" and "2" should be in subscript

For (-7,-6) we use x2 and y2 (because this point can be anywhere along a line):

x2 = -7, y2 = -6

Plug these values into the formula above:

m = (y-(-6)) ÷ (x-(-7))

m = (y+6) ÷ (x+7)

At this stage, the equation can't be solved as there are two unknowns. Therefore, the gradient must be found another way. Two lines are parallel if they have the same gradient - in their y=mx+c equations, m will be equal.

x - y=7 is the line alluded to in the question. Rearranging this equation into the line equation format gives:

y = x-7 ---> The gradient (coefficient of x) is 1.

Therefore, the gradient of the other parallel line must also be 1.

This can be substituted into the previous equation to give:

1 = (y+6)÷(x+7)

x+7 = y+6

x+1 = y

Therefore, the answer is y=x+1

5 0

2 years ago

What value makes the Inequality n > -5

Crank

Answer:

Any numbers above -5 make this inequality true. Values that will make the inequality true: -4, -3, -2, -1, 0, 1...

6 0

3 years ago

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