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4 years ago

14

What is the limit of the infinite series?

1 answer:

Ipatiy [6.2K]4 years ago

7 0

Answer:

does not exist

Step-by-step explanation:

According to the divergence test, for an infinite series ∑ aₙ, the series diverges if lim(n→∞) aₙ ≠ 0.

Applying the divergence test here:

lim(n→∞) (3n⁵ / (4n⁵ + 1)) = 3/4

This is not 0, so the series does not converge, meaning the sum does not exist.

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Last one I promise peeps

AleksandrR [38]

Answer:

Ello mate, the answer to your question/problem is "J"

Step-by-step explanation: Look at image provided below :)

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3 years ago

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What is the graph of the equation? y=-2

Marrrta [24]

Your answer would look like this

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3 years ago

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Consider the quadratic function f(x)=−x^2+x+30

yaroslaw [1]

Answer:

The smallest xx-intercept is $x = -5$

The largest xx-intercept is $x = 6$

The yy-intercept is $y = 30$ .

Step-by-step explanation:

Given a quadratic function in the following format:

$f(x) = ax^{2} + bx + c = 0, a \neq 0$

The x-values of the x-intercepts are $x_{1}, x_{2}$ , given by the following formulas.

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

We have that:

$f(x) = -x^{2} + x + 30$

This is not in the format above. I will multiply by (-1), so we have:

$f(x) = x^{2} - x - 30$

So, $a = 1, b = -1, c = -30$

$\bigtriangleup = b^{2} - 4ac = (-1)^{2} - 4*(1)*(-30) = 1 + 120 = 121$

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a} = \frac{-(-1) + \sqrt{121}}{2(1)} = \frac{12}{2} = 6$

$x_{2} = \frac{-b + \sqrt{\bigtriangleup}}{2*a} = \frac{-(-1) - \sqrt{121}}{2(1)} = \frac{-10}{2} = -5$

This means that

The smallest xx-intercept is $x = -5$

The largest xx-intercept is $x = 6$

The y-intercept is the value of f(x) when x = 0. So

$f(x)=−x^{2}+x+30$

$f(0) = 30$

The yy-intercept is $y = 30$ .

5 0

3 years ago

Order of Operations

emmasim [6.3K]

Given:

$4 + 9 + 16 \div 4 - 8 -3 \times 5$

To find:

The number of possible solutions, if the Order of Operations did not exist.

Solution:

We have,

$4 + 9 + 16 \div 4 - 8 -3 \times 5$

Now,

Case 1:

$[4 + 9 + 16] \div [4 - 8 -(3 \times 5)]=29\div (4 - 8 -15)$

$4 + 9 + 16 \div 4 - 8 -(3 \times 5)=-\dfrac{29}{19}$

Case 2:

$[4 + 9 + 16] \div (4 - 8 -3) \times 5=29\div (-7\times 5)$

$4 + 9 + 16 \div (4 - 8 -3) \times 5=-\dfrac{29}{35}$

Case 3:

$4 + 9 + (16 \div 4) - 8 -(3 \times 5)=4+9+4-8-15$

$4 + 9 + (16 \div 4) - 8 -(3 \times 5)=-6$

Many more possibilities are there.

Therefore, there are more than 3 possible solutions.

6 0

4 years ago

Marcel received an award for community service. The award included a check for $2,265. Three businesses each contributed the sam

natima [27]

$755 each this is so simple you just divide 2265 by 3.

8 0

3 years ago

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