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likoan [24]
3 years ago
15

Which expression is equivalent to *picture attached*

Mathematics
1 answer:
Sedbober [7]3 years ago
7 0

Answer:

\sum_{n=3}^{20}(n(n+1))=\sum _{n=3}^{20} n^2+ \sum _{n=3}^{20} n

Step-by-step explanation:

Given expression : \sum_{n=3}^{20}(n(n+1))

Solving further :

\Rightarrow \sum_{n=3}^{20}(n^2+n)

\Rightarrow \sum _{n=3}^{20} n^2+ \sum _{n=3}^{20} n

So, \sum_{n=3}^{20}(n(n+1))=\sum _{n=3}^{20} n^2+ \sum _{n=3}^{20} n

So, The given expression is equivalent to Option A

So, Option A is the answer

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-2 or -2/1

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2 years ago
Solve the given equation by completing the square.
Dmitriy789 [7]

Answer:

a = -4, b = 3 and c = 6 (OR)

a = -4, b = -3 and c = 6

Step-by-step explanation:

x^{2} +8x = 38

Add 16 to both sides.

x^{2} +8x+16 = 38 + 16

(x+4)^{2} =54

x + 4 = \sqrt{54} or x + 4 = -\sqrt{54}

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Therefore, a = -4, b = 3 and c = 6 (OR)

a = -4, b = -3 and c = 6

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3 years ago
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The eccentricity e of an ellipse is defined as the number c/a, where a is the distance of a vertex from the center and c is the
Anna71 [15]

Answer:

Check below, please.

Step-by-step explanation:

Hi, there!

Since we can describe eccentricity as e=\frac{c}{a}

a) Eccentricity close to 0

An ellipsis with eccentricity whose value is 0, is in fact, a degenerate one almost a circle. An ellipse whose value is close to zero is almost a degenerate circle. The closer the eccentricity comes to zero, the more rounded gets the ellipse just like a circle. (Check picture, please)

\frac{x^2}{a^2} +\frac{y^2}{b^2} =1 \:(Ellipse \:formula)\\a^2=b^2+c^2 \: (Pythagorean\: Theorem)\:a=longer \:axis.\:b=shorter \:axis)\\a^2=b^2+(0)^2 \:(c\:is \:the\: distance \: the\: Foci)\\\\a^2=b^2 \\a=b\: (the \:halves \:of \:each\:axes \:measure \:the \:same)

b) Eccentricity =5

5=\frac{c}{a} \:c=5a

An eccentricity equal to 5 implies that the distance between the Foci has to be five (5) times larger than the half of its longer axis! In this case, there can't be an ellipse since the eccentricity must be between 0 and 1 in other words:

If\:e=\frac{c}{a} \:then\:c>0 , and\: c>0 \:then \:1>e>0

c) Eccentricity close to 1

In this case, the eccentricity close or equal to 1 We must conceive an ellipse whose measure for the half of the longer axis a and the distance between the Foci 'c' they both have the same size.

a=c\\\\a^2=b^2+c^2\:(In \:the\:Pythagorean\:Theorem\: we \:should\:conceive \:b=0)

Then:\\\\a=c\\e=\frac{c}{a}\therefore e=1

7 0
2 years ago
I need help with a question
postnew [5]

Solution

The range is defined as:

Range = Max - Min

For this case we can conclude that the best solution would be:

B, C

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