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Monica [59]
3 years ago
8

Write the sum in sigma notation. 5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 + 50

Mathematics
2 answers:
alisha [4.7K]3 years ago
8 0
The sum in sigma notation for the sequence will be as follows:
From
<span>5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 + 50
first term=5
common difference=5
number of terms=10
n=nth term
thus the sum will be:
(i=2 to 10)</span>∑(5(n-1)+5)
DochEvi [55]3 years ago
5 0

Please mark brainliest

The sum in sigma notation for the sequence as follows:

From

5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 + 50

first term=5

common difference=5

number of terms=10

n=nth term

therefore the total will be:

(i=2 to 10)∑(5(n-1)+5

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How do you simplify, explain<br> -3√7 -3√7 -2√32
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Since the first two numbers have the same radical, they can be combined. 
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Simplify -2</span>√32 to get -8<span>√2.

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8 0
3 years ago
How many and what type of solutions does the equation have?
Natali [406]
Let's solve the equation 2k^2 = 9 + 3k
First, subtract each side by (9+3k) to get 0 on the right side of the equation
2k^2 = 9 + 3k
2k^2 - (9+3k) = 9+3k - (9+3k)
2k^2 - 9 - 3k = 9 + 3k - 9 - 3k
2k^2 - 3k - 9 = 0

As you see, we got a quadratic equation of general form ax^2 + bx + c, in which a = 2, b= -3, and c = -9.
Δ = b^2 - 4ac
Δ = (-3)^2 - 4 (2)(-9)
Δ<u /> = 9 + 72
Δ<u /> = 81

Δ<u />>0 so the equation got 2 real solutions:
k = (-b + √Δ)/2a = (-(-3) + √<u />81) / 2*2 = (3+9)/4 = 12/4 = 3
AND
k = (-b -√Δ)/2a = (-(-3) - √<u />81)/2*2 = (3-9)/4 = -6/4 = -3/2

So the solutions to 2k^2 = 9+3k are k=3 and k=-3/2

A rational number is either an integer number, or a decimal number that got a definitive number of digits after the decimal point.

3 is an integer number, so it's rational.
-3/2 = -1.5, and -1.5 got a definitive number of digit after the decimal point, so it's rational.

So 2k^2 = 9 + 3k have two rational solutions (Option B).

Hope this Helps! :)
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