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vagabundo [1.1K]
2 years ago
7

Help finding the sum

Mathematics
1 answer:
Mamont248 [21]2 years ago
5 0
It is asking you to find the sum of k^2 - 1 from k=1 to k=4. Since that is only 4 numbers, calculating the sum by hand wouldn’t be that bad.

(1^2 - 1) + (2^2 - 1) + (3^2 - 1) + (4^2 - 1) = 26

The easier way to find the sum is to use a few simple formulas.

When we have a term that is just a constant c, the formula is c*n.

When we have a variable k, the formula is k*n*(n+1)/2.

When we have a squared variable, the formula is k*n*(n+1)*(2n+1)/6.

In this case, we have a squared variable k^2 and a constant of -1.

So plug in n=4 to the formulas:

4*5*9/6 - 1*4 = 26

The answer is 26







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What is the value of x in the equation 3 minus 2 x = negative 1.5 x? –6 –2 2 6
olga_2 [115]

Answer:

x=118.25

Step-by-step explanation:

3−2x=−1.5−6−226

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4 0
3 years ago
Read 2 more answers
Abigail's mom asked her to go to the store and purchase 5 kilograms (kg) of potatoes. When she got to the store, she saw the pot
Over [174]
1 kilogram = 2.205 pounds

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5 0
3 years ago
PLEASE HELP ASAP!
Mamont248 [21]

Answer:

(x-4)^2+(y+7)^2=43

Step-by-step explanation:

The standard equation of a circle is given by:

(x-h)^2+(y-k)^2=r^2

Where (<em>h, k</em>) is the center and <em>r</em> is the radius.

We are given that the center is (4, -7) and that the radius is √(43).

So, <em>h</em> = 4, <em>k</em> = -7, and <em>r</em> = √(43). Substitute:

(x-(4))^2+(y-(-7))^2=(\sqrt{43})^2

Simplify. Hence, our equation is:

(x-4)^2+(y+7)^2=43

7 0
3 years ago
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