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2 years ago

Find the sum of odd numbers which are divisible by 3 and lie between 1 and 500

Mathematics

1 answer:

Zolol [24]2 years ago

7 0

Answer:

20,667.

Step-by-step explanation:

This is an arithmetic series:

3 + 9 + 15 + 21 + 27+ ...........+ 495

First term = 3 and common difference = 6.

Number of terms = ( 495 - 3 )/6 + 1

= 83.

So Sn = n/2(a1 + l) where a1 = first term and l = last term

S83 = 83/2(3 + 495)

= 41.5 * 498

= 20,667.

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2 years ago

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liubo4ka [24]

Answer:

$9k^{3} -3k^{2}-29k-12$

Step-by-step explanation:

Step 1: Expand by distributing sum groups.

$3k(3k^{2} -5k-3)+4(3k^{2} -5k-3)$

Step 2: Expand by distributing terms.

$9k^{3} -15k^{2} -9k+4(3k^{2} -5k-3)$

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Step 4: Collect like terms.

$9k^{3}+(-15k^{2} +12k^{2} )+(-9k-20k)-12$

Step 5: Simplify.

$9k^{3} -3k^{2}-29k-12$

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2 years ago

Evaluate the expression for p = -1<br> -38p2 - 78p

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Answer:

$- 38 {p}^{2} - 78p \\ p = - 1 \\ - 38 \times { (- 1)}^{2} - 78 \times - 1 \\ - 38 + 78 \\ = 40 \\ thank \: you$

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2 years ago

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Evaluate: 4+8 divided by 2x (6-3)

Volgvan

Answer:

Step-by-step explanation:

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2 years ago

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