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2 years ago

-5 > -5 - 8w algebra answer

Mathematics

2 answers:

Gennadij [26K]2 years ago

5 0

Answer:

w > 0 --> solution

wariber [46]2 years ago

4 0

Answer:

w > 0

Step-by-step explanation:

-5 > -5 - 8w

+5 +5

--------------------

0 > -8w

/-8 /-8

--------------------

0 > w

w > 0

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How can i prove this property to be true for all values of n, using mathematical induction.

chubhunter [2.5K]

Proof -

So, in the first part we'll verify by taking n = 1.

$\implies \: 1 = {1}^{2} = \frac{1(1 + 1)(2 + 1)}{6}$

$\implies{ \frac{1(2)(3)}{6} }$

$\implies{ 1}$

Therefore, it is true for the first part.

In the second part we will assume that,

$\: { {1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} = \frac{k(k + 1)(2k + 1)}{6} }$

and we will prove that,

$\sf{ \: { {1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k + 1)(k + 1 + 1) \{2(k + 1) + 1\}}{6}}}$

$\: {{1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k + 1)(k + 2) (2k + 3)}{6}}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{k (k + 1) (2k + 1) }{6} + \frac{(k + 1) ^{2} }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{k(k+1)(2k+1)+6(k+1)^ 2 }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)\{k(2k+1)+6(k+1)\} }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)(2k^2 +k+6k+6) }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)(2k^2+7k+6) }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)(k+2)(2k+3) }{6}$

Henceforth, by using the principle of mathematical induction 1²+2² +3²+....+n² = n(n+1)(2n+1)/ 6 for all positive integers n.

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8 0

1 year ago

Please help to 10. Thank you

lubasha [3.4K]

Answer: i dont know is the ans

Step-by-step explanation:

7 0

2 years ago

Help please and thanks will mark as brainliest

algol13

11)
3x = 180
x = 60

12)

6x + 18 = 180
6x = 162
x = 27

3 0

2 years ago

A candy store owner wants to mix some candy costing $1.25 a pound

maria [59]

The candy store owner should use 37.5 pounds of the candy costing $1.25 a pound.

Given:

Candy costing $1.25 a pound is to be mixed with candy costing $1.45 a pound
The resulting mixture should be 50 pounds of candy
The resulting mixture should cost $1.30 a pound

To find: The amount of candy costing $1.25 a pound that should be mixed

Let us assume that the resulting mixture should be made by mixing 'x' pounds of candy costing $1.25 a pound.

Since the total weight of the resulting mixture should be 50 pounds, 'x' pounds of candy costing $1.25 a pound should be mixed with ' $50-x$ ' pounds of candy costing $1.45 a pound.

Then, the resulting mixture contains 'x' pounds of candy costing $1.25 a pound and ' $50-x$ ' pounds of candy costing $1.45 a pound.

Accordingly, the total cost of the resulting mixture is $1.25x+1.45(50-x)$

However, the resulting mixture should be 50 pounds and should cost $1.30 a pound. Accordingly, the total cost of the resulting mixture is $1.30 \times 50$