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2 years ago

Find the 107 term of the sequence -9, -5, -1, 3, 7, ....

Mathematics

1 answer:

Alex17521 [72]2 years ago

5 0

the sequence is 4n-13, so substitute 107 into n and it will become 4(107)-13 =415

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Find the value of -8 - 12 - (-20). 24 -40 16 0

TiliK225 [7]

Answer:

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you have to use the steps of maths

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2 years ago

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*The sum of two numbers is 400. If the first number is decreased by 20% and the second number is decreased by 15%, then the sum

Dahasolnce [82]

Answer:

The two numbers are .8*160=128 and .85*240=204

Step-by-step explanation:

First sentence: x+y=400

Second sentence .8x+.85y=400-68

Solve y in the first sentence: y=400-x

Plug first into second: .8x+.85(400-x)=332

Distribute: .8x+.85(400)-.85x=332

Combine like terms: -.05x+.85(400)=332

Simplify(multiply): -.05x+ 340=332

Subtract 340 on both sides: -.05x =332-340

Simplify(subtract): -.05x =-8

Divide both sides by -.05: x =-8/-.05

Simplify (division): x = 160

So y=400-x=400-160=240

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2 years ago

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What's the least common factor of 45 and 18?

maw [93]

Factor them
45=3*3*5
18=2*3*3

lcf=2*3*3*5=90

5 0

3 years ago

Multiply or divide, write your answer in scientific notation.

Sidana [21]

1.44/2.4 • 10^10/10^2
= 6 • 10^7 in scientific notation

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2 years ago

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