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3 years ago

How do I solve these?

Mathematics

1 answer:

Illusion [34]3 years ago

8 0

Answer: x= (95/11. 82/11. -24/11)

e= ( 2 4 5 )

( 0 -2 -1/2)

(0 0 -23/4)

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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

2 years ago

Please help, I will mark!!

Alik [6]

Answer:

$d=28.28$

Step-by-step explanation:

To calculate the lenght of the diagonal d across the square, we can assume that the square it is compound of two right triangles. So, we can resolve this exercise using The Pythagorean Theorem.

The Pythagorean theorem states that in every right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the respective lengths of the legs. It is the best-known proposition among those that have their own name in mathematics.

If in a right triangle there are legs of length a and b, and the measure of the hypotenuse is c, then the following relation is fulfilled:

$a^{2} +b^{2} =c^{2}$ a is the height, b is the base, and c is

the hypotenuse.

To obtain the value of the hypotenuse

$c= \sqrt{a^{2} +b^{2} }$

To find the value of the lenght of the diagonal d across the square, we have:

$d=\sqrt{a^{2} +b^{2} }$ Where a = b = 20

Substituting the values

$d=\sqrt{(20)^{2} +(20)^{2} }\\d=\sqrt{400+400} =\sqrt{800} \\d=28.284$

Round the answer to 2 decimal places

$d=28.28$

6 0

3 years ago

Help quickly please!!!

vagabundo [1.1K]

Answer:

1/3 B h

1/3(50.24)(10)

plug in into a calculator

6 0

3 years ago

Ava buys a book for 8.58. She pays with a $10 bill. How much charge will she get

navik [9.2K]

1.42 because you subtract 10 from the price

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3 years ago

Read 2 more answers

What is 100 + 4200 + 50 rounded to the nearest ten thousand

Finger [1]

100+4200 is 4,300 4,300+50 is 4,350 if you rounded that to the nearest 10,000 it's 10,000 Explain cause I think you're missing something from the question

4 0

3 years ago