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shepuryov [24]
3 years ago
7

The measure of C if a=6 and b=9.

Mathematics
1 answer:
kirill115 [55]3 years ago
3 0

Answer:

c=3\sqrt{13}

Step-by-step explanation:

Pythagorean Theorem: In the right triangle the square of longer side is equal to the sum of the squares of other sides.

c^2=a^2+b^2\\\\c^2=6^2+9^2\\\\c^2=36+81\\\\c^2=117\\\\c^2=3\times 3\times 13\\\\c=\sqrt{3\times 3\times 13}\\\\c=3\sqrt{13}

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Help pleasee I don't know how to do this
andreyandreev [35.5K]

Answer:

Just factorize the given plynomials.

Step-by-step explanation:

like in first example 7x+49

7(x+7)

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2 years ago
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Zinaida [17]

Answer: 43.988

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3 years ago
The sequence$$1,2,1,2,2,1,2,2,2,1,2,2,2,2,1,2,2,2,2,2,1,2,\dots$$consists of $1$'s separated by blocks of $2$'s with $n$ $2$'s i
kicyunya [14]

Consider the lengths of consecutive 1-2 blocks.

block 1 - 1, 2 - length 2

block 2 - 1, 2, 2 - length 3

block 3 - 1, 2, 2, 2 - length 4

block 4 - 1, 2, 2, 2, 2 - length 5

and so on.


Recall the formula for the sum of consecutive positive integers,

\displaystyle \sum_{i=1}^j i = 1 + 2 + 3 + \cdots + j = \frac{j(j+1)}2 \implies \sum_{i=2}^j = \frac{j(j+1) - 2}2

Now,

1234 = \dfrac{j(j+1)-2}2 \implies 2470 = j(j+1) \implies j\approx49.2016

which means that the 1234th term in the sequence occurs somewhere about 1/5 of the way through the 49th 1-2 block.

In the first 48 blocks, the sequence contains 48 copies of 1 and 1 + 2 + 3 + ... + 47 copies of 2, hence they make up a total of

\displaystyle \sum_{i=1}^48 1 + \sum_{i=1}^{48} i = 48+\frac{48(48+1)}2 = 1224

numbers, and their sum is

\displaystyle \sum_{i=1}^{48} 1 + \sum_{i=1}^{48} 2i = 48 + 48(48+1) = 48\times50 = 2400

This leaves us with the contribution of the first 10 terms in the 49th block, which consist of one 1 and nine 2s with a sum of 1+9\times2=19.

So, the sum of the first 1234 terms in the sequence is 2419.

8 0
2 years ago
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goldenfox [79]

Answer:

See below.

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