3 years ago

Round 1,995,298 to the nearest 10,000

Mathematics

2 answers:

harkovskaia [24]3 years ago

6 0

Um... It's actually 2,000,000. Sorry, grammar mistakes or any kind of mistake is my pet peeve. :P

snow_tiger [21]3 years ago

3 0

The correct answer is 2,00,000 because

^is the ten thousands and every thing changes because of that

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worty [1.4K]

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Step-by-step explanation:

4 0

3 years ago

Check down the image that’s give below

gregori [183]

Solution:

$\frac{3x}{ {x}^{2} + 6x + 9} + \frac{x + 3}{ {x}^{2} - 9 }$

In the first fraction, we have to factorise the denominator using (a + b)² = a² + 2ab + b². And in the second fraction, we have to factorise the denominator using a² - b² = (a - b)(a + b) identity.

$= \frac{3x}{ {(x)}^{2} + 2(x)(3) + ( {3)}^{2} } + \frac{x + 3}{(x)^{2} - (3)^{2} } \\ = \frac{3x}{(x + 3) ^{2} } + \frac{x + 3}{(x + 3)(x - 3)}$

From the second fraction, cancel out from both sides (x + 3), the we get:

$= \frac{3x}{(x + 3) ^{2} } + \frac{1}{(x - 3)} \\ = \frac{3x(x - 3) + 1(x + 3) ^{2} }{(x + 3)^{2} (x - 3)} \\ = \frac{3x ^{2} - 9 x+ {x}^{2} + 6x + 9}{ {(x + 3)}^{2} (x - 3)} \\ = \frac{ {4x}^{2} - 3x + 9}{(x + 3)(x + 3)(x - 3)}$