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2 years ago

Whats the squareroot of 98

Mathematics

1 answer:

Taya2010 [7]2 years ago

3 0

Answer: 7√2

Step-by-step explanation: To simplify a square root where the number inside the radical is not a perfect square like the square root of 98, we start by making a factor tree for the number inside.

98 factors as 2 · 49 and if you know your perfect squares,

you should be able to recognize 49 as 7 · 7.

What we are looking for in our factor tree

are pairs of factors that are the same.

If a factor pairs up, it will come out of the radical.

If a factor does not pair up, then it stays inside the radical.

So here, since our 7's pair up, a 7 will come out of the radical.

Since the 2 does not pair up, it stays inside the radical.

So our answer is 7√2.

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A polynomial asymptote is a function $p(x)$ such that

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Since this equation defines a hyperbola, we expect the asymptotes to be lines of the form $p(x)=ax+b$ .

Ignore the negative root (we don't need it). If $y=2x-1+2\sqrt{x^2-x}$ , then we want to find constants $a,b$ such that

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since $x\to\infty$ forces us to have $x>0$ . And as $x\to\infty$ , the $\dfrac1x$ term is "negligible", so really $\sqrt{x^2-x}\approx x$ . We can then treat the limand like

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The other asymptote is obtained similarly by examining the limit as $x\to-\infty$ .

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