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Bad White [126]

4 years ago

8

29849 (3 sig. Fig )

1 answer:

Slav-nsk [51]4 years ago

6 0

3 sig fig means 3 significant figures. This means that only the first 3 digits in a number are above 0.
The first 3 digits in 29849 are 298 and, because 4 is less than 5, the 8 is not round up, and remains the same. The 4 and the 9, however, become 0.

<span>The answer is 29800</span>

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2. Find the product (2x + 3)(3x - 2).

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If a/b < c/d with b > 0, d > 0, prove that a+c / b+d lies between the two fractions a/b and c/d .

Tems11 [23]

Divide through everything by <em>b</em> :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ab + \dfrac cb}{1 + \dfrac db}$

Since <em>a/b</em> < <em>c/d</em>, it follows that

$\dfrac{a+c}{b+d} < \dfrac{\dfrac cd+\dfrac cb}{1 + \dfrac db}$

Multiply through everything on the right side by <em>b/d</em> to get

$\dfrac{a+c}{b+d} < \dfrac{\dfrac{bc}{d^2}+\dfrac cd}{\dfrac bd+1} = \dfrac{\dfrac cd\left(\dfrac bd+1\right)}{\dfrac bd+1} = \dfrac cd$

and so (<em>a</em> + <em>c</em>)/(<em>b</em> + <em>d</em>) < <em>c/d</em>.

For the other side, you can do something similar and divide through everything by <em>d</em> :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ad + \dfrac cd}{\dfrac bd + 1}$

and <em>a/b</em> < <em>c/d</em> tells us that

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ad + \dfrac ab}{\dfrac bd + 1}$

Then

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ab + \dfrac{ad}{b^2}}{1 + \dfrac db} = \dfrac{\dfrac ab\left(1+\dfrac db\right)}{1 + \dfrac db} = \dfrac ab$

and so (<em>a</em> + <em>c</em>)/(<em>b</em> + <em>d</em>) > <em>a/b</em>.

Then together we get the desired inequality.

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