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vovikov84 [41]
3 years ago
9

Can someone help me do this

Mathematics
1 answer:
Tomtit [17]3 years ago
6 0

Step-by-step explanation:

the angle is greater than 180 degrees so it makes it an obtuse triangle

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which one of the following uses estimation to check 267-154=113? a. 300-200=100. b. 113+154=267. c. 300-150=150. d. 270-150=120
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A. 300-200=100
You round 267 to 300, 154 to 200, and 113 to 100.
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In right triangle ABC shown below, what is the<br> degree measure of ZC ?
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90+30=120

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If a/b &lt; c/d with b &gt; 0, d &gt; 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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3 years ago
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