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Y_Kistochka [10]
3 years ago
6

A bar of metal is cooling from 1000°C to room temperature, 22°C. The temperature, H, of the bar t minutes after it starts coolin

g is given, in °C, by H = 22 + 978e−0.1t.
Mathematics
1 answer:
NemiM [27]3 years ago
8 0
<span>Answer:
       I assume you're looking for a time-weighed average rather than just the [H(0) + H(60)]/2 average.
       Havg = (1/60)â«[20+980e^(-0.1t)]dt (evaluated from 0 to 60)
       Havg = (1/60)[20t - 98e^(-0.1t)] evaluated from 0 to 60
       Havg = (1/60)[20(60) - 98e^(-6) - 20(0) + 98e^(0)] = 1297.76/60 = 21.629 °C</span>
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8 0
3 years ago
Please answer correctly !!!!!!!!!!!!!!!! Will mark brainliest !!!!!!!!!!!!!!!!!!!!
bija089 [108]

Answer:

14

Step-by-step explanation:

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5 0
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Answer:

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Given: ∠N≅∠S, line l bisects TR at Q.

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From  ΔNQT and ΔSQR

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Hence proved.

Statement                                                 Reason

1. ∠N≅∠S                                                    given

2. ∠NQT≅∠SQR                            Vertical angles are congruent

3.  line l bisects TR at Q.                            given

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5. ΔNQT≅ΔSQR                           AAS theorem

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