If the temperature after 4.4 degree drop was -2.5 degrees Celsius what was the temperature at noon
1 answer:
Answer:
1.9°C
Step-by-step explanation:
Let X be the temperature at noon and -2.5°C be the current temperature after the drop.
-We calculate temperature at noon as:
Hence, the temperature at noon was 1.9°C
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Answer:
<h3>Given</h3>
m∠REG = 78° mAR = 46° ER ≅ GA <h3>Solution</h3>
m∠GAR = 180° - m∠REG = 180° - 78° = 102° (supplementary angles sum to 180°) m∠TAR = 1/2mAR = 1/2(46°) = 23° (tangent chord angle is half the size of intercepted arc) m∠GAN = 180° - (m∠TAR + m∠GAR) = 180° - (23° + 102°) = 55° (straight angle is 180°) mAG = 2m∠GAN = 2(55°) = 110° mRE = mAG = 110° (as ER ≅ GA) mGE = 360° - (mAG + mAR + mRE) = 360° - (110° + 46° + 110°) = 94° (full circle is 360°)
Step-by-step explanation:
Clue: (I'm not giving the answer this is just clues) The word "absolute" helps you understand not to estimate so you want to get the RIGHT answer not the estimation.