![\bf tan(x^o)=1.11\impliedby \textit{taking }tan^{-1}\textit{ to both sides} \\\\\\ tan^{-1}[tan(x^o)]=tan^{-1}(1.11)\implies \measuredangle x=tan^{-1}(1.11)](https://tex.z-dn.net/?f=%5Cbf%20tan%28x%5Eo%29%3D1.11%5Cimpliedby%20%5Ctextit%7Btaking%20%7Dtan%5E%7B-1%7D%5Ctextit%7B%20to%20both%20sides%7D%0A%5C%5C%5C%5C%5C%5C%0Atan%5E%7B-1%7D%5Btan%28x%5Eo%29%5D%3Dtan%5E%7B-1%7D%281.11%29%5Cimplies%20%5Cmeasuredangle%20x%3Dtan%5E%7B-1%7D%281.11%29)
plug that in your calculator, make sure the calculator is in Degree mode
18 birds are left then regroup a ten 28 I hope it’ helps
Answer:
Sue travels for 3 hours and a half
Sue stays stationary for 2 hours and a half
Step-by-step explanation:
Everytime the curve climbs, it means Sue is travelling. Everytime the curve stays flat, it means Sue is stays stationary. It remains to count up the total for both categories, with one square being 30 minutes.
<h3>
Answer: 12.5 (choice C)</h3>
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We apply the pythagorean theorem to find this answer.
a = 11 and b = 6 are the given legs
c = unknown hypotenuse
So,
a^2+b^2 = c^2
c = sqrt( a^2+b^2 )
c = sqrt( 11^2 + 6^2 )
c = sqrt( 121 + 36 )
c = sqrt( 157 )
c = 12.52996 approximately
c = 12.5
Side note: once you replace 'a' and b with 11 and 6, you can compute everything with a calculator in one single step more or less. The steps above are shown if you wanted to find the exact value sqrt(157).
<u>Part 1) </u>To find the measure of ∠A in ∆ABC, use
we know that
In the triangle ABC
Applying the law of sines
in this problem we have
therefore
<u>the answer Part 1) is</u>
Law of Sines
<u>Part 2) </u>To find the length of side HI in ∆HIG, use
we know that
In the triangle HIG
Applying the law of cosines
In this problem we have
g=HI
G=angle Beta
substitute
therefore
<u>the answer Part 2) is</u>
Law of Cosines
