In this problem, you apply principles in trigonometry. Since it is not mentioned, you will not assume that the triangle is a special triangle such as the right triangle. Hence, you cannot use Pythagorean formulas. The only equations you can use is the Law of Sines and Law of Cosines.
For finding side a, you can answer this easily by the Law of Cosines. The equation is
a2=b2 +c2 -2bccosA
a2 = 11^2 + 8^2 -2(11)(8)(cos54)
a2 = 81.55
a = √81.55
a = 9
Then, we use the Law of Sines to find angles B and C. The formula would be
a/sinA = b/sinB = c/sinC
9/sin54° = 11/sinB
B = 80.4°
9/sin54° = 8/sinC
C = 45.6°
The answer would be: a ≈ 9, C ≈ 45.6, B ≈ 80.4
C. non linear this will be the answer
Ok so
x - the smallest
The other 4 are : x+1, x+2, x+3, x+4
We know that sum is 120
So
x+x+1+x+2+x+3+x+4=120
Combine like terms
5x+10=120
Subtract 10 on both sides
5x=120-10
5x=110
Divide by 5 on both sides
X=22
The third number is x+2 = 22+2=24
<span>If you plug in 0, you get the indeterminate form 0/0. You can, therefore, apply L'Hopital's Rule to get the limit as h approaches 0 of e^(2+h),
which is just e^2.
</span><span><span><span>[e^(<span>2+h) </span></span>− <span>e^2]/</span></span>h </span>= [<span><span><span>e^2</span>(<span>e^h</span>−1)]/</span>h
</span><span>so in the limit, as h goes to 0, you'll notice that the numerator and denominator each go to zero (e^h goes to 1, and so e^h-1 goes to zero). This means the form is 'indeterminate' (here, 0/0), so we may use L'Hoptial's rule:
</span><span>
=<span>e^2</span></span>
---|--/--/--|--/--/--|--/--/--|--/--/--|--/--/--|--/--/--|--/--/--|--/--/--|--/--/--|--/--/--|--------------
-5 -4-11/3-3 -2 -1 0 1 2 8/3 3 4 5
8/3≈ 2,6
-11/3 ≈-3,6
