![\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
Answer:
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Answer:
See explanation
Step-by-step explanation:
Assuming the given inequality is 
Then the corresponding linear equation is 
When x=0, we have 

When y=0, we have 

The T-table is:
<u>x | y</u>
0 | 4
6 | 12
We plot this points and draw a solid straight line as shown in the attachment.
Now let us test the origin: (0,0) by plugging x=0 and y=0 into the inequality.

....This is true so we shade the lower half plane as shown in the attachment.
Answer:
p=2
Step-by-step explanation:
4.05p+14.40=4.50(p+3) < equation
4.05p+14.40=4.50p+13.50 < multiply
14.40=.45p+13.50 < subtract
.9=.45p < subtract
2=p < divide
SA: 2pi (r^2)+2pi(r)(h)
In order to find radius (r) of the base you must divide 10 by pi then find the square root of the result
√(10/pi)=r: 1.784cm
h: a) 8.0cm
b) 6.5cm
c) 9.4cm
2pi (3.183)+2pi(1.784)(8)=109.673cm^2
2pi (3.183)+2pi(1.784)(6.5)=92.859cm^2
2pi (3.183)+2pi(1.784)(9.4)=125.366cm^2