First solve for the trig function 'cot'

Next take the sqrt of both sides (include plus/minus)

Now take reciprocal of both sides, this will change trig function to 'tan'
(cot = 1/tan)

Finally use the unit circle or inverse tan on your calculator to find x.
There will be 4 solutions, one for each quadrant.
Answer:
Option B - 
Step-by-step explanation:
Given : Expression 
To find : Expand each expression ?
Solution :
Using logarithmic properties,

and 
Here, A=4y^5 and B=x^2



Using logarithmic property, 

Therefore, option B is correct.
Answer:
75%
42/56 = 42/56
Convert to decimal form:
42÷56 = 0.75
= 75%
Can you mark as brainliest if its right?
Answer:
The solution is 8=r
Step-by-step explanation:
When the question states, "What is the distance from the edge of the dart board to the exact center of the bullseye (to the nearest inch)," it is basically asking to find the radius rounded to the nearest whole number given the circumference quantity.
Thus, we need to substitute the numbers already given into the radius equation.
53.3=2 x 3.14 x r
Now, divide both sides by 3.14.
16.97=2r
Again, use division property of equality and divide both sides of the equation by 2.
8.4=r ... then rounding to the nearest inch... 8=r
All things said, the solution is 8=r
I hope this helped!!
~Penny

- Find square root of 144 by factorisation method.

We can find the square root of 144 using the factorisation method. In this method, you need to factorise 144 first. Then, you'll get your answer in the form of prime factors. In this case, it's ⇨ 2 × 2 × 2 × 2 × 3 × 3.
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To find the square root using factorisation, we need to group the same number in pairs. That is, 2 × 2 × 2 × 2 × 3 × 3 by grouping same numbers in pairs will become ⇨ (2, 2), (2, 2), (3, 3).
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Now, you should take only 1 of the number from these groups. So, (2, 2), (2, 2), (3, 3) will change to ⇨ 2 × 2 × 3.
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Finally, just multiply this set of numbers to find the square root of 144. 2 × 2 × 3 = 4 × 3 =
⇨ square root of 144.
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- The square root of 144 is <em><u>1</u></em><em><u>2</u></em><em><u>.</u></em>