Hi there!

We can begin by multiplying by its conjugate:

Simplify using the identity:


Take the square root of the expression:

Multiply again by the conjugate to get a SINGLE term in the denominator:

Simplify:

Use the above trig identity one more:

Cancel out sinA:

Split the fraction into two:

Recall:

Simplify:

Answer:

Step-by-step explanation:
Hello!
because if you multiply numbers with exponents you add the exponents.
Now we've got
.
Sadly it has to be in exponential form. You can substitute
into the equation if you know a.
It would be the third choice if ur rotating from point A
<h2>
Answer:</h2>
<u>, "six and one over two ft".</u>
<h2>
Step-by-step explanation:</h2>
Let's considerate the fact that the garden has a <u>square shape</u>.
<h3>1. Finding values of interest.</h3>
Amount of fence that the gardener already has:
ft.
Length of one side:
ft.
If one side measures
ft, and the square garden has 4 sides of equal length, because it's a square, then we must multiply the measure of one side by 4 to find the total length of fence needed:

<h3>
2. How much more does he need?</h3>
The gardener already has
, which equals
. Hence, the difference between the amount needed and the amount that the gardeneralready has will give us the remaining amount required. Let's do that:

<h3>3. Express your result.</h3>

Answer: There are no solutions.
Step-by-step explanation: