3)

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6)

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7)

Given:
Consider the below figure attached with this question.
To find:
The perimeter of the polygon UVWXYZ.
Solution:
All sides and vertical and horizontal. So, we can easily find the side lengths of the given polygon by counting the boxes between two points.
From the figure it is clear that,
UV = 9 units
VW = 2 units
WX = 3 units
XY = 3 units
YZ = 6 units
ZU = 5 units
Now, perimeter is the sum of all the sides.



Therefore, the perimeter of UVWXYZ is 28 units.
By "which is an identity" they just mean "which trigonometric equation is true?"
What you have to do is take one of these and sort it out to an identity you know is true, or...
*FYI: You can always test identites like this:
Use the short angle of a 3-4-5 triangle, which would have these trig ratios:
sinx = 3/5 cscx = 5/3
cosx = 4/5 secx = 5/4
tanx = 4/3 cotx = 3/4
Then just plug them in and see if it works. If it doesn't, it can't be an identity!
Let's start with c, just because it seems obvious.
The Pythagorean identity states that sin²x + cos²x = 1, so this same statement with a minus is obviously not true.
Next would be d. csc²x + cot²x = 1 is not true because of a similar Pythagorean identity 1 + cot²x = csc²x. (if you need help remembering these identites, do yourslef a favor and search up the Magic Hexagon.)
Next is b. Here we have (cscx + cotx)² = 1. Let's take the square root of each side...cscx + cotx = 1. Now you should be able to see why this can't work as a Pythagorean Identity. There's always that test we can do for verification...5/3 + 3/4 ≠ 1, nor is (5/3 + 3/4)².
By process of elimination, a must be true. You can test w/ our example ratios:
sin²xsec²x+1 = tan²xcsc²x
(3/5)²(5/4)²+1 = (4/5)²(5/3)²
(9/25)(25/16)+1 = (16/25)(25/9)
(225/400)+1 = (400/225)
(9/16)+1 = (16/9)
(81/144)+1 = (256/144)
(81/144)+(144/144) = (256/144)
(256/144) = (256/144)
3x - 5 = -26
<u>step 1:</u>
subtract +5 on each side because you have to do the inverse operation
<u>step 2:</u>
3x= -21
<u>step 3:</u>
now divide 3 on each side
the answer is -7
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Hope this helped!
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<em>TheOneAndOnlyLara~</em>