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marysya [2.9K]
3 years ago
8

Simplify the expression tan(-x)csc(-x)/sec(-x)cot(-x)

Mathematics
2 answers:
erik [133]3 years ago
6 0
The answer is csc(x)cot(x)tan(x) over sec(x)

goldfiish [28.3K]3 years ago
5 0
\frac{ tan(-x) cosec (-x)}{sec (-x) cot (-x)} ≡ \frac{ [- tan(x)]  [-cosec (-x)]}{[sec (x)] [-cot (x)]}

                                 ≡ \frac{ tan(x) cosec (-x)}{-sec (x) cot (x)} 

                                 ≡ \frac{ (\frac{sin (x)}{cos (x)})   (\frac{1}{sin (x )}) }{(- \frac{1}{cos (x)}) ( \frac{cos (x)}{sin (x)})  }

                                 ≡ \frac{ \frac{1}{cos (x)} }{-  \frac{1}{sin (x)} }

                                 ≡ - \frac{sin (x)}{cos (x)}

                                 ≡ - tan (x)<span>
</span>
⇒ \frac{ tan(-x) cosec (-x)}{sec (-x) cot (-x)} ≡ tan (-x)

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stich3 [128]

The other equality which must be stated by Jose is that angles BAC and BAE are congruent and their measures are equal.

<h3>What other congruence statements must Jose state?</h3>

It follows from the task content that Jose is trying to prove the congruence of both triangles by means of the Side-Angle-Side congruence theorem.

It therefore follows that since, Jose has identified that the ratio of corresponding sides are equal as indicated in the task content, the equality which Jose has to state is the angle congruence equality.

Read more on congruence theorem;

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7 0
2 years ago
Solve <img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B3%7D%20%20%3D%20%20%5Cfrac%7Bh%7D%7Bh%20-%203000%7D%20" id="TexFormula1" ti
dezoksy [38]
<span>\sqrt{3}(h-3000)=h
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8 0
3 years ago
At an airport, it cost 7 to park for one hour and 5 per hour for each additional hour.Let x represent the number hours parked.wr
kompoz [17]
The cost of parking is an initial cost plus an hourly cost.
The first hour costs $7.
You need a function for the cost of more than 1 hour,
meaning 2, 3, 4, etc. hours.
Each hour after the first hour costs $5.

1 hour: $7
2 hours: $7 + $5                  = 7 + 5 * 1          = 12
3 hours: $7 + $5 + $5          = 7 + 5 * 2          = 17
4 hours: $7 + $5 + $5 + $5  = 7 + 5 * 3          = 22

Notice the pattern above in the middle column.
The number of $5 charges you add is one less than the number of hours.
For 2 hours, you only add one $5 charge.
For 3 hours, you add two $5 charges.

Since the number of hours is x, according to the problem, 1 hour less than the number of hours is x - 1.

The fixed charge is the $7 for the first hour.
Each additional hour is $5, so you multiply 1 less than the number of hours,
x - 1, by 5 and add to 7.

C(x) = 7 + 5(x - 1)

This can be left as it is, or it can be simplified as

C(x) = 7 + 5x - 5

C(x) = 5x + 2

Answer: C(x) = 5x + 2

Check:
For 2 hours: C(2) = 5(2) + 2 = 10 + 2 = 12
For 3 hours: C(3) = 5(3) + 2 = 15 + 2 = 17
For 4 hours: C(3) = 5(4) + 2 = 20 + 2 = 22
Notice that the totals for 2, 3, 4 hours here
are the same as the right column in the table above.
6 0
4 years ago
Use the graph of f '(x) below to find the x values of the relative maximum on the graph of f(x):
Lana71 [14]

Answer:

You have relative maximum at x=1.

Step-by-step explanation:

-Note that f' is continuous and smooth everywhere. f therefore exists everywhere on the domain provided in the graph.

f' is greater than 0 when the curve is above the x-axis.

f' greater than 0 means that f is increasing there.

f' is less than 0 when the curve is below the x-axis.

f' is less than 0 means that f is decreasing there.

Since we are looking for relative maximum(s), we are looking for when the graph of f switches from increasing to decreasing. That forms something that looks like this '∩' sort of.

This means we are looking for when f' switches from positive to negative. At that switch point is where we have the relative maximum occurring at.

Looking at the graph the switch points are at x=0, x=1, and x=2.

At x=0, we have f' is less than 0 before x=0 and that f' is greater than 0 after x=0.  That means f is decreasing to increasing here. There would be a relative minimum at x=0.

At x=1, we have f' is greater than 0 before x=1 and that f' is less than 0 after x=1. That means f is increasing to decreasing here. There would be a relative maximum at x=1.

At x=2, we have f' is less than 0 before x=2 and that f' is greater than 0 after x=2. That means f is decreasing to increasing here. There would be a relative minimum at x=2.

Conclusion:

* Relative minimums at x=0 and x=2

* Relative maximums at x=1

3 0
3 years ago
Use the following graph for questions 1 - 4. Rise is the vertical change of
aksik [14]

Answer:

C 2

Step-by-step explanation:

because every time you go 1 to the side you also go 2 up

7 0
3 years ago
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