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Pavlova-9 [17]
4 years ago
7

Prove csc(pi/2 -x) = sec x

Mathematics
1 answer:
kifflom [539]4 years ago
6 0

Answer:

see explanation

Step-by-step explanation:

Consider the left side

Using the trigonometric identities

csc x = \frac{1}{sinx} and sec x = \frac{1}{cosx}

Note that sin( \frac{\pi }{2} - x) = cos x

Hence

csc(\frac{\pi }{2} - x)

= \frac{1}{sin(\frac{\pi }{2}-x) }

= \frac{1}{cosx}

= sec x = right side ⇒ proven

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40 tickets were sold for a concert, some at 75 cents and the rest at $1.25. if the total raised was $42, how many had the cheape
Murljashka [212]

Answer:

34

Step-by-step explanation:

$1.25 = 125 cents.

$42 = 4200 cents

Tickets sold at 75 cents = x

Tickets sold at 125 cents = y

x + y = 40

75x + 125y = 4200

Multiply the first equation by 75

75x + 75y = 3000

75x + 125y = 4200

Subtract the the second equation from the first.

75x + 75y = 3000

- 75x + 125y = 4200

-------------------------------

0 - 150y = - 1200

Divide both sides by - 150

-150y/-150 = -1200/-150

y = 8

Substitute y = 8 into the first equation

x + y = 42

x + 8 = 42

x = 42 - 8

x = 34

34 tickets were sold for 75 cents

8 tickets were sold for $1.25

3 0
3 years ago
Your friend says that there are 36 feet in 12 yards is your friend correct? explain your reasoning.
galben [10]

Answer:

Yes

Step-by-step explanation:

1 yard is 3 feet. so 12 yards is 36 by 12*3

7 0
3 years ago
2. Which rule best represents the
hoa [83]

Answer:

The answer is A!

Step-by-step explanation: I’m pretty sure it’s A!
Unfortunately I haven’t seen this question on any of my tests but I’m pretty sure It is A! Trust me!

7 0
2 years ago
the sum of the digits of a two digit number is 10. if the digits are reversed, the new number will be 18 more than the original
bearhunter [10]
<span>t+u = 10
10u+t= 10t+u-54
Rearrange:
t+u=10
9t-9u=54
Simplify:

1st; t+u=10
2nd: t-u=6
Add 1st and 2nd and solve for t,as follows:
2t = 16
t = 8
Substitute in t+u=10 to solve for u:
8+u=10
u = 2
Original Number = 82</span>
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3 years ago
Answer ASAP ASAP please
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