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hjlf
4 years ago
15

Simplify the expression. csc^2x sec^2x/sec^2x + csc^2x

Mathematics
1 answer:
Svetlanka [38]4 years ago
3 0
2csc^2x + 1 is the correct answer to simplify this expression
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NEED HELP ALGEBRA 1
tigry1 [53]
<h3>Answer: E) 1/5</h3>

========================================================

Method 1

Use your calculator to find the decimal version of each fraction

  • 4/10 = 0.40
  • 12/25 = 0.48
  • 14/20 = 0.70
  • 28/50 = 0.56
  • 1/5 = 0.20

We see that 0.20 is the smallest decimal of the list, so 1/5 is the smallest fraction of the list.

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

Method 2

To compare fractions, we need to get all of the denominators to the same value. A good target is the lowest common denominator (LCD)

In this case, the LCD is 100

  • 4/10 = 40/100 .... multiply top and bottom by 10
  • 12/25 = 48/100 .... multiply top and bottom by 4
  • 14/20 = 70/100 .... multiply top and bottom by 5
  • 28/50 = 56/100  .... multiply top and bottom by 2
  • 1/5 = 20/100  .... multiply top and bottom by 20

The last fraction has the smallest denominator, so 1/5 is the smallest.

4 0
3 years ago
Read 2 more answers
Leila bought 2 pairs of shoes that were the same price. Including the $3 sales tax, she paid a total of $57. What was the cost o
nirvana33 [79]
Each pain was 15 dollars
3 0
3 years ago
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An airplane leaves an airport and flies due west 150 miles and then 170 miles in the direction S 49.17°W. How far is the plane f
Sphinxa [80]

Answer:

Distance between plane and airport is 134.4 miles.

Step-by-step explanation:

Given : An airplane leaves an airport and flies due west 150 miles and then 170 miles in the direction S 49.17°W.

To find : How far is the plane from the airport.

Solution : Distance from airport to west is 150 miles and then 170 miles in the direction south  and angle form is S 49.17° W

Refer the attached picture for clearance.

Applying law of cosines

c^2=a^2+b^2-2ab Cos(C)

c=\sqrt{a^2+b^2-2ab Cos(C)}

where a= 150 miles

b=170 miles

C=   49.17° angle in degree

c = distance between plane from the airport

Put values in the formula,

c=\sqrt{a^2+b^2-2ab Cos(C)}

c=\sqrt{150^2+170^2-2(150)(170) Cos(49.17^{\circ})}

c=\sqrt{22500+28900-51000(0.653)}

c=\sqrt{51400-33303}

c=\sqrt{18057}

c=134.37

Therefore, Distance between plane and airport is 134.4 miles.

5 0
3 years ago
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What is the inverse of f(x)=-x-1
weeeeeb [17]

Answer:nugget

Step-by-step explanation: Idk...sorry for wasting ur time!!!237482929

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3 years ago
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Find the surface area of the figure. Round to the nearest hundredth if needed.
Damm [24]
280 your welcome good luck
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