Wednesday

September 17, 2014

September 17, 2014

Total # Posts: 39

**Pre-Calculus**

Find the inverse of the following quadratic equation. Hint: Complete the square first. y= 2x²+16x-5
*March 7, 2013*

**Pre-Calculus**

Find the inverse of the following quadratic equation. Hint: Complete the square first. y= x²+14x+50
*March 7, 2013*

**Trigonometry**

Solve the equation for all values of x. 2sin(2x)-√3=0, on the interval [0,2π).
*February 28, 2013*

**Trigonometry**

Solve the equation for all values of x. -2cos²x-sin x+1=0, on the interval [0,2π).
*February 28, 2013*

**Trigonometry**

1.) Apply the following method you used in #2! You should notably understand this already! Goodness grief! 2.) let A = arctan x let B = arccos x then we are looking for sin(A + B) . now, tan A = x , sin A = x/√(x^2+1) cos A = 1/√(x^2+1) ... create a right triangle ...
*February 21, 2013*

**Trigonometry**

Write the trigonometric expression as an algebraic expression. 1.) SIN(ARCSIN X+ARCCOS X) ANSWER: 1 2.) SIN(ARCTAN 2X-ARCCOS X) ANSWER: 2x²-SquareRoot of 1-x²/Square Root of 4x²+1.
*February 21, 2013*

**Trigonometry**

AWESOME! SUPERB! I had it right when you provided support. THANKS!
*February 21, 2013*

**Trigonometry**

Verify the identities please. 1.) TAN(X+π)-TAN(π-X)= 2 TAN X 2.) SIN(X+Y)+SIN(X-Y)= 2 SIN X COS Y
*February 21, 2013*

**Trigonometry**

Directions: Use a graphing utility to approximate the solutions of the equation in the interval [0,2π) by setting the equation equal to zero, graphing the new equation, and using the ZERO or ROOT feature to approximate the x-intercepts of the graph. (Note: These problems ...
*February 20, 2013*

**Trigonometry**

Verify the identities algebraically. 1.) TAN^5 X= TAN³X SEC²X-TAN³X 2.) COS³X SIN²X= (SIN²X-SIN^4X)COS X
*February 18, 2013*

**Trigonometry**

Verify the identity algebraically. This problem is very intriguing and awesome at the same time. It's wonderfully amazing! 1.) TAN³α-1/TAN α-1= TAN²α + TAN α + 1
*February 17, 2013*

**Trigonometry**

State the quadrant in which θ lies. 1.) CSC θ is greater than 0 and TAN θ is less than 0. 2.) SEC θ is greater than 0 and SIN θ is less than 0.
*February 17, 2013*

**Trigonometry**

2.) SEC X COS X= 1 1/COS X*COS X/1= 1 BINGO! 3.) CSC²X-CSC X SIN X+SIN X/SINX-TAN X+ COS X/SIN X 1/SIN²X-1/SIN X*SIN X/1-TAN X+COS X/ SIN X 1/SIN²X-1-SIN X/COS X+COS X/SIN X 1/SIN²X-1-TAN X*1/TAN X= CSC²X COOL! Need help on #6 and 7 please. Thank you!
*February 17, 2013*

**Trigonometry**

Verify the identities. 1.) √1-COSθ/1+COSθ= 1+SINθ/SINθ 2.) SEC X SIN(π/2-X)= 1 3.) CSC X(CSC X-SIN X)+SIN X-COS X/SIN X + COT X= CSC²X 4.) CSC^4 X-2 CSC²X+1= COT^4 X 5.) CSC^4 θ-COT^4 θ= 2 CSC²θ-1 6.) TAN^5 X= TAN&...
*February 17, 2013*

**Trigonometry**

Verify the identity algebraically. TAN X + COT Y/TAN X COT Y= TAN Y + COT X
*February 17, 2013*

**Trigonometry**

1.) 1/TAN[(π/2)-X]=COT X BINGO! SOLVED! 2.) SEC X/-CSC X 1/COS X ÷ -1/COS X 1/COS X * -SIN X/1 -TAN X YES BINGO! WOW!
*February 14, 2013*

**Trigonometry**

Verify the identities. 1.) SIN[(π/2)-X]/COS[(π/2)-X]=COT X 2.) SEC(-X)/CSC(-X)= -TAN X 3.) (1 + SIN Y)[1 + SIN(-Y)]= COS²Y 4.) 1 + CSC(-θ)/COS(-θ) + COT(-θ)= SEC θ (Note: Just relax through verifying/solving these nice fun looking math ...
*February 14, 2013*

**Trigonometry**

Verify/Solve the identities. 1.) SIN^1/2 X COS X-SIN^5/2 X COS X 2.) Long problem, but it's fun to solve! SEC^6 X(SEC X TAN X)-SEC^4 X(SEC X TAN X)
*February 14, 2013*

**Trigonometry**

Solve the equation. 1.) COS X CSC X
*February 10, 2013*

**Trigonometry**

Google this: 5.1 using fundamental identities Comcast Click and open the first result that appears on your search (Result= PDF Document). Textbook pages will appear and go to page 383 on problem #129. The directions will mention "Solve the right triangle shown in the ...
*February 10, 2013*

**Trigonometry**

Let's solve this fun trigonometric fun math problem! 1.) SEC²X-1/SIN²X
*February 10, 2013*

**Trigonometry**

Solve the trigonometric equations. 1.) SIN²x(CSC²x-1) 2.) COT x SEC X 3.) COS²[(π/2)-x]/COS X (Note: These mathematical problems are somewhat tricky, but useful for students as of learning how to their fundamental identities. Let's enjoy solving these ...
*February 10, 2013*

**Trigonometry**

A ship is 50 miles east and 35 miles south of port. If the captain wants to sail directly to port, what bearing should be taken?
*February 4, 2013*

**Trigonometry**

Two fire towers are 30 km apart, tower A being due west of tower B. A fire is spotted from the towers, and the bearings from A and B are E 14° N and W 34° N, respectively. Find the distance d of the fire from the line segment AB.
*February 4, 2013*

**Trigonometry**

Since I have also forgot to label the conversion for the solution, it is ultimately measured in FEET.
*February 4, 2013*

**Trigonometry**

BP= 3091.79 AB/sin 2.5° = 3091.79/sin 4° 3091.785015 sin 2.5° = 134.8617682 134.8617682/sin 4° = 1933.33 Awesome! Thanks for the help!
*February 4, 2013*

**Trigonometry**

An observer in a lighthouse 350 feet above sea level observes two ships directly offshore. The angles of depression to the ships are 4° and 6.5°. How far apart are the ships?
*February 4, 2013*

**Trigonometry**

A plane is 160 miles north and 85 miles east of an airport. If the pilot wants to fly directly to the airport, what bearing would be taken?
*February 4, 2013*

**Trigonometry**

Reiny thanks a lot for the help! I welcome you with a nice smile! Here is the website where I can show how the problems look/appear like... Google this: 4.8 applications and models Comcast Click the first result that appears on the web search page. Go to page 8 in the PDF ...
*February 4, 2013*

**Trigonometry**

Surveying A surveyor wishes to find the distance across a swamp. The bearing from A to B (Segment AB is opposite side of triangle) is N 32° W. The surveyor walks 50 meters from A to C, and at the point C the bearing to B is N 68° W. (Segment AC is adjacent side of ...
*February 4, 2013*

**Trigonometry**

Wave Motion A buoy oscillates in simple harmonic motion as waves go past. At a given time it is noted that the buoy moves a total of 3.5 feet from its low point to its high point, and that it returns to its high point every 10 seconds. Write an equation that describes the ...
*February 4, 2013*

**Trigonometry**

Mountain Descent A road sign at the top of a mountain indicates that for the next 4 miles the grade is 12%. Find the angle of the grade and the change in elevation for a car descending the mountain.
*February 4, 2013*

**Trigonometry**

Height of a Kite A 100-foot line is attached to a kite. When the kite has pulled the line taut, the angle of elevation to the kite is approximately 50°. Now everybody lets approximate the height of the kite!
*February 4, 2013*

**Trigonometry**

Angle of Depression Find the angle of depression from the top of the lighthouse 250 feet above water level to the water line of a ship 2.5 miles offshore.
*February 4, 2013*

**Trigonometry**

Navigation A ship leaves port at noon and has a bearing of S 29° W. If the ship sails at 20 knots, how many nautical miles south and how many nautical miles west will the ship have traveled by 6:00 P.M.?
*February 4, 2013*

**Trigonometry**

Height of a mountain While traveling across flat land, you notice a mountain directly in front of you. The angle of elevation to the peak is 2.5°. After you drive 18 miles closer to the mountain, the angle of elevation is 10°. Approximate the height of the mountain.
*February 4, 2013*

**Trigonometry**

A passenger in an airplane flying at an altitude of 10 kilometers sees two towns directly to the left of the plane. The angles of depression to the towns are 28° and 55°. How far apart are the towns?
*February 4, 2013*

**Trigonometry**

The length of a shadow of a tree is 130 feet when the angle of elevation of the sun is θ°. a) Write the height h of the tree as a function of θ. b) θ=10° What's height? θ=15° Height? θ=20° How tall is the height? θ=25° What...
*February 4, 2013*

**mathematics**

Stop cheating on Gauss you loser
*July 15, 2009*

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