In a right triangle ABC, sin(C)=3/5 and cos(C)=4/5. Find tan(C).
A, B, and C represent angles of the right triangle. We are given the sine and cosine values for this triangle. SOH CAH TOA is a simple way to remember how to find sine, cosine, and tangent. This acronym stands for Sine=Opposite ÷ Hypotenuse, Cosine=Adjacent ÷ Hypotenuse, Tangent=Opposite÷Adjacent. Opposite, Adjacent, and Hypotenuse refer to the sides of the triangle (opposite is the side across from the angle being referenced, in this case angle C, hypotenuse is the side that is not connected to the right angle, and adjacent is the side touching the angle being reference and the right angle). With this information, we can quickly see that the given information tells us the length of each side of the triangle. From sin(C) we know opposite=3 and hypotenuse=5, and from cos(C) we know adjacent=4 and of course, the hypotenuse is still the same. To find tangent we need opposite ÷ adjacent, so the final answer is: tan(C) = 3/4
6x - 4y + 7 = 4x + 2y + 9, solve x.
This problem is solved by combining terms. Since we are not given a value for y, the answer will have a y component. We begin combining terms with the x components on one side of the equal sign and all else on the other by adding or subtracting from each side, remember what we do to one side of the equal sign must be done to both. Let's start by combining x terms onto the left side of the equation by subtracting 4x from each side as follows: 6x - 4y + 7 = 4x + 2y + 9 -4x -4x We now have: 2x - 4y + 7 = 2y + 9 and we need only x terms on the left side, so we combine the like terms on the left side of the equation: 2x - 4y + 7 = 2y + 9 + 4y - 7 +4y - 7 Now we have, 2x = 6y + 2. Since x is multiplied by 2, we need to divide both sides of the equation by two to get our final answer. Since there are two terms on the right side of the equation, we will distribute the division and divide both terms by 2. 6y ÷ 2 = 3y and 2 ÷ 2 = 1, so our final answer is x = 3y + 1
A new highway on-ramp is being designed. If the average acceleration of a car is 3 meters per second squared and the speed limit is 55 miles per hour, how long should the on-ramp be in order to allow cars sufficient distance to reach the speed limit before merging onto the highway.
Since we are given two measurements (acceleration and speed) in different systems, we must first convert one value. 1 mph = 0.447 m/s, so 55 mph = 24.6 m/s. Now with the numbers in the same measurement system (SI), we can focus on the question. Assuming a car starts from 0 m/s, with an acceleration of 3 m/s^2 we can find the time it will take the car to reach the speed limit by using a kinematic equation: Vf = Vi + a*t, (Vf = final velocity, Vi = initial velocity, a = acceleration, t = time) Solving this equation for time, we have: t = (Vf - Vi) / a, which we can substitute our values into. For this problem, t = (24.6 m/s - 0 m/s) ÷ (3 m/s^2), so t = 8.2 s. Using another kinematic equation, we can solve this problem for the sufficient distance using given information and the solved time. d = (Vf + Vi) * t / 2 (d = distance, Vf = final velocity, Vi = initial velocity, t = time). For this problem, d = (24.6 m/s + 0 m/s) * (8.2 s) ÷ 2, so is d = 100.9 m. The final answer is, the on-ramp should be 100.9 m long in order to allow cars to reach the speed limit before merging onto the highway.