1. let h(t) be in the A*sin(bt + c) + M form, where A = amplitude, b = 2π/k, k = 1/f, f = frequency, t = time (in seconds), c = the horizontal translation, and M = the midline. 2. A = the diameter/2 = 7 3. b = 2π/k, k = 1/3 (because the frequency is 3 rev/sec), so b = 2π/(1/3)...

1. "the sum of the number and 6" = x + 6 2. "the square of the sum of the number and 6" = (x + 6)2 3. "the square of the sum of the number and 6 is 169" = (x + 6)2 = 169 4. x + 6 = ±√169 5. x + 6 = ±13 6. x + 6 =...

1. let m∠ S = x and the m∠ L = 5x + 12 2. since the two are complementary, we know that their sum equals 90° 3. therefore, (x) + (5x + 12) = 90 4. solve for x: 6x = 78 5. x = 13 6. hence, m∠ S = 13° and m∠ L = 5(13) + 12 = 77°

1. we will have to use a few trig identities for this one 2. sin(2x) = sin x 3. using the double-angle identity: 2sinxcosx = sin x 4. cosx = sinx/2sinx = 1/2 5. x = cos-1(1/2) = π/3 or -π/3 6. therefore, the solution for #1 is π/3 or -π/3 7. if #2 is cos2x +...

1. velocity = displacement/elapsed time 2. therefore, the velocity = 4000/16 m/s = 250 m/s 3. since velocity is a vector, we must include the direction with its speed of 250 m/s 4. hence, the velocity = 250 m/s, 35° north of east, assuming that it is traveling in the positive x dire...

Let's use this formula: B = A(1+i)n - (P/i)*([1+i]n - 1), where B is the balance after n payments, A is the initial amount of the loan, P is the monthly payment amount, and i is the interest rate compounded monthly.   1. A = 30000, P = 400, i = .06/12 = .005, and n = 2 years*(12...

1. let's use the law of cosines for this :) 2. let a = 70 and b = 175 3. also, 21° is the angle that is opposite to the side of the triangle that we are looking for 4. the law of cosines = c2 = a2 + b2 - 2*a*b*cos(21°) 5. substitute the values in: c2 = (70)2 + (175)2...

1. the angle of elevation formed by the ground and the bottom of the antenna is 62.6° 2. the side opposite to this angle is 1000 3. to find the distance from the building to the point of observation, we do: tan(62.6°) = 1000/a 4. a = 1000/tan(62.6°) ≈ 518.351 5. the angle of...

1. let t(x) = 5 and b(x) = 5/6*t(x) 2. substitute the values: 5/6*5 = 25/6 3. simplify: 4 1/6 mi

1. this is a proportions problem 2. since y and x vary directly, y = k*x 3. thus, 10 = k*2, and 5 = k 4. now, let's use our new value for x to solve for y 5. y = k*x, so y = 5*x 6. let's plug in x = 8 7. y = 5*8 = 40 8. note, the proportions are equal: y/x = k, 10/2 =...

1. we can use the displacement formula for this, d = r*t 2. 1/2 = (1/5)*t 3. solve for t 4. (5)(1/2) = t 5. 5/2 = t 6. 2 1/2 = t 7. t = 2.5 hours

1. to find the answer for A), we must find the x-intercepts 2. set f(t) = 0 to do this :) 3. 0 = -16t2 + 96t = -16t*(t-6), so t = 0 or 6 4. since t = 0 is the initial time, we know that t = 6 is our final time, i.e. when it hits the ground. 5. to find B), we must find the vertex,...

For the nominative case, use the perspective chart :)   1. first person singular = I 2. second person singular = you 3. third person singular = he, she, or it 4. first person plural = we 5. second person plural = you 6. third person plural = they   The...

1. we will need to use some basic trigonometry for this one :) 2. first, we must divide the base by 2 so that we get a right triangle with a base of 15 and a hypotenuse of 17 3. now, let's call the angle that is adjacent to the base angle A 4. to find the angle, we do: cos(A) = adjacent/hypotenuse...

1. this is called dimensional analysis 2. 1 mi = 5280 ft 3. 1 hr = 60 min  4. 1 min = 60 sec 5. therefore,   65 mi/ 1 hr * (1 hr/60 min) * (1 min/60 sec) = 65 mi/ (60*60) sec = 65 mi/3600 sec 65 mi/3600 sec * 5280 ft/1 mi = (65*5280) ft/3600 sec = 343200 ft/3600...

1. let's call the red flare r(x) and the green flare g(x) 2. r(x) is in the vertex form, so (15, 175) is the vertex point for this projectile. 3. since the vertex occurs at the maximum, the red flare reaches a height of 175 meters once it has gone a horizontal distance of 15 meters in the...

1. "the sum of x and 4" = x+4 2. "all divided by 3" = (x+4)/3

1. it seems beneficial to use vectors for this problem. (Remember, velocity is a vector, but speed is a magnitude.) 2. let rb = the velocity of Train B and ra = the velocity of Train A. 3. the absolute value of (rb) = the absolute value of (ra) + 15 4. sb, the speed of B, =...

1. L = 5W 2. 120 = 2L + 2(5W) 3. 120 = 2(5W) + 2(5W) 4. 120 = 20W 5. 120/20 = W 6. 6 = W 7. L = 5(6) 8. L = 30    W = 6 ft and L = 30 ft