Marika solved the equation (6x + 15)2 + 24 = 0. Her work is below. 1. (6x + 15)2 = –24 2. StartRoot (6 + 15) squared EndRoot = S
2 answers:
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Answer:
110in
Step-by-step explanation:
You are given the angle and the length of one leg of the triangle. Therefore, you will want to use one of the trigonometric formulae you've learnt. The available ones are:
where <em>O</em> represents the leg opposite the angle, <em>A </em>represents the leg Adjacent to the angle, and <em>H</em> represents the hypotenuse. The lamppost in this problem is the leg <em>opposite the angle</em>, so you want a formula that has <em>O </em>in the equation, namely <em>sin </em>or <em>tan</em>. Because you were given the leg <em>adjacent </em>to the angle, it makes sense here to use <em>tan.</em> Set up the equation as such to solve:
For some reason, it wanted you to convert 70 degrees to radians, so that may have been where you could have run into issues. If you plugged in 70 as degrees, you would get 48 which isn't an option. Converting 70 to 1.22 rad gets you the answer of 110 in.
Answer:
- (a) C(d) = 270 + 0.20d
- (b) C(1800) = 630
- (c) see attached (lower right graph). The slope represents the cost (in dollars) per mile.
- (d) It represents the fixed cost (amount she pays even if she does not drive)
- (e) the cost increases as the number of miles driven increases
Step-by-step explanation:
(a) You are given two points, (d, C) = (360, 342) and (780, 426). The two-point form of the equation for a line can be helpful.
... y = (y2 -y1)/(x2 -x1)(x -x1) +y1
Using the given points and variables, ...
... C = (426 -342)/(780 -360)(d -360) +342
... C = (84/420)(d -360) +342
... C = 0.20d +270 . . . . the equation for cost as a function of miles driven
(b) Fill in the value for d and do the arithmetic.
... C = 0.20·1800 +270 = 360 +270 = 630
(c) The slope of the curve is (change in dollars)/(change in miles) = $/mile, the cost of each mile driven. (It is $0.20 per mile.)
(d) The C-intercept is C(0) = cost for 0 miles driven. (It is $270.)
(e) Both your own experience and the problem statement tell you that the cost of vehicle operation varies with miles driven. If nothing else, you need to pay more for gas the farther you drive.
(In real life, you don't expect the model to be exactly linear, becasue you get a large jump in cost when you reach certain maintenance interval miles: 15,000, 30,000, and so on.)
