A sample of seawater contains 6.277 g of sodium chloride per liter of solution. How many mg of sodium chloride would be contained in 15.0 mL of this solution?
We can easily solve this problem using a technique known as dimensional analysis. What's really nice about this technique is that it takes advantage of the multiplicative identity property (a number times 1 is that number) by creating fractions equal to 1 out of conversion factors. As a result, we can perform multiple conversions between units, all in one step, without changing the actual value of the original quantity. All we need to know are the conversion factors for the specific problem we're working with. In this example, those conversion factors are 1 g = 1000 mg and 1 L = 1000 mL. Before we start the calculations, let's remind us what units we want to be converting. We start with grams and liters and want to end with milligrams and milliliters. The following calculation yields the unit conversions we are looking for: (6.277 g NaCl / 1 L solution) x (1000 mg NaCl / 1 g NaCl) x (1 L solution / 1000 mL solution) = 6.277 mg NaCl/ 1 mL solution But we're not done yet! We've only come up with another conversion factor that we can use to solve the actual problem. 15.0 mL solution x (6.277 mg NaCl / 1 mL solution) = 94.2* mg NaCl [ANSWER] *taking into consideration significant figures QUICK TIP: If you get stuck trying to figure out how to orient your conversion fraction, just remember that you want unwanted units to cancel. For example, in converting 300 cm/s to units of m/s, since centimeters comprises the numerator, you want to orient your conversion fraction 1 m / 100 cm so that the centimeters cancel, leaving meters in the numerator.
Determine if the following function is continuous at x = –2: f(x) = x^2 + 2x, if x ≤ –2 5x + 10, if x > –2
This exercise deals with the definition of continuity for a single variable function. Once you understand this definition, you'll be able to tackle these types of problems in no time! In calculus, continuity at a point x = a on a function y = f(x) guarantees the satisfaction of three conditions: (1) the limit of the function as x approaches a exists, (2) the function evaluated at a exists, and (3) the results of (1) and (2) are equivalent. Let's test the first condition of this definition. In order for the first condition to be true, (a) the limit of the function as x approaches a from the left must exist, (b) the limit of the function as x approaches a from the right must exist, and (c) the results of (a) and (b) must be equivalent. (a) The limit as x approaches –2 from the left of f(x) = (–2)^2 + 2(–2) = 0. (b) The limit as x approaches –2 from the right of f(x) = 5(–2) + 10 = 0. (c) The limits as x approaches –2 from the left and right of f(x) are both equal to 0. Therefore, the limit as x approaches –2 of f(x) equals 0. Onto the second condition! This one's pretty simple: f(–2) = (–2)^2 + 2(–2) = 0. And finally, we reach the third condition. The limit as x approaches –2 of f(x) equals f(–2), since both expressions evaluate to 0. Therefore, the function y = f(x) at the point x = –2 is, indeed, continuous by definition.
Solve by completing the square: x^2 – 8x + 5 = 0
Completing the square is one of many ways to solve a quadratic equation. It requires a bit of algebraic manipulation, but with practice, you can master these types of problems! Let's begin! Our primary goal is to have a perfect square on one side of the equation and a constant on the other side. A perfect square simply refers to a quadratic expression that factors into two identical binomials, for example: x^2 – 2x + 1 = (x – 1)(x – 1) = (x – 1)^2. First, let's deal with the constant. x^2 – 8x + 5 = 0 x^2 – 8x = 0 – 5 The constant is now on the right side of the equation. However, the left side of the equation isn't quite a perfect square yet. To fix this, we can follow this simple rule of thumb: (1) find the coefficient of the first-degree variable x, (2) divide that coefficient by two, (3) square the result. Therefore, in this particular problem, we take our first-degree coefficient 8, divide it by 2 (8/2 = 4), and square the result (4^2 = 16). We then add this number to both sides of the equation to maintain the equality. x^2 – 8x + 16 = –5 + 16 Now, since the quadratic expression on the left side of the equation is a perfect square, we can rewrite the equation. (x – 4)^2 = 11 To finish off this problem, we only need to perform some basic algebra. And there you have it! x – 4 = sqrt (11) x = sqrt (11) + 4 [ANSWER]