Tutor profile: Rebecca R.
Examine the excerpt from the following poem, El Sueño by Pablo Neruda. Respond to the questions at the end. Andando en las arena yo decidí dejarte Pisaba un barro oscuro que temblaba Y hundiendome y saliendo decidí que salieras de mí, que me pesabas como piedra cortante. 1) ¿En qué tiempo verbal está este párrafo?¿Cómo lo sabes? 2) Hay una palabra que falta un acento. ¿Cual es? ¿Por qué lo necesita? 3) ¿Cuál de estos verbos son irregulares?
1) Este párrafo está en el tiempo pasado. Determinamos esto porque todos los verbos están en tiempo pasado. Por ejemplo, se refiere a sí mismo como "andando", y sabemos que está diciendo que estaba caminando en algún momento antes del presente. This paragraph is in the past tense. We can determine this because all of the verbs are in the past tense. For example, he refers to himself by saying "I was walking," and we know that he's saying that he was walking in some moment before the present. 2) La palabra que falta un acento es hundiéndome. Si una palabra termina en una vocal, entonces se estresa la penúltima sílaba. The word that is missing an accent is "hundiéndome". If a word ends in a vowel, then you stress the penultimate syllable. 3) No hay verbos irregulares. There are no irregular verbs.
Find the domain and range for f(x) = x^2 - 4x + 4
Let's recall what the domain and range of a function are. The domain is the set of values for which the independent variable, x, is defined. The range, however, is the set of values of the function itself; f(x), over the domain of x. Essentially, the domain is the set of possible input values that x can be, whereas the range is the set of possible outputs. On an x,y axis, the domain is shown on the x-axis and the range is shown on the y-axis. To determine the domain of a function like this, we look for values of x which we know would be excluded from the equation. One method is by graphing the equation, or avoiding common mathematical mistakes like a 0 in the denominator or negative value under the square root. For this function, we can recognize that it will be an upward facing parabola (due to the x^2). Input a few values into the function, lets say 0, 2, -1, 3: f(0) = 0^2 - 4(0) + 4 = 0 - 0 + 4 = 4 f(2) = 2^2 - 4(2) + 4 = 4 - 8 + 4 = 0 f(-1) = (-1)^2 - 4(-1) + 4 = 1 + 4 + 4 = 9 f(3) = (3)^2 - 4(3) + 4 = 9 - 12 + 4 = 1 Based on these results, we can start to visualize a parabola with a vertex at (2,0), which extends infinitely on either side. Because there are no real x-values for which this equation will not produce a real product, we know that the domain is all possible values. This is denoted as (-∞,∞). Now we must find the range, or all possible outputs for the given domain (which is all real numbers). We previously determined that the lowest y-value, the lowest point on the graph, occurs at the point (2,0). On either side of this, we have increasingly large y-values, which extend towards infinity. Given this information we know that the range, or all possibly outputs, is [0,∞). Remember, the hard bracket means that the 0 is included in the range. Here is our final answer: Domain - (-∞,∞) Range - [0,∞)
Subject: Basic Chemistry
Balance the following chemical equation: SnO2 + H2 → Sn + H2O
First, let's recall what it means for an equation to be unbalanced. Chemical reactions can be demonstrated mathematically by writing out the reactants (left side) and products (right side), and amounts of every time of atom involved can be determined using basic math and the Law of Conservation of Mass. Because we know that the mass must be the same before and after the reaction, we can determine the amount of each reactant used to create a product by balancing the two sides. When the number of atoms of each element in the reactant does not match the number of atoms of each element in the product, an equation is said to be unbalanced. To balance the above equation, we first must determine the number of atoms of each element on each side of the equation. Here, this is as follows: Reactant: 1Sn, 2O, 2H Product: 1Sn, 1O, 2H Immediately we see that we must add another O to the products. This can be accomplished by adding a 2 coefficient in front of the H20 (remember, we can't break up the atoms in product compounds). Here is our new equation: SnO2 + H2 → Sn + 2 H2O We're getting closer, but now our hydrogen atoms are out of balance: Reactant: 1Sn, 2O, 2H Product: 1Sn, 2O, 4H Because we know we need 4H in the product, we can surmise that we need to put a 2 coefficient in front of the H2 in the reactant. The 2H in the reactant are multiplied by 2, giving us 2X2=4 hydrogen atoms. Here is our new equation: SnO2 + 2 H2 → Sn + 2 H2O Reactant: 1Sn, 2O, 4H Product: 1Sn, 2O, 4H As you can see, there are equal numbers of each atom on both the reactant and product sides. The equation is now balanced!
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