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Ariadna A.
Tutor for 5 months and Honours student
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Spanish
TutorMe
Question:

Conjugate the following verbs in the correct tense acoording to the context: 1) Mi madre se____(llamar) María y es muy buena al ajedrez. 2)Mañana nos ____(levantar) a las 8:00 y ____(estudiar) todo el día. 3)No____(encontrar) mis zapatos, los has visto?

Ariadna A.
Answer:

1)This sentence is in the present tense, therefore the right verb to use in this case is "llama". 2)This sentence is in the future, therefore all verbs used have to be in the future: levataremos, estudiaremos. 3)This sentence is in the present tense, so the right verb to use is "encuentro".

Basic Math
TutorMe
Question:

What is the mass of 38cm$$^3$$ of cooper?

Ariadna A.
Answer:

This question can be solved in two different ways: 1)Mathematical way: We know that we want to get as final result a value in grams, therefore we can look for the density of cooper which is 8.96 g/cm$$^3$$. So using conversion factors we obtain: $$38\;cm^3\cdot\frac{8.96\;g}{1\;cm^3}=\boxed{340.48\;g}$$ 2)Physical way: We know that the density is given by $$\rho=M/V$$, where M is the mass and V is the volume. Isolating the mass and knowing that the the density of cooper is 8.96 g/cm$$^3$$: $$M=\rho V=(8.96\;g/cm^3)(38\;cm^3)=\boxed{340.48\;g}$$

Calculus
TutorMe
Question:

Find the equation for a parabola with vertex (2,0) and directrix x=6.

Ariadna A.
Answer:

In order to find the equation of this parabola, we will need to find 2 distances: the distance from an arbitrary point on the parabola ($$x_0,y_0$$) to the focus (2,0) and then the distance from that same arbitrary point on the directrix x=6. The distance from ($$x_0,y_0$$) to (2,0) can be calculated as follows: $$\sqrt{(x_0-2)^2+(y_0-0)^2}=\sqrt{(x_0-2)^2+y_0^2}$$ The distance from ($$x_0,y_0$$) to x=6 is given by: $$|x_0-6|$$ Now we equal both expressions and simplfy: $$\sqrt{(x_0-2)^2+(y_0-0)^2}=\sqrt{(x_0-2)^2+y_0^2}=|x_0-6|$$ $$(x_0-2)^2+y_0^2=(x_0-6)^2$$ $$x_0^2+2^2-2(2\cdot x_0)+y_0^2=x_0^2+6^2-2(6\cdot x_0)$$ $$x_0^2+4-4x_0+y_0^2=x_0^2+36-12x_0$$ $$4-4x_0+y_0^2=36-12x_0$$ $$y_0^2=32-8x_0=-8(x_0-4)$$ This equation is true for all points on the parabola, so we can change ($$x_0,y_0$$) into (x,y) and obtain the final equation for this parabola, which is: $$\boxed{y^2=-8(x-4)}$$

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