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Jordan C.
Student at Columbia University, Private Tutor
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Spanish
TutorMe
Question:

Conjugate the "poder" in the subjunctive imperfect and use it in a sentence:

Jordan C.

yo-pudiera nosotros-pudiéramos tú-pudieras vosotros-pudierais él ellos ella-pudiera ellas-pudieran Ud Uds (Note the change from the o to the u. This is an irregular verb in the imperfect. Note that in -er ending very, the accent goes over the e in the nosotros form. Example sentence: Si yo pudiera ser Superman, volaría a Fiji. Translation: If I could be Superman, I would fly to Fiji. (I would explain how the imperfect subjunctive is often used in "If/then" statements.)

English as a Second Language
TutorMe
Question:

Please identify the subject, verb, parts of the following sentence: The dog ran to me.

Jordan C.

Dog: Subject: the person or thing who is doing the action. ran: Verb: the action to me: prepositional phrase. We know this because it is indicated by the preposition to. (Emphasize the subject precedes the verb--the person/thing before the action.)

Algebra
TutorMe
Question:

Solve for solution(s) of x: (x+5)^(1/2) - (5-2x)^(1/4)=0

Jordan C.

1. First, we want to turn this expression into an equality so that when we manipulate one side, we know to do the exact same thing to the other side. (x+5)^(1/2) - (5-2x)^(1/4)=0 -Since the right-hand side of the equation is just zero, we can add (5-2x)^(1/4) to both sides. This will eliminate from the left-hand side and give us the equation we want. Now we have: (x+5)^(1/2) = (5-2x)^(1/4) -This allows us to better see the different parts of the equations for manipulating them and solving for x. (This is a general manipulation learned in Algebra. I explained it in case.) 2. Now, we want to try to eliminate the fractional exponents in order to isolate x from it's parenthetical expressions: -To eliminate the 1/2 exponent on x+5, I would, ideally, want to raise the 1/2 exponent to the second. This would cancel the values to be one, and any value raised to the first is itself. [(x+5)^(1/2)]^2 So this works, for now... -However, what we do to one side, we must do to the other: [(5-2x)^(1/4)]^2 -When we raise the right-hand side to the second degree, we still are left with the phrase raised to the half: (5-2x)^(1/2) -To fix this, then, we can raise this phrase to the second, again: [(5-2x)^(1/2)]^2 -This cancels the exponent completely. So, we have really raised this expression to the fourth:[(5-2x)^(1/4)]^4 This means that we have to raise the left-hand side to the fourth as well: [(x+5)^(1/2)]^4 -This leaves us with (x+5)^2. So, for our full equation right now, we have: (x+5)^2=5-2x 3. We can expand, or factor, the binomial [(x+5)^2] using FOIL: (x+5)^2= (x+5)(x+5) -Multiply the Firsts: x*x= x^2 -Multiply the Outside: x*5=5x -Multiply the Inside: 5*x=5x -Multiply the Lasts: 5*5=25 -So, when we factor completely, we are left with: x^2+5x+5x+25= x^2+10x+25 (This topic is typically covered in Algebra, but I explained it in detail in case.) -We can't forget that this is an equation, so this expression is set equal to the left-hand side we simplified earlier, giving us: x^2+10x+25=5-2x 4. Now, this is a typical Algebra problem where you would solve for x by combining like terms, giving us: x^2+12x+20=0 -We can see that this is factorable: (x+10)(x+2)=0 -We can, then, solve for x, and we get two solutions: x+10=0, therefore x=-10 x+2=0, therefore x=-2 5. We must check to make sure these solutions work in the original problem by substituting our two values for x: ((-10)+5)^(1/2) - (5-2(-10))^(1/4) =4.204. Since x=-10 does not hold in the equation we used, the solution is extraneous. ((-2)+5)^(1/2) - (5-2(-2))^(1/4) =0. Since x= -2 holds true in the equation we used, it is a solution!

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