Conjugate the "poder" in the subjunctive imperfect and use it in a sentence:
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.)
Please identify the subject, verb, parts of the following sentence: The dog ran to me.
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.)
Solve for solution(s) of x: (x+5)^(1/2) - (5-2x)^(1/4)=0
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!