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Tutor profile: Andre A.

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Andre A.
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Subject: Calculus

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Question:

$$y = \frac{sin(x)}{x^{2}},$$ Find $$\frac{\mathrm{d}y }{\mathrm{d} x}$$.

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Andre A.
Answer:

Upon looking on the function, we notice that we have one function of x being divided by another function of x. This implies that we will be using the quotient rule to solve the differential. By definition, the quotient rule is: Given that $$y = \frac{u}{v} $$, then $$\frac{\mathrm{d}y }{\mathrm{d} x} =\frac{\frac{\mathrm{d} u}{\mathrm{d} x}\times v-\frac{\mathrm{d} v}{\mathrm{d} x}\times u}{v^{2}} $$. By using this formula let us declare u and v. $$ u = sin(x)$$ and $$v = x^{2}$$. Solving for $$\frac{\mathrm{d} u}{\mathrm{d} x}$$ and $$\frac{\mathrm{d} v}{\mathrm{d} x}$$, we obtain - $$\frac{\mathrm{d} u}{\mathrm{d} x} = cosx $$ (from definition that the differential of sin(x) = cos(x) $$\frac{\mathrm{d} v}{\mathrm{d} x} = 2x$$ (using the power rule - $$\frac{\mathrm{d}}{\mathrm{d} x} x^{n} = nx^{n-1}$$ Now that we have solved for our unknowns, we can substitute the values into the formula. $$\frac{\mathrm{d}y }{\mathrm{d} x} =\frac{\frac{\mathrm{d} u}{\mathrm{d} x}\times v-\frac{\mathrm{d} v}{\mathrm{d} x}\times u}{v^{2}} $$ $$\frac{\mathrm{d}y }{\mathrm{d} x} =\frac{x^{2}cos(x)-2xsin(x)}{x^{4}} $$ $$\frac{\mathrm{d}y }{\mathrm{d} x} =\frac{xcos(x)-2sin(x)}{x^{3}} $$ (dividing the numerator and denominator by x to simplify)

Subject: Statistics

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Question:

Let $$P(A) = 0.7, P(B) = 0.4$$ and $$P(A \cap B)$$ = 0.3. Find the probability of $$P(AUB)$$.

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Andre A.
Answer:

We are given that the probability of event A happening is 0.7 and that the probability of event B happening is 0.4. Also, we given that the probability of both events occurring is 0.3. The $$P(AUB)$$, which is the probability of either A or B happening, is equal to $$ P(A) + P(B) - P(A \cap B)$$ Using this equation we can now solve for $$P(AUB)$$ $$P(AUB) = P(A) + P(B) - P(A \cap B)$$ $$P(AUB) = 0.7 + 0.4 - 0.3$$ $$P(AUB) = 0.8$$

Subject: Algebra

TutorMe
Question:

Solve for x and y. $$7x + 3y = 24$$ $$21x - 5y = 58$$

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Andre A.
Answer:

Since we were given two equations we will solve this simultaneously. There are two methods to solve simultaneous equations, the substitution method and the elimination method. Using the substitution method: Let us call $$7x + 3y = 24$$, equation (1) and let us call $$21x - 5y = 58$$, equation (2). Now using equation (1), let us form a 3rd equation by isolation x on one side of the equation. (1) $$7x + 3y = 24$$ $$7x + 3y - 3y = 24 - 3y$$ (Subtracting 3y from both sides) $$7x = 24 - 3y$$ ($$3y - 3y = 0$$) $$7x \div 7 = (24 - 3y) \div 7$$ (Dividing both sides by 7) $$x = (24 - 3y) \div 7$$ - Let us call this equation (3) Now let us substitute equation (3) into equation (2) by replacing x in eq. 2 with eq. 3. $$21(24 - 3y) \div 7 - 5y = 58$$ (This is our newly formed equation) $$3(24 - 3y) - 5y = 58$$ (21 divided by 7 = 3) $$72 - 9y - 5y = 58$$ (We multiplied the numbers in the bracket by 3) $$72 - 14y = 58$$ ($$-9y - 5y = -14y$$) $$72 - 58 - 14y = 58 - 58$$ (Subtracting both sides by 58. 72 - 58 = 14) $$14 -14y + 14y = 14y$$ (Adding 14y to both sides) $$14 \div 14 = 14y \div 14$$ (Dividing both sides by 14) $$y = 1$$ Now that we have solved for y, we can substitute y = 1 into equation (3) to solve for x. $$x = (24 - 3 \times 1) \div 7$$ (Replacing y with 1) $$x = (24 - 3)/7$$ $$(3 \times 1 = 3)$$ $$x = 21 \div 7$$ $$(24-3 = 21)$$ $$x = 3$$ $$(21 \div 7 = 3)$$ In conclusion, x = 3 and y = 1. To prove this we can substitute both equations into equation (1). $$7 \times 3 +3\times 1 = 24$$ $$21 + 3 = 24$$ $$24 = 24$$ Since both sides are equal, therefore our answers are correct and we have solved this question. Using the elimination method: When using the elimination method, our goal is to either subtract or add both equations so that we can eliminate one of the unknown variables. Eliminating x: Let us compare the coefficients of x and find the least common multiple between them. The coefficients of x are 7 and 21. We realize that 21 \div 7 = 3 which means that the least common multiple is 21. Now we want the coefficients of x in both equation to be equal to 21. To do that we will multiply the first equation by 3 and the second equation by 1. $$(7x + 3y = 24) \times 3$$ $$(21x - 5y = 58) \times 1$$ $$21x + 9y = 72$$ (Multiply the equations) $$21x - 5y = 58$$ $$21x - 21x + 9y - (-5y) = 72 - 58 $$ (Subtracting the equations) $$14y = 14$$ $$(9y - (-5y) = 14y$$ because of law of negation - Subtracting by a negative number is the same as adding the number) $$14y \div 14 = 14 \div 4$$ $$y = 1$$ Now that we have solved for y we can solve for x. Using the first equation we can substitute y=1. $$7x + 3 \times 1 = 24$$ $$7x + 3 -3 = 24-3$$ (Subtracting 3 from both sides) $$7x = 21$$ $$7x \div 7 = 21 \div 7$$ (Dividing both sides by 7) $$x = 3$$

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