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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.

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.

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

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Question:

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

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Andre A.

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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