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# Tutor profile: Christian F.

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Christian F.
Mathematics Teacher for 7 years
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## Questions

### Subject:Basic Math

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Question:

calculate $$. 33 \times .59$$

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Christian F.

- To solve a question of this nature, we take note of the number of decimal places (4, that's 2 each) - Next we multiply the two as whole numbers: i. e $$33 \times 59 = 1947$$ - Finally we count back the number of decimal places recorded above and fix the decimal point to the left: ==> $$. 33 \times .59 = .1947$$

### Subject:Calculus

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Question:

Find the derivative of $$g(y)=(y^2−4) \sin(2y)$$

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Christian F.

To solve a problem of this nature, we have to recognise that $$g(y)$$ is a product of two functions: $$y^2 - 4$$ and $$\sin (2y)$$. Hence we make use of the product rule: - we introduce two intermediate functions u and v such that: $$u = y^2 - 4$$ and $$v = \sin (2y)$$ then we differentiate them. - To differentiate u, we use the rule of power functions... i.e $$\frac {d}{dy} (y^n) = n (y^{n-1})$$ ==> $$\frac {du}{dy} = 2y$$ - To differentiate v, we make use of the chain rule for composite functions i. e $$\frac {dy}{dx} = \frac {dy}{df} . \frac {df}{dx}$$ where y is the "outer" function and f is the "inner" function. ==> $$\frac {dv}{dy} = 2 \cos(2y)$$ - the product rule states that if $$y(x) = u(x) . v(x)$$ then $$\frac {dy}{dx} = v \frac {du}{dx} + u \frac {dv}{dx}$$ Since we have all our functions ( u and v) with their differentials $$\frac {du}{dx} and \frac {dv}{dx}$$ we can substitute them in the rule to have: ==> $$\frac {dy}{dx} = (y^2 - 4) (2 \cos(2y)) + (\sin (2y)) (2y)$$

### Subject:Algebra

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Question:

One positive number is 8 times another positive number, and the difference between them is 63. what are the 2 numbers?

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Christian F.

Dealing with problems of this nature, it is helpful to use variables (or letters) to keep track of the numbers we don't know and use them to create an equation from each given condition. Hence we proceed thus: - let the two numbers be assigned letters 'a' and 'b'. - the first condition says "One number is 8 times the other" : ==> a = 8b ........................................ eq 1 - the second condition says "the difference between them is 63: ==> a - b = 63 ................................... eq 2 Now we have derived two equations from the conditions and all we need to do is solve them by substituting "8b" from "eq1" in the place of "a" in "eq2" since they are equal: ==> 8b - b = 63 ...... this is an equation consistent with the two above and has just 'b' as the unkown which can be found. ==> 7b = 63 ....... to get the value of b, we divide both sides of this equation by 7. ==> b = 9. - to get the value of 'a' , we substitute the value of b in 'eq1' : ==> a = 8 * 9 = 72. Therefore a = 72 and b = 9.

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