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Tutor profile: Neal L.

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Neal L.
Math Tutor
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Questions

Subject: Trigonometry

TutorMe
Question:

Using trigonometric identities, simplify $$\frac{\cos{t}}{1-\sin^2{t}}$$.

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Neal L.
Answer:

First, recall the Pythagorean identity $$\sin^2{t}+cos^2{t}=1$$. With this, $$cos^2{t}=1-\sin^2{t}$$. Substitute this into the original expression. $(\frac{\cos{t}}{1-\sin^2{t}}=\frac{\cos{t}}{\cos^2{t}}$) Simplify $$\frac{\cos{t}}{\cos^2{t}}$$. $(\frac{\cos{t}}{\cos^2{t}}=\frac{1}{\cos{t}}=\sec{t}$) Thus, the simplified expression is $$\sec{t}$$.

Subject: Calculus

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Question:

Find the slope of the function $$f(x)=a(x-b)^2(x-c)^2$$ at $$x=0$$ given that $$a,b,\text{and}~c$$ are constants.

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Neal L.
Answer:

To find the slope of a function at any given point, find its first derivative. $(f(x)=a(x-b)^2(x-c)^2$)$(f'(x)=\frac{d}{dx}[a(x-b)^2(x-c)^2]$)To differentiate $$a(x-b)^2(x-c)^2$$, use the product rule. $(\frac{d}{dx}[a(x-b)^2(x-c)^2]=a[\frac{d}{dx}(x-b)^2](x-c)^2+a(x-b)^2\frac{d}{dx}[(x-c)^2]$)$(\frac{d}{dx}[a(x-b)^2(x-c)^2]=a[2(x-b)(1)](x-c)^2+a(x-b)^2[2(x-c)(1)]$)$(\frac{d}{dx}[a(x-b)^2(x-c)^2]=2a(x-b)(x-c)^2+2a(x-b)^2(x-c)$)Factorize $$2a(x-b)(x-c)^2+2a(x-b)^2(x-c)$$. $(2a(x-b)(x-c)^2+2a(x-b)^2(x-c)=2a(x-b)(x-c)[(x-c)+(x-b)]$)$(2a(x-b)(x-c)^2+2a(x-b)^2(x-c)=2a(x-b)(x-c)(2x-b-c)$)Substitute $$x=0$$ into the expression $$2a(x-b)(x-c)(2x-b-c)$$.$(2a(0-b)(0-c)[2(0)-b-c]$)Simplify the expression. $(2a(0-b)(0-c)[2(0)-b-c]=2a(-b)(-c)[-b-c]=-2abc(b+c)$) Thus, the slope of $$f(x)$$ at $$x=0$$ is $$-2abc(b+c)$$.

Subject: Algebra

TutorMe
Question:

The sum of two numbers is $$-30$$ while their product is $$-1000$$. Find the sum of their squares.

Inactive
Neal L.
Answer:

Let $$x$$ and $$y$$ be the two numbers. Rewrite the given conditions and requirement algebraically. $(x+y=-30$)$(xy=-1000$)$(x^2+y^2=~?$)Note that, for convenience, the expression $$x^2+y^2$$ can be rewritten in terms of the expressions $$x+y$$ and $$xy$$. $(x^2+y^2=x^2+2xy+y^2-2xy$)$(x^2+y^2=(x+y)^2-2xy$)Substitute $$x+y=-30$$ and $$xy=-1000$$ into the expression $$(x+y)^2-2xy$$. $((x+y)^2-2xy=(-30)^2-2(-1000)$)Simplify the expression. $((-30)^2-2(-1000)=2900$)Thus, the sum of the squares of the two numbers is $$2900$$.

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