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# Tutor profile: Joseph B.

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Joseph B.
Math and Science tutor for 2 years
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## Questions

### Subject:Linear Algebra

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Question:

Find the solutions for the given system $$x+2y=1$$ $$3x+2y+4z=7$$ $$-2x+y-2z = -1$$

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Joseph B.

We can solve this using a mixture of Gaussian elimination and substitution First we will multiply row 1 by -3 and add it to row 2 $$-3[x+2y] +[3x +2y+4z] = -3(1) + 7$$ Then multiply row 1 by 2 and add it to row 3 $$2[x+2y] + [-2x +y - 2z] = 2(1) + (-1)$$ Which gives us a new system: $$x+2y = 1$$ $$-4y +4z =4$$ $$5y -2z = 1$$ We can divide row 2 by 4 $$x+2y = 1$$ $$-y +z =1$$ $$5y -2z = 1$$ We will repeat the above process using row 2, multiplying it by 2 and 5, then adding to row 1 and row 3 respectively. After doing so our new system should look like this: $$x = -1$$ $$-y +z =1$$ $$3z = 6$$ Solving row 3 gives us $$z = 2$$, we can see that $$x=-1$$, and substituting $$z=2$$ into row 2 and solving tells us that $$y=1$$

### Subject:Trigonometry

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Question:

Using double angle identities, prove that $$1-\sin^{2}{x} = \cos^{2}{x}$$

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Joseph B.

We will start with the left side of the equation, and work it through to equivalency. First we will use a double angle identity $$\cos(2x) = \cos^{2}(x)-\sin^{2}(x)$$ then replace the $$\sin^{2}(x)$$ $$1-[\cos^{2}(x)-\cos(x)]$$ Then using another double angle identity $$\cos(2x) = 2\cos^{2}(x) - 1$$ we make another substitution $$1 -[\cos^{2}(x) - (2\cos^{2}(x) - 1)]$$ Redistribute and combine like terms $$1-[\cos^{2}(x) -2\cos^{2}(x) +1]$$ $$1-[-\cos^{2}(x) +1]$$ $$\cos^{2}(x)$$ Thus $$1 -\sin^{2}(x) = \cos^{2}(x)$$

### Subject:Calculus

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Question:

Find the foci, vertices, center, eccentricity, and asymptotes of the conic section: $$9x^{2} - 16y^{2} -36x -32y - 92 = 0$$ (This is a calculus III problem)

Inactive
Joseph B.

First we will want to rearrange the function in the form of: $$\frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1$$ To do so we reorganize the function as so: $$9x^{2} - 36x -16y^{2} - 32y = 92$$ From here we factor, then complete the square: $$9(x^{2} - 4x) -16 (y^{2} -2y) = 92$$ $$\frac{(x-2)^{2}}{9} + \frac{(y+1)^{2}}{16} = 1$$ Looking at this equation we can see that the center is at (h,k) = (2, -1) We also know that the verticies are $$a$$ distance away, symmetric about the center = (-1,-1), (5, -1) Similarly the foci are $$b$$ distance away, symmetric about the center = (-3,-1), (7,-1) Eccentricity is found using the formula $$\frac{\sqrt{a^{2}+b^{2}} }{a}$$ = $$\frac{5}{3}$$ Lastly, using $$c^{2} = a^{2} + b^{2}$$ we find that the equations of the asymptotes to be $$y = \frac{4x}{3}-\frac{11}{2}$$ and $$y = \frac{5}{3} - \frac{4x}{3}$$

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