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# Tutor profile: Rohan S.

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Rohan S.
Classroom Teacher and 1v1 Math Tutor
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## Questions

### Subject:Pre-Calculus

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Question:

Verify the following Identity. You may only manipulate the Left Hand side. \$\$(cosx)/(1+sinx) + (1+sinx)/(cosx) = 2secx\$\$

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Rohan S.

Since we are only allowed to play around with the Left Hand side, let us begin with our focus there. We see a sum of 2 fractions, and both of those fractions seem fully simplified, since we can not possibly factor or reduce either. So let's go ahead and combine the two fractions, using Least Common Denominator. Since the 2 fractions have distinct denominators with nothing in common, we can safely say the LCD is the product of the two. So, our first step is to multiply each denominator to the other fraction (Top and Bottom) as shown: \$\$((cosx)(cosx))/((cosx)(1+sinx)) + ((1+sinx)(1+sinx))/(1+sinx)(cosx)\$\$ By multiplying across, we get \$\$(cos^2x+1+2sinx+sin^2x)/((cosx)(1+sinx))\$\$ In the numerator, we clearly see a sum of \$\$cos^2x\$\$ and \$\$sin^2x\$\$, allowing us to use the Pythagorean Identity \$\$cos^2x+sin^2x = 1\$\$. This leaves us with \$\$(1+1+2sinx)/((cosx)(1+sinx)\$\$, and eventually \$\$(2+2sinx)/((cosx)(1+sinx)\$\$. By factoring out a 2 in the numerator, we are left with \$\$2(1+sinx)/((1+sinx)(cosx)\$\$ allowing us to reduce the fraction by cancelling out the \$\$1+sinx\$\$ out, and leaving us with \$\$2/(cosx)\$\$. We know that \$\$secx\$\$ is the reciprocal of \$\$cosx\$\$, allowing us to say that \$\$2/cosx\$\$ can be rewritten as \$\$2secx\$\$, proving our Identity true.

### Subject:Geometry

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Question:

A Frustum has a small radius of 4in, a big radius of 8in and a height of 10in. Find the Volume of the Frustum.

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Rohan S.

A Frustum is a somewhat irregular shape that is made when a smaller volume in the shape of a cone is cut out from the top of a larger cone. Real-Life examples of a Frustum include a lamp, a bucket or a traffic cone, for a better idea of imagining the shape itself. In order to find the volume of the entire Frustum, we would need to look at the volume of the larger cone, and then subtract the volume of the smaller cone. To start things off, we need to find the heights of both the smaller and larger cones. The "height" of 10in is neither the height of the larger or smaller cones. The sum of 8 and the smaller height makes the height of the larger cone. In order to find both heights, we need to set up a proportionality system, since both cones are similar. \$\$(x/4) = ((x+10)/8)\$\$. Where x is the small height and "x+10" is the larger height. By cross multiplying, we get \$\$8x = 4x+40\$\$. Using simple Algebra, we get x to equal 10. So we know, our small height is 10in and our large height is 20in. Now, since we have our radii and heights. We have enough information to solve for the volume of both cones as we know the volume of a cone is \$\$1/3πr^2h\$\$ Larger Cone = \$\$(1/3)π(8^2)(20)\$\$ If we calculate correctly, we should get \$\$1340.4in^2\$\$ Smaller Cone = \$\$(1/3)π(4^2)(10)\$\$ If we calculate correctly, we should get \$\$167.6in^2\$\$ Final Volume = Larger Volume - Smaller Volume = \$\$1340.4 - 167.6 = 1172.8in^2\$\$

### Subject:Algebra

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Question:

Using Polynomial Division, simplify the quotient: \$\$(12x+18x^3-8) / (3x^2+2)\$\$

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Rohan S.

When it comes to Polynomial Division, we have to keep in mind that it is usually something we simplify algebraic expressions with, if the numerator is not factorable. In this problem, that is exactly the case. We're going to start off with re arranging our numerator in Standard Form. Have all of our terms rewritten with the greatest degree leading the expression, and the constant at the tail end, giving us \$\$18x^3+12x-8\$\$. Next step is where the actual Division happens. It is in a way similar to the Long Division we've been doing since Elementary School, in that, we are looking for a factor that multiplies with our Divisor to equal our Dividend. Our main focus is always the first term in each. Our first term in the Divisor is \$\$3x^2\$\$ and our first term in the Dividend is \$\$18x^3\$\$. We know that 3 multiplies with 6 to equal 18, and we know that \$\$x^2\$\$ multiplies with only \$\$x\$\$ to equal \$\$x^3\$\$. So that gives us the first term of our quotient "\$\$6x\$\$". Now when we multiply \$\$6x\$\$ to \$\$3x\$\$, we have to keep in mind that the entire Divisor consists of two terms, meaning we will have to distribute it out in order to go further, giving us \$\$18x^3+12x\$\$. Luckily, this worked out pretty nicely for us since the second term of the quotient worked itself out! We will now go ahead and subtract this product from our Dividend. \$\$(18x^3+12x-8)-(18x^3+12x)\$\$ gives us just -8. Now, if you notice, we can not go further than here, as we are only left with a constant, and there is nothing that multiplies with \$\$3x^2\$\$ to equal -8. Our final answer, including our remainder will be written as \$\$6x-8/(3x^2+2)\$\$.

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