Alexis C.

Mathematics teacher in high school and college for over 10 years

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Pre-Calculus

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Question:

Obtain the center of the following ellipse: 9x^{2}+y^{2}-36x+2y+28=0

Alexis C.

Answer:

Since the ellipse is not in its standard form, we need to transform it. Just as a reminder, the standard form of an ellipse is : \frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1 ; where the center is (h,k). Now back to the problem. We need to convert it to its standard form. The first step is to complete the square, so we are going to set up the problem the following way: 9x^{2}-36x+(.)+ y^{2}+2y+(.)=-28 Going by parts, first were going to complete the square of the "x" section. In order to complete the square, we need the coefficient of x^{2} to be 1, so we are going to set it up in the form: 9(x^{2}-4x+[ .]). Ignoring the 9 on the outside for now, we can complete the square with the rule that says that you take half of the x coefficient and then square it, so (\frac{4}{2})^{2} = 4. Therefore the x part is going to become 9(x^{2}-4x+4), or 9(x^{2}-2)^{2}. Now with the "y" section. The coefficient on the y^{2} is already 1, so we can complete the square directly, making it y^{2}+2y+1. Now, whatever we added on the left side, also has to be added on the right side. On the left side we added a 36 (Why 36?. Because the 4 that we added has a 9 multiplying it) and a 1 for a total of 37. This gives us 9(x^{2}-2)^{2}+(y+1)^{2}=-28+37 ; 9(x^{2}-2)^{2}+(y+1)^{2}=9 Finally, we need to divide everything by 9 since the standard form is equal to 1 on the right side. The equation is (x^{2}-2)^{2}+\frac{(y+1)^{2}}{9}=1 With this we can determine that the center is at (2,-1)

Calculus

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Question:

Solve the following integral: \int 9x^{3}\sqrt{6x^{4}+2}dx

Alexis C.

Answer:

In all integrals, first lets remove the constant if there is one. In this case, we can put the 9 outside of the integral to simplify things, leaving us with 9\int x^{3}\sqrt{6x^{4}+2} dx Now, this integral can´t be solved directly, therefore we need to use the u-substitution method. This method consists of creating a u and a du (derivative of that same u) and then substituting these values into the integral in order to make it "solvable". We need to think what wer are going to make "u" equal to. The whole idea is that "u" is going to be part of the original integral, and "du" is going to contain the rest of the original integral. Usually, when there is a sqrare root involve, it is best to set u equal to whatever is inside this root. We are going to se u = 6x^{4}+2. Then du is going to be the derivative of u; du = 24x^{3}dx. Now we need to se if susbtituting u and du into the original equation takes care of everything or if there is something extra or missing. We notice that u does include everything inside the square root and du does include everything else (x^{3} and dx) except for the 24 that is not in the original integral. Therefore we need to remove this 24 by putting it as a division on the left side, therefore leaving us with: \frac{du}{24} = x^{3}dx Now we are ready to substitute these new variables into the original integral. The \frac{1}{24} will go outside of the integral since it´s a constant. Since we already have a 9 out there, the constant outside of the integral will be \frac{9}{24}. This will leave us with the integral : \frac{9}{24} \int \sqrt{u} This is the same as \frac{9}{24}\int u^{\frac{1}{2}}du. The integral of u\frac{1}{2} is \frac{u^{\frac{3}{2}}}{\frac{3}{2}}, which is equal to \frac{2u^{\frac{3}{2}}}{3}. So, now we have \frac{9}{24}\left [ \frac{2u^{\frac{3}{2}}}{3} \right ] du Multiplying and simplifying the constants, we are left with \frac{1}{4}u^{\frac{3}{2}} + C Now we just need to put the answer back into the original terms since the original integral didn´t have a "u" variable. Since we had said previously that u =6x^{4}+2, our final answer is going to be: \frac{1}{4}(6x^4+2)^{\frac{3}{2}}+C

Algebra

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Question:

John wants to buy a new computer. The model that he likes costs $300. His job at a restaurant earns him $40 a day. If he currently has $60 saved in his bank account, what is the minimum amount of days that John must work in order to be able to buy the computer that he wants?

Alexis C.

Answer:

Since the computer costs $300, we know that John must have at least that amount available. Therefore, it must be true that his total money (saved plus earned) must be \geq 300. His total money is the amount he currently has saved ($60) plus the amount he will have by working ($40 per day). Putting the following two paragraphs together, we have the equation in the form: money saved + money earned \geq 300. Money saved is just the $60 in his bank account. Money earned is the $40 per day. Since this amount depends on the number of days worked, we are going to represent it as $40d, where "d" represents days. Therefore we have: $60 + $40d \geq $300. Nos using algebra, we want to solve for d (number of days needed to work). Since we have $60 on he left side, we can transfer it over the right side with the opposite sign: $40d \geq $300 - $60 $40d \geq $240; now we will divide by $40 in order to have the "d" all by itself. d \geq $240 / $40; Therefore d \geq 6. John has to work at least 6 days in order to have enough money to purchase the computer.

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