Enable contrast version

# Tutor profile: Victor B.

Inactive
Victor B.
Experienced Tutor and Physician-in-Training!
Tutor Satisfaction Guarantee

## Questions

### Subject:Chemistry

TutorMe
Question:

If one-eighth of an original sample of krypton-74 remains unchanged after 34.5 minutes, what is the half-life of the isotope?

Inactive
Victor B.

The half-life is the amount of time that it takes for a sample of an element to lose half of its current composition. If there is one-eighth of a sample remaining, then three half-lives have passed (one-half remains after one half-life, one-fourth remains after two half-lives). So if three half-lives have passed after 34.5 minutes, then one half-life must be 11.5 minutes (34.5/3=11.5).

### Subject:Biology

TutorMe
Question:

The only nitrogenous base that is missing in RNA is: A Adenine B Guanine C Cytosine D Thymine E Uraci

Inactive
Victor B.

We know that Uracil replaces one of the bases (A, G, C, T) in RNA, but which one? We know that the two types of bases are purines and pyrimidines, and Uracil will most likely replace one of the bases that share its type. If you remember that pyramid(ines) are sharp and therefore "CUT," you know that "U" will probably replace either "C" or "T." Since C is first in the anagram, it must be super-important, so the one that is most likely to be replaced is "T." The answer is Thymine. Thymine is missing in RNA :)

### Subject:Algebra

TutorMe
Question:

Find the standard form of a circle that lies on three points: (2,1), (0,5), and (−1,2)

Inactive
Victor B.

First, we acknowledge the general equation for a circle can be written as: x^2 + y^2 + Cx + Dy + E = 0 From this, we create a system of three equations, one for each given point on the circle: (2,1): 2C + D + E = -5 (0,5): 5C + D = -25 (-1,2): C - 2D - E = 5 We can use substitution and/or elimination to solve the system of equations for the variables C, D, and E. In the interest of brevity, the solution will be: C = -2 D = -6 E = -5 We can substitute these values into the general equation of a circle in order to obtain: x^2 + y^2 -2x - 6y + 5 = 0 Next we group the variables in preparation of completing the square: (x^2 - 2x) + (y^2 - 6y) = -5 Finally, complete the square for both variables. (x^2 - 2x + 1) + (y^2 - 6y +9) = -5 + 1 + 9 Now factor to find the equation of the circle in standard form! (x -1)^2 + (y - 3)^2 = 5

## Contact tutor

Send a message explaining your
needs and Victor will reply soon.
Contact Victor