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Pre Calculus Homework Help

  • Our company offers quality PreCalculus Homework Help.

  • We offer personalized help, for any level

  • We can provide fully detailed step-by-step solutions

  • Our rate starts $25/hour. We provide a free quote in hours. Quick turnaround!

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For you guys taking high school or pre-calculus classes we have a lot to offer. We know how challenging the transition from high school to college can be. There's no need to struggle with incomplete study material.

One of the things students always complain is that they don't have enough solved problems to study with. With us, that's not a problem anymore!

Submit by e-mail your problems to us for a free quote.

Having a well explained list of solved problems can improve your learning 100 percent. Regrettably, current textbooks come mostly with the solutions to the sample problems, but they don't show detailed solutions.

Make sure to visit our store. There you'll find study guides for sale with hundreds of solved problems on pre-calculus, trigonometry, algebra, etc.

Pre Calculus Help

This is a list of subjects you may be interested on

  • Algebra
    • Systems of Equations.
    • Word problems.
    • Factorization.
    • Polynomials
    • Quadratic Equations

  • Geometry
    • Triangles.
    • Areas and Volumes.
    • Circles.
    • Lines
    • Planes

  • Trigonometry
    • Circles.
    • Sine, Cosine, Tangent, Cotangent, Secant, Cosecant.
    • Pythagoras Theorem.
    • De Moivre's Formula.
    • etc....

We know how it feels to deal with those hard problems. With our help, you'll get that extra edge you need! No need to sweat with poorly explained problems from the books. We can provide solutions that suit your own needs. See some examples:

Problem 1. Solve the following system of equations:

$\displaystyle 3x+2y=11

$\displaystyle 5x-4y=11

Solution. We have to find $ x$ and $ y $ in such a way that (1) and (2) are satisfied simultaneously. The method consists on eliminating one of the variables using both equation, multiplying them by the right constants. More precisely, if we multiply the first equation by 2 we obtain

$\displaystyle \tag{3}

$\displaystyle \tag{4}

and now we sum both equations (3) and (4) to get $ 11x=33$ , which means that $ x=3$ . Now that we have $ x$ , we can find $ y $ simply replacing any of the equations (1) or (2). We choose (1), so we get $ 9+2y=11$ , which implies that $ 2y=2$ , and then $ y=1$ . Summarizing

$\displaystyle x=3 \quad , \quad y=1$

is the solution.$ \Box$

Problem 2. Rationalize the denominator

$\displaystyle \frac{\sqrt{2}}{\sqrt 3-\sqrt 2} $

Solution. We have the factor $ \sqrt 3-\sqrt 2$ in the denominator, so we multiply both the numerator and denominator by the conjugate $ \sqrt 3+\sqrt 2$ , to obtain:

$\displaystyle \frac{\sqrt{2}}{\sqrt 3-\sqrt 2}= \frac{\sqrt{2} (\sqrt 3+\sqrt 2...
...rt 3+\sqrt 2)}
= \frac{\sqrt 2 \sqrt 3 + (\sqrt 2)^2}{(\sqrt 3)^2-(\sqrt 2)^2} $

$\displaystyle = \frac{\sqrt 2 \sqrt 3 + (\sqrt 2)^2}{3-2}=\sqrt 6-2 $

This way, we have found an equivalent expression with no radicals at the denominator, i.e.

$\displaystyle \frac{\sqrt{2}}{\sqrt 3-\sqrt 2}=\sqrt 6-2 $

but this last expression is rationalized.$ \Box$

Check more examples here.

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