TOP 10 MUST KNOW BARUCH MTH 1030 FINAL PROBLEMS

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Baruch MTH 1030 Final Overview

MTH 1030 is an algebra course in Baruch College. This course covers most basic quantitative courses at the college, including linear equations, rates of change, rational expressions, circles, functions and their graphs, inverse functions, exponential and logarithmic functions, the geometric series, an introduction to annuities, non-linear systems of equations and related applications. Therefore, the students are required to master TI 89 or TI 92 graphical calculator for this course.

To view the syllabus for Baruch College MTH 1030 CLICK HERE

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To view sample final exam questions for Baruch College MTH 1030 CLICK HERE.

Baruch MTH 1030 Final Topic 1. Exponents and Polynomials.

Rule 1.

Turn both parts of the equation to the same base.

Rule 2.

Set exponents on both sides equal to each other and solve for x.

MTH 1030, Sample Final FA16, Problem 10

TOP 10 MUST KNOW BARUCH MTH 1030 FINAL PROBLEMS (1)
We need to solve for x from the following equation 8^{x-1}=4^{2x-3}

Step 1.

Get the same base on both sides:

2^{3(x-1)}=2^{2(2x-3)}

2^{3x-3}=2^{4x-6}

Step 2.

Set exponents on both sides equal to each other and solve for x:

3x-3=4x-6

x=3

The answer is D.

Baruch MTH 1030 Final Topic 2. Circle.

Equation of a circle with radius R and origin (h,k) is \left(x-h\right)^2+\left(y-k\right)^2=R^2. Complete the square for x and y to turn the equation into the form \left(x-h\right)^2+\left(y-k\right)^2=R^2.

MTH 1030, Sample Final FA18, Problem 2

TOP 10 MUST KNOW BARUCH MTH 1030 FINAL PROBLEMS (2)
In Problem 2 of the Final Test FA18 we need to find the center and the radius of the circle x^2-6x+y^2-y=2

Step 1.

Complete the squares for x and y:

x^2-6x+\left(\frac{-6}{2}\right)^2+y^2-y+\left(\frac{1}{2}\right)^2=2+\left(\frac{-6}{2}\right)^2+\left(\frac{1}{2}\right)^2

x^2-6x+9+y^2-y+\frac{1}{4}=2+9+\frac{1}{4}

\left(x-3\right)^2+\left(y-\frac{1}{2}\right)^2=11\frac{1}{4}

\left(x-3\right)^2+\left(y-\frac{1}{2}\right)^2=\frac{45}{4}

The center is \left(3,\frac{1}{2}\right) and the radius is \sqrt{\frac{45}{4}}=\frac{\sqrt{45}}{2}=\frac{3\sqrt5}{2}

The answer is C.

Baruch MTH 1030 Final Topic 3. Rationalizing.

Rationalizing is often connected to the concept of conjugate.

The conjugate of a+b is a-b.

The conjugate of a-b is a+b.

MTH 1030, Sample Final FA16, Problem 3

TOP 10 MUST KNOW BARUCH MTH 1030 FINAL PROBLEMS (3)
In Problem 3 of the Final Test FA16 we need to rationalize the denominator of the following fraction \frac{\sqrt3+3}{\sqrt3-1}

Step 1.

It is clear that in order to rationalize the denominator, we need to identify the conjugate of the denominator and multiply both top and bottom by that very conjugate.

Denominator is \sqrt3-1.

The conjugate to the denominator is \sqrt3+1. Let’s multiply both top and bottom by \sqrt3+1:

\frac{\sqrt3+3}{\sqrt3-1}\ast\frac{\sqrt3+1}{\sqrt3+1}=

\frac{(\sqrt3+3)(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}

Step 2.

Multiply out both top and bottom and simplify:

\frac{(\sqrt3+3)(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}=\frac{3+\sqrt3+3\sqrt3+3}{3+\sqrt3-\sqrt3-1}=\frac{6+4\sqrt3}{2}=3+2\sqrt3

The answer is C.

Baruch MTH 1030 Final Topic 4. Complex Numbers.

Problems containing fractions with complex numbers are often solved in similar fashion described in the previous topic of rationalizing.

The conjugate of a+b is a-b.

The conjugate of a-b is a+b.

Also, let’s not forget that i^2=-1

MTH 1030, Sample Final FA16, Problem 4

TOP 10 MUST KNOW BARUCH MTH 1030 FINAL PROBLEMS (4)
In Problem 4 of the Final Test FA16 we need to rationalize the denominator of the following fraction \frac{12-5i}{2+3i}

Step 1.

It is clear that in order to rationalize the denominator, we need to identify the conjugate of the denominator and multiply both top and bottom by that very conjugate.

Denominator is 2+3i.

The conjugate to the denominator is 2-3i. Let’s multiply both top and bottom by 2-3i:

\frac{12-5i}{2+3i}\ast\frac{2-3i}{2-3i}=

\frac{(12-5i)(2-3i)}{(2+3i)(2-3i)}

Step 2.

Multiply out both top and bottom and simplify:

\frac{(12-5i)(2-3i)}{(2+3i)(2-3i)}=\frac{24-36i-10i+15i^2}{4-6i+6i-9i^2}=\frac{24-46i-15}{4+9}=\frac{9-46i}{13}=\frac{9}{13}-\frac{46i}{13}

The answer is E.

Baruch MTH 1030 Final Topic 5. Domain and Range.

Domain can be found by examining the denominator of a fraction.

Step 1.

Set the denominator to 0 and solve for x

Step 2.

Exclude all solutions from Step 1. The remainder of x values will be included in the domain.

MTH 1030, Sample Final FA16, Problem 1

TOP 10 MUST KNOW BARUCH MTH 1030 FINAL PROBLEMS (5)
In Problem 1 of the Final Test FA16 we need to find the domain of the fraction \frac{2x^2+2x}{x^2-4x}

Step 1.

Let’s set the denominator to 0 and solve for x:

x^2-4x=

x\left(x-4\right)=0\rightarrow x=0 and x=4

Step 2.

We need to know exclude x=0 and x=4 from the set of all values.

Thus, the domain is (-\infty,0)\cup(0,4)\cup(4,\infty).

The answer is A.

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Baruch MTH 1030 Final Topic 6. Completing the Square.

To solve a quadratic equation by completing the square follow these steps:

Step 1.

Complete the square

Step 2.

Take square root of both sides and solve for x

MTH 1030, Sample Final FA18, Problem 12

TOP 10 MUST KNOW BARUCH MTH 1030 FINAL PROBLEMS (6)
In Problem 12 of the Final Test FA18 we need to solve the quadratic equation x^2+8x=7 by completing the square:

Step 1.

Complete the square first

x^2+8x=7

x^2+8x+\left(\frac{8}{2}\right)^2=7+\left(\frac{8}{2}\right)^2

x^2+8x+16=7+16

\left(x+4\right)^2=23

Step 2.

Take square root of both sides and solve for x

\sqrt{\left(x+4\right)^2}=\pm\sqrt{23}

x+4=\pm\sqrt{23}

x=-4\pm\sqrt{23}

The answer is E.

Baruch MTH 1030 Final Topic 7. Radical Equations.

To solve a radical equation, follow these steps:

Step 1.

Move radical part of the equation to one side and everything else to the other side

Step 2.

Square both sides of the equation, move everything to one side and solve for x

Step 3.

Make sure to double check all answers from Step 2 by plugging into the original equation

MTH 1030, Sample Final FA18, Problem 5

TOP 10 MUST KNOW BARUCH MTH 1030 FINAL PROBLEMS (7)
In Problem 5 of the Final Test FA18 we need to solve the radical equation \sqrt x=6-x:

Step 1.

The equation is already properly prepared. The radical part of the equation is already on the left-hand side and the rest is on the right-hand side

Step 2.

Let’s now square both sides of the equation and solve for x:

\sqrt x=6-x

\left(\sqrt x\right)^2=\left(6-x\right)^2

x=36-12x+x^2

x^2-12x+36-x=0

x^2-13x+36=0

x^2-13x+36=0

\left(x-4\right)\left(x-9\right)=0\rightarrow x=4 and x=9.

Step 3.

It is time to double check whether x=4 and x=9 are qualified answers.

When x=4: \sqrt4=6-4\rightarrow2=2. Thus, x=4 works.

When x=9: \sqrt9=6-9\rightarrow-3=3. Thus, x=9 does not work.

The answer is C.

Baruch MTH 1030 Final Topic 8. Rational Exponents.

Rational exponential expressions become easy with the following rules:

Rule 1:

x^a\ast x^b=x^{a+b}

Rule 2:

\frac{x^a}{x^b}=x^{a-b}

Rule 3:

\left(x^a\right)^b=x^{ab}

Rule 4:

x^{-n}=\frac{1}{x^n}

MTH 1030, Sample Final FA16, Problem 11

TOP 10 MUST KNOW BARUCH MTH 1030 FINAL PROBLEMS (8)
In Problem 11 of the Final Test FA16 we need to simplify \left(\frac{8xy^5}{2x^5y^3}\right)^{-\frac{1}{2}}:

Step 1.

Let’s simplify the expression within the brackets first:

\frac{8xy^5}{2x^5y^3}=\frac{8x^1y^5}{2x^5y^3}=\frac{4y^2}{x^4}=\frac{2^2y^2}{x^4}

Step 2.

Let’s now incorporate the outside exponent:

\left(\frac{2^2y^2}{x^4}\right)^{-\frac{1}{2}}=\frac{\left(2^2\right)^{-\frac{1}{2}}\ast\left(y^2\right)^{-\frac{1}{2}}}{\left(x^4\right)^{-\frac{1}{2}}}=\frac{2^{-1}y^{-1}}{x^{-2}}=\frac{x^2}{2^1y^1}=\frac{x^2}{2y}

The answer is D.

Baruch MTH 1030 Final Topic 9. Logarithmic Exponents.

Logarithmic equations are also quite easy with the following rules:

Rule 1:

\log{a}+\log{b}=\log{ab}

Rule 2:

\log{a}-\log{b}=\log{\frac{a}{b}}

Rule 3:

\log{a^n}=n\log{a}

MTH 1030, Sample Final FA18, Problem 18

TOP 10 MUST KNOW BARUCH MTH 1030 FINAL PROBLEMS (9)
In Problem 18 of the Final Test FA18 we need to solve for x from the following equation:

\log{(x+2)}+\log{(x+4)}=\log{3}

Step 1.

Let’s combine logarithms together:

\log{(x+2)(x+4)}=\log{3}

Step 2.

We can now set expressions within logarithms equal to one another and solve for x:

\left(x+2\right)\left(x+4\right)=3

x^2+2x+4x+8=3

x^2+6x+8=3

x^2+6x+5=0

\left(x+1\right)\left(x+5\right)=0\rightarrow x=-1 and x=-5.

Step 3.

Let’s test both answers by plugging them back into the original equation:

x=-1:

\log{(-1+2)}+\log{(-1+4)}=\log{1}+\log{3}=\log{3}

Thus, x=-1 works.

x=-5:

\log{(-5+2)}+\log{(-5+4)}=\log{-3}+\log{-1}=\emptyset. There are no solutions, since expression inside the logarithm must be always greater than 0.

Thus, x=-5 must be rejected.

The final answer is A.

Baruch MTH 1030 Final Topic 10. Inverse Functions.

The following two steps will help you solve any Inverse Function problem.

Step 1.

y=g(x) and x and y values need to be inter-changed.

Step 2.

Solve for y

Let’s show how inverse problems can be easily solved with the aforementioned two steps.

MTH 1030, Sample Final FA16, Problem 23

TOP 10 MUST KNOW BARUCH MTH 1030 FINAL PROBLEMS (10)
The inverse of g\left(x\right)=\frac{2x-2}{x+6} should start with the realization that g\left(x\right)=y

Re-writing the equation y=\frac{2x-2}{x+6} always helps with the inverse functions.

Step 1.

Replacing x and y values results in the following:

x=\frac{2x-2}{x+6}

Step 2.

Solve for y:

\frac{x}{1}=\frac{2y-2}{y+6}

x\ast\left(y+6\right)=2y-2

xy+6x=2y-2

xy-2y=-6x-2

y(x-2)=-6x-2

y=\frac{-6x-2}{x-2}

Step 3.

Let’s now plug in x=4:

y\left(4\right)=\frac{-6\ast4-2}{4-2}=\frac{-26}{2}=-13

The answer is B.

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