TABLE OF CONTENTS

## Baruch College MTH 1030 – Final Test FA18 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__.

## Problem 1

### Step 1.

Let’s split the following equation into two parts

### Step 2.

The answer is .

## Problem 2

Equation of a circle with radius and origin is .

### Step 1.

Rearrange the equation to move and terms together.

### Step 2.

Find the coefficients to complete the square for and :

Origin

The answer is .

## Problem 3

### Step 1.

Let’s get both parts of the equation to the same base:

### Step 2.

Get exponents equal to each other:

The answer is .

## Problem 4

### Step 1.

Let’s simplify this logarithm into separate elements:

### Step 2.

The answer is .

## Problem 5

### Step 1.

Square both sides of the equation:

### Step 2.

Move everything to one side and solve for :

and

### Step 3.

Test both answers by plugging into the original equation:

:

. Thus, works!

:

. Thus, we must reject .

The answer is .

## Problem 6

### Step 1.

Flip and :

### Step 2.

Solve for :

The answer is .

## Problem 7

### Step 1.

To find :

Thus,

The answer is .

## Problem 8

### Step 1.

To find the vertex, compute :

### Step 2.

To find -intercepts, set and solve for :

and

x-intercepts are and .

### Step 3.

To find -intercepts, plug in into :

-intercept is

The answer is .

## Problem 9

### Step 1.

Set both top and bottom to :

and

### Step 2.

Mark all of these points on a number line and test them:

Thus, the expression is less than or equal to 0 on the following intervals: .

The answer is .

## Problem 10

### Step 1.

Discriminant of , can tell everything about roots:

two complex Non-real answers.

The answer is .