Destroy Baruch College MTH 1030 – Final Test FA18 In 30 Minutes Or LESS

1 Step 1

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

To get your hands on Baruch College MTH 1030 Last Minute Cram Course, CLICK HERE

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

Problem 1

Destroy Baruch College MTH 1030 – Final Test FA18 In 30 Minutes Or LESS (1)
$i^1=i$

$i^2=-1$

$i^3=-i$

$i^4=1$

Step 1.

Let’s split the following equation $i^{30}-i^{34}$ into two parts

$i^{30}=i^{28+2}=i^{28}\ast i^2=\left(i^4\right)^7\ast i^2=1^7\ast\left(-1\right)=-1$

$i^{34}=i^{32+2}=i^{32}\ast i^2=\left(i^4\right)^8\ast i^2=1^8\ast\left(-1\right)=-1$

Step 2.

$i^{30}-i^{34}=-1-\left(-1\right)=-1+1=0$

The answer is $C$ .

Problem 2

Destroy Baruch College MTH 1030 – Final Test FA18 In 30 Minutes Or LESS (2)
Equation of a circle with radius $R$ and origin $(h,k)$ is $\left(x-h\right)^2+\left(y-k\right)^2=R^2$ .

Step 1.

Rearrange the equation to move $x$ and $y$ terms together.

$x^2-6x+y^2-y=2$

$x^2-6x+__+y^2-y+__=2$

Step 2.

Find the coefficients to complete the square for $x$ and $y$ :

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

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

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

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

Origin $\left(h,k\right)=\left(3,\frac{1}{2}\right)$

$R^2=\frac{45}{4} and R=\frac{\sqrt{45}}{2}=\frac{3\sqrt5}{2}$

The answer is $C$ .

Problem 3

Destroy Baruch College MTH 1030 – Final Test FA18 In 30 Minutes Or LESS (3)

Step 1.

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

$5^{3x-1}=\left(\frac{1}{25}\right)^{2x-1}$

$5^{3x-1}=\left(5^{-2}\right)^{2x-1}$

$5^{3x-1}=5^{-4x+2}$

Step 2.

Get exponents equal to each other:

$3x-1=-4x+2$

$7x=3$

$x=\frac{3}{7}$

The answer is $A$ .

Problem 4

Destroy Baruch College MTH 1030 – Final Test FA18 In 30 Minutes Or LESS (4)

Step 1.

Let’s simplify this logarithm into separate elements:

$\ln{\frac{\sqrt[3]{x}y^5}{z^9}}=\ln{\sqrt[3]{x}y^5}-\ln{z^9}=\ln{\sqrt[3]{x}}+\ln{y^5}-\ln{z^9}=\ln{x^\frac{1}{3}}+\ln{y^5}-\ln{z^9}$

Step 2.

$\ln{x^\frac{1}{3}}+\ln{y^5}-\ln{z^9}=\frac{1}{3}\ln{x}+5\ln{y}-9\ln{z}$

The answer is $B$ .

Problem 5

Destroy Baruch College MTH 1030 – Final Test FA18 In 30 Minutes Or LESS (5)

Step 1.

Square both sides of the equation:

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

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

Step 2.

Move everything to one side and solve for $x$ :

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

$x^2-13x+36=0$

$x=4$ and $x=9$

Step 3.

Test both answers by plugging into the original equation:

$x=4$ :

$\sqrt x=6-x\rightarrow\sqrt4=6-4\rightarrow2=2$ . Thus, $x=4$ works!

$x=9$ :

$\sqrt9=6-9\rightarrow\sqrt9=6-9\rightarrow3=-3$ . Thus, we must reject $x=9$ .

The answer is $C$ .

1 Step 1

Problem 6

Destroy Baruch College MTH 1030 – Final Test FA18 In 30 Minutes Or LESS (6)

Step 1.

Flip $x$ and $y$ :

$x=\sqrt[3]{y+1}$

Step 2.

Solve for $y$ :

$[x]3=3y+13$

$x^3=y+1$

$y=x^3-1$

The answer is $B$ .

Problem 7

Destroy Baruch College MTH 1030 – Final Test FA18 In 30 Minutes Or LESS (7)

Step 1.

To find $g(f\left(x\right)) plug in f\left(x\right) for every x in function g\left(x\right)$ :

$f\left(x\right)=3x$

$g\left(x\right)=4x^2$

Thus, $g\left(f\left(x\right)\right)=4\ast\left(3x\right)^2=4\ast9x^2=36x^2$

The answer is $A$ .

Problem 8

Destroy Baruch College MTH 1030 – Final Test FA18 In 30 Minutes Or LESS (8)

Step 1.

To find the vertex, compute $x_{vertex}=-\frac{b}{2a} and then find f(x_{vertex})$ :

$f\left(x\right)=2x^2-4x-6$

$x_{vertex}=-\frac{-4}{2\ast2}=1$

$f\left(x_{vertex}\right)=2\ast1^2-4\ast1-6=2-4-6=-8$

$Vertex=(1,-8)$

Step 2.

To find $x$ -intercepts, set $f\left(x\right)=0$ and solve for $x$ :

$2x^2-4x-6=0$

$2(x^2-2x-3)=0$

$2\left(x-3\right)\left(x+1\right)=0\rightarrow x=3$ and $x=-1$

x-intercepts are $(3,0)$ and $(-1,0)$ .

Step 3.

To find $y$ -intercepts, plug in $x=0$ into $f\left(x\right)$ :

$f\left(0\right)=2\ast0^2-4\ast0-6=-6$

$y$ -intercept is $(0,-6)$

The answer is $A$ .

Problem 9

Destroy Baruch College MTH 1030 – Final Test FA18 In 30 Minutes Or LESS (9)

Step 1.

Set both top and bottom to $0$ :

$2-3x=0\rightarrow x=\frac{2}{3}$

$x^2-x=x(x-1)\rightarrow x=0$ and $x=1$

Step 2.

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

$----- 0 ----- \frac{2}{3} ----- 1 -----$

$f\left(-\frac{1}{2}\right)=\frac{2-3\left(-\frac{1}{2}\right)}{\left(-\frac{1}{2}\right)^2-\left(-\frac{1}{2}\right)}=\frac{2+\frac{3}{2}}{\frac{1}{4}+\frac{1}{2}}=\frac{3\frac{1}{2}}{\frac{3}{4}}\rightarrow(+)$

$f\left(\frac{1}{3}\right)=\frac{2-3\left(\frac{1}{3}\right)}{\left(\frac{1}{3}\right)^2-\left(\frac{1}{3}\right)}=\frac{2-1}{\frac{1}{9}-\frac{1}{3}}=\frac{1}{-\frac{2}{9}}\rightarrow(-)$

$f\left(\frac{3}{4}\right)=\frac{2-3\left(\frac{3}{4}\right)}{\left(\frac{3}{4}\right)^2-\left(-\frac{1}{2}\right)}=\frac{2+\frac{3}{2}}{\frac{1}{4}+\frac{1}{2}}=\frac{3\frac{1}{2}}{\frac{3}{4}}\rightarrow(+)$

$f\left(\frac{5}{4}\right)=\frac{2-3\left(\frac{5}{4}\right)}{\left(\frac{5}{4}\right)^2-\left(\frac{5}{4}\right)}=\frac{2-\frac{15}{4}}{\frac{25}{16}-\frac{5}{4}}=\frac{-\frac{7}{4}}{\frac{5}{4}}\rightarrow(-)$

Thus, the expression is less than or equal to 0 on the following intervals: $\left(0,\frac{2}{3}\right]\cup\left(1,\infty\right)$ .

The answer is $A$ .

Problem 10

Destroy Baruch College MTH 1030 – Final Test FA18 In 30 Minutes Or LESS (10)

Step 1.

Discriminant of $y=ax^2+b+c$ , $D=b^2-4ac$ can tell everything about roots:

$D={(-5)}^2-4\ast4\ast10=25-160=-135<0\rightarrow$ two complex Non-real answers.

The answer is $D$ .

1 Step 1

Destroy Baruch College MTH 1030 – Final Test FA18 In 30 Minutes Or LESS

Baruch College MTH 1030 – Final Test FA18 Overview

Problem 1

Step 1.

Step 2.

Problem 2

Step 1.

Step 2.

Problem 3

Step 1.

Step 2.

Problem 4

Step 1.

Step 2.

Problem 5

Step 1.

Step 2.

Step 3.

Problem 6

Step 1.

Step 2.

Problem 7

Step 1.

Problem 8

Step 1.

Step 2.

Step 3.

Problem 9

Step 1.

Step 2.

Problem 10

Step 1.

GET ALL OF OUR

VIDEO SOLUTIONS

FOR JUST $7.99/MONTH!

Share this content

Baruch College MTH 1030 – Final Test FA18 Overview

Problem 1

Step 1.

Step 2.

Problem 2

Step 1.

Step 2.

Problem 3

Step 1.

Step 2.

Problem 4

Step 1.

Step 2.

Problem 5

Step 1.

Step 2.

Step 3.

Problem 6

Step 1.

Step 2.

Problem 7

Step 1.

Problem 8

Step 1.

Step 2.

Step 3.

Problem 9

Step 1.

Step 2.

Problem 10

Step 1.

Share this content

You Might Also Like

TOP 10 MUST KNOW BARUCH MTH 2207 FINAL PROBLEMS

TOP 10 MUST KNOW BARUCH MTH 2205 FINAL PROBLEMS

Master Baruch College MTH 2003 – Final Test SP17 In 30 Minutes Or LESS

GET ALL OF OUR

VIDEO SOLUTIONS

FOR JUST $7.99/MONTH!