Dismantle Baruch College CSTM 0120 – Final Test F019A In 20 Minutes Or LESS

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Baruch College CSTM 0120 – Final Test F019A Overview

CSTM 0120 is an prealgebra course in Baruch College. This is a non-credit course containing those topics from intermediate algebra that are needed prior to taking MTH 1030, College Algebra.

To view the syllabus for Baruch College CSTM 0120 CLICK HERE

To get your hands on Baruch College CSTM 0120 Last Minute Cram Course, CLICK HERE

To view sample final exam questions for Baruch College CSTM 0120 CLICK HERE.

Baruch College CSTM 0120 Final Test F019A Problem 1

Baruch College CSTM 0120 – Final Test F019A In 20 Minutes Or LESS (1)

Step 1.

Factor $x^2-9=(x-3)(x+3)$

Step 2.

Factors $x^2-64x=x(x-64)$

Thus, we have the following factors: $\left(x-3\right),\ \ \left(x+3\right),\ x\ and\ (x-64)$

The answer is $D$ .

Baruch College CSTM 0120 Final Test F019A Problem 2

Baruch College CSTM 0120 – Final Test F019A In 20 Minutes Or LESS (2)

Step 1.

Factor the greatest common monomial factor $315x^6+18x^5-9x^4=9x^4(35x^2+2x-1)$

Step 2.

Let’s now factor the quadratic equation $35x^2+2x-1$ :

$D=b^2-4ac=2^2-4\ast35\ast\left(-1\right)=4+140=144$

$x_{1,2}=\frac{-b\pm\sqrt D}{2a}=\frac{-2\pm\sqrt{144}}{2\ast35}=\frac{-2\pm12}{70}$

$x_1=\frac{-2+12}{70}=\frac{10}{70}=\frac{1}{7}$

$x_2=\frac{-2-12}{70}=\frac{-14}{70}=-\frac{1}{5}$

Step 3.

Plug in the answers in the formula $ax^2+bx+c=a(x-x_1)(x-x_2)$

$9x^4\left(35x^2+2x-1\right)=9x^4\ast35\left(x-\frac{1}{7}\right)\left(x+\frac{1}{5}\right)=9x^4\left(7x-1\right)\left(5x+1\right)$

The answer is $C$ .

Baruch College CSTM 0120 Final Test F019A Problem 3

Baruch College CSTM 0120 – Final Test F019A In 20 Minutes Or LESS (3)

Step 1.

Find the greatest common factor for the constants $16$ and $72: 8$

Step 2.

Find the greatest common factor for the $x$ -terms $x^5and\ x^3: x^3$

Step 3.

Find the greatest common factor for the $y$ -terms $y^2and\ y^9: y^2$

The greatest common factor is then $8x^3y^2$

The answer is $A$ .

Baruch College CSTM 0120 Final Test F019A Problem 4

Baruch College CSTM 0120 – Final Test F019A In 20 Minutes Or LESS (4)

Step 1.

Factor out the expression $6x^2-47x+35$ using the quadratic formula:

$D=b^2-4ac=\left(-47\right)^2-4\ast6\ast35=2209-840=1369$

$x_{1,2}=\frac{-b\pm\sqrt D}{2a}=\frac{-(-47)\pm\sqrt{1369}}{2\ast6}=\frac{47\pm37}{12}$

$x_1=\frac{47+37}{12}=\frac{84}{12}=7$

$x_2=\frac{47-37}{12}=\frac{10}{12}=\frac{5}{6}$

Step 2.

Plug in the answers in the formula $ax^2+bx+c=a(x-x_1)(x-x_2)$

$6x^2-47x+35=6\left(x-7\right)\left(x-\frac{5}{6}\right)=\left(x-7\right)(6x-5)$

The answer is C.

Baruch College CSTM 0120 Final Test F019A Problem 5

Baruch College CSTM 0120 – Final Test F019A In 20 Minutes Or LESS (5)

Step 1.

Factor out the expression $x^2-4x-5$ using the quadratic formula:

$D=b^2-4ac=\left(-4\right)^2-4\ast1\ast\left(-5\right)=16+20=36$

$x_{1,2}=\frac{-b\pm\sqrt D}{2a}=\frac{-(-4)\pm\sqrt{36}}{2\ast1}=\frac{4\pm6}{2}$

$x_1=\frac{4+6}{2}=\frac{10}{2}=5$

$x_2=\frac{4-6}{2}=\frac{-2}{2}=-1$

Step 2.

Plug in the answers in the formula $ax^2+bx+c=a(x-x_1)(x-x_2)$

$x^2-4x-5\ =\left(x-5\right)\left(x-\left(-1\right)\right)=\left(x-5\right)(x+1)$

Step 3.

Factor out the expression $x^2+x-2$ using the quadratic formula:

$D=b^2-4ac=\left(1\right)^2-4\ast1\ast\left(-2\right)=1+8=9$

$x_{1,2}=\frac{-b\pm\sqrt D}{2a}=\frac{-1\pm\sqrt9}{2\ast1}=\frac{-1\pm3}{2}$

$x_1=\frac{-1+3}{2}=\frac{2}{2}=1$

$x_2=\frac{-1-3}{2}=\frac{-4}{2}=-2$

Step 4.

Plug in the answers in the formula $ax^2+bx+c=a(x-x_1)(x-x_2)$

$x^2+x-2\ =\left(x-1\right)\left(x-\left(-2\right)\right)=\left(x-1\right)(x+2)$

The answer is $A$ .

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Baruch College CSTM 0120 Final Test F019A Problem 6

Baruch College CSTM 0120 – Final Test F019A In 20 Minutes Or LESS (6)

Step 1.

Plug in $x=-8$ into $f(x)$ :

$f\left(-8\right)=3\ast\left(-8\right)^2-2\ast\left(-8\right)+6=3\ast64+16+6=192+16+6=214$

The answer is $D$ .

Baruch College CSTM 0120 Final Test F019A Problem 7

Baruch College CSTM 0120 – Final Test F019A In 20 Minutes Or LESS (7)

Step 1.

Move all parts to one side:

${10x}^2-3=13x$

${10x}^2-13x-3=0$

Step 2.

Factor out the expression $10x^2-13x-3=0$ using the quadratic formula:

$D=b^2-4ac=\left(-13\right)^2-4\ast10\ast\left(-3\right)=169+120=289$

$x_{1,2}=\frac{-b\pm\sqrt D}{2a}=\frac{-(-13)\pm\sqrt{289}}{2\ast10}=\frac{13\pm17}{20}$

$x_1=\frac{13+17}{20}=\frac{30}{20}=\frac{3}{2}$

$x_2=\frac{13-17}{20}=\frac{-4}{20}=-\frac{1}{5}$

The answer is B.

Baruch College CSTM 0120 Final Test F019A Problem 8

Baruch College CSTM 0120 – Final Test F019A In 20 Minutes Or LESS (8)

Step 1.

Factor out the bottom for the first fraction $\frac{1}{12y-30}$

$\frac{1}{12y-30}=\frac{1}{6(2y-5)}=\frac{1}{2\ast3\ast(2y-5)}$

Step 2.

Factor out the bottom for the first fraction $\frac{1}{30y-75}$

$\frac{1}{30y-75}=\frac{1}{15(2y-5)}=\frac{1}{3\ast5\ast(2y-5)}$

Step 3.

Both denominators have a common factor of $3\ast(2y-5)$ . So the common denominator is:

$2\ast3\ast5\ast\left(2y-5\right)=30(2y-5)$

The answer is $E$ .

Baruch College CSTM 0120 Final Test F019A Problem 9

Baruch College CSTM 0120 – Final Test F019A In 20 Minutes Or LESS (9)

Step 1.

Factor out the bottom for the first fraction $\frac{5}{72n-45}$

$\frac{5}{72n-45}=\frac{5}{9(8n-5)}$

Step 2.

Factor out the bottom for the first fraction $\frac{2n-7}{56n-35}$

$\frac{2n-7}{56n-35}=\frac{2n-7}{7(8n-5)}$

Step 3.

Both denominators have a common factor of $(8n-5)$ .

The lowest common denominator is then $7\ast9\ast(8n-5)$ .

$\frac{5}{72n-45}+\frac{2n-7}{56n-35}=$

$\frac{5}{9(8n-5)}+\frac{2n-7}{7(8n-5)}=$

$\frac{5}{9\left(8n-5\right)}\ast\frac{7}{7}+\frac{2n-7}{7\left(8n-5\right)}\ast\frac{9}{9}=$

$\frac{35}{63\left(8n-5\right)}+\frac{18n-63}{63\left(8n-5\right)}=$

$\frac{35+18n-63}{63\left(8n-5\right)}=\frac{18n-28}{63\left(8n-5\right)}$

The answer is $E$ .

Baruch College CSTM 0120 Final Test F019A Problem 10

Baruch College CSTM 0120 – Final Test F019A In 20 Minutes Or LESS (10)

Step 1.

Get every term of the equation to a common denominator:

$\frac{3}{t-1}=\frac{5}{t-1}+9$

$\frac{3}{t-1}=\frac{5}{t-1}+9\ast\frac{t-1}{t-1}$

$\frac{3}{t-1}=\frac{5}{t-1}+\frac{9}{1}\ast\frac{t-1}{t-1}$

$\frac{3}{t-1}=\frac{5}{t-1}+\frac{9t-9}{t-1}$

Step 2.

Get rid of denominators and solve for $\ t$ :

$3=5+9t-9$

$9t=3-5+9$

$9t=7\rightarrow t=\frac{7}{9}$

The answer is $B$ .

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