math exam and answer key

8/3/2019 Math Exam and Answer Key

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tafi 1010 bram #2 J$ame:EaIt 2011

Graphu the ineguality: 5y 5x

Solve the equation : 12- 5"1 = -12

3.) Let r(")

- lzr-4

r(*) :- e(*).

1.)

2.)

and g(")

- l+5"1 . Find all values of x for which



6)

1.1 Sohrcfte lrrcquaffiry -5<4r+l<3 ild grryhfiesdutbn set-

5.) Solve ffle irqualtty' 1(3x-l)-5 26r-4(3-r) and urite the solution as an

interval.

Factor 45y3x2 -5y*o

Factor: 9yt -72

8") Sofue - 2x3 +2lx - -L7x2

7.)

9.) Sohrc fiZr-{<ll ard write the solution set as an interval.



10-) Solre ffF ilrequalrtf 3r-3>12 or 4x+6>1 (Graph ffre solutiffi set ard

wrib the sohrtion usittg interval notation.)

11.) Letfl = {-2,-1 ,1 ,0,4,6,8,9}, B = {-3,1 ,2,6,8}, and C = {-2,-4,A,2,4,6}find(arrc)v B .

12.) Solve lZ*+31 -5 >6 and write the solution using interval notation.

13.} Factor yt + 5y' - 4y -2A



14-) U\rtrat b the dornain d ,f(r)- f -5 ,12 +16

-

15.) A rectangular gard,gn is 25 m longer then it is wide. lf the garden is to containan area of 7500 m', what are the dimensions?

Bonus: Solve the systemfz

1l=x13 2'{

I 1x+ 3v -3lz r'



M. ath 1010f a:- - 2i::

Exarm #2 key Narne:

nct

Solve the equation: l?-5rl = -12

No solution because absolute value is never negative.

3.) Let f {*) _ ltx-tl and e(*) _ l+ s*l . Find all values of x for which

r t")You are looking for where the two graphs intersect. Therefore, set them

equal andsolve.

lZx - 1 | = 14 - Sxl there are two cases!

divide by 7.

4+5x3 = 3x now d ivide by 3

2.)

iase 1 2x-1=S -5xSin-piirving gives 7x = 5 now

-;{-/J



S3'i**''p, f€ n,g:elq.6,fiir[* -j < i--,, - I < j &nro g€p,h -inig so;.,iiilo,n sgt

isc,iate x lm ilne rr't#ddle S'lbtmd -l[frrc r ell

A -ji ^^.b LiiES"

-5<-[x-l<] +

each side by 4 +

-6<-[r<] nowdlvftde

-6 .{.r 7

14.1

Don't need to switch directions ofinequality!

-3 I<x

Take out GCF+ 451'3xt - sJo'

squares)

5-r.tt {9-t" - rt) =+ 5,r-*'(3.1'- rX3"},' + Jr)

7") Factor: 9],t -72

as an

- t') look for special forrns (difference of

5.) Solve the inequality -2(3x - l) - 5 > 6x -1(3 - r) and write the solution

interval.

-2(3r "1) - 5 > 6r - 4(3 -;) + distribution gives

-6x +2-5 > 6.u- 17+4.r * collect like terms * -6x-3> l0x -LZNow isolate x. Add 3 to both sides and subtract 10x frorn both sides.

-16:r > -9 now divide both sides by -16, and remember to switch directions ofinequality.

e { e \x 3 - now write as an interval, ,r € i -*-: I16 \ 'l6J

6. ) Factor 45 7,3 xt - 5J**

- 5,,x'{g-rr'

- 72 - g(]'' - s) look for special forms (difference of cubes)



lll.A

-.\.t-rt

'n't;

-':

---s+---?\- r '-

a +A|ii4ra-: b/r.{ar '.\d.LJ 3 g"

t/{r"F +. i '-: i'

Ar,

F

F

ZgrO 3,n 'C"ine Sr-dgl ,\g','* Fam0fi ]-anqg O;i

h;is':,u*: fe'dcr surulirer 2* 21 = 42. What are fadors of 42 whlch cornbine to give 17?

!2 = ''l ,+ n 3 and 14 + 3 ='[ 7 thls is the pairof numbers thatworks. Now rewrite€xpr"e$sncn repilacing 17 by that pair of numbers.

lr: - l7r -?l =+ 2.rr +l4x*3r+?t pair up first two terms and factor out 2x. Pair upsecond two terms and factor out a 3.

2.r: -14r+3r +?l = 2x(x+7)+3{x+7) now factor out the x + 7.

2r(; *7)+3(x -7) = (x+7X2;+3)

Therefore,

Now set each factor to zero!

x = 0 or 2x + 3 = 0 orx + 7 - 0 and solve each.

X = 0 Or X = -312 Or X = -l

g.) Solve lZr-31 <11 and write the solution set as an interval.

Since this is "less than", the 2x - 3 must be trapped between -1 1 and 11.

lzt-31 < l1=-11< 2x-3<ll now solve by isolating x in the middle.

-1 I S 2:'-3 < I I= -8< 2x <14 add 3 to all sides.

-t lr 14

l?2

1 C" ) Solve the inequality: 3r - 3 > 1? ot" 4x + 6 > -? (Graph the solution set and

write the solution usins interval notation.Sofve each lnequaliU:

:-:^-:l' ii--, li--+r':i

-.&'-<-a/

t. _i

t'q,[o'*$hliTjD'n EEon "Cf' ils the union.

rn'*'he: ':S Sn,&CeC? FfOm -2 tC ilmf;nity.-*'_

2x3 +17 x2 +?lx = 0 =+ r(zxt + 17 x +21) = o =+ x(Zr+3Xx + 7) - 0



1l-l lxtfi= l.L t"l ,O,4,6,qgl, $ = {-3,1.2"6,q}, rld Q = {-2,4,03'46}find{'lncl-fr-

ffiinsnagm: AnC-t-L0,4,61 uffi&Aard C have h mnrnpn? (lntersection)

ilorr ur*rn iliswi& B- lhbn='or'

i-2-o4-6!, .r{-3,1.,?"6"E} - {-3, a.0,1* 2-4,6,9}

12"1 Solve lZ= * 3l - 5 > 6 and write the solution using interval notation.

SimpFfy by adding 5 to both sides!

lz* * 3l > I Irhis is "greater than" which means "extreme ends" further away from the

origin.

l2x+31 > 11 means2x+3> 11 or2x+3<-11

Solve each of these:

2x +3 > 11 + 2x > I + divide bV2+ x > 4

or

2x+3<-1 1+ 2x<-14+divideby2*x <-7

Extreme ends:

r € (-'rc.-7)

u'{4^ oc,}

1.t

!

(r.J

4*t-r:l;{6e

:J.

13-) Facttr _1,-=+ 5-1-= - 47,_ZA

Fqn brrls m palr-wlse factor! Pair up first two terms take out t' and pair up secondhobnnsatd talre 0ut4r-t *5J'-4r, -Zg=+ rlb-+s)- 4$1,+S)

lftrtare;(r5) + J,'(J,+5)-4{},+5}=(-u= -4}{}'+5} bok for special forrn

{ffidsqurm}+(l'=_4}tl,+5}+(y+2N,},-2}(:,+5)stop.

math exam and answer key

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