DIST1L2
FILE INFORMATION
FILENAME(S): DIST1L2
FILE TYPE(S): PRG
FILE SIZE: 7.3K
FIRST SEEN: 2025-10-19 22:48:55
APPEARS ON: 1 disk(s)
FILE HASH
affad1f0651cbe172fef33c4e177f68967eedcc1fe064ee4ed8b563e1ba6cf16
FOUND ON DISKS (1 DISKS)
| DISK TITLE | FILENAME | FILE TYPE | COLLECTION | TRACK | SECTOR | ACTIONS |
|---|---|---|---|---|---|---|
| HHM 100785 43S1 | DIST1L2 | PRG | Radd Maxx | 15 | 1 | DOWNLOAD FILE |
FILE CONTENT & ANALYSIS
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(7,I,{})@H{}.@HTHE RATE OF SPEED FOR {}
IS {}.@I(8,I,{})@H{}@H{} IS {} TIME.@I(1
2,I,{})@H{}@H{} IS {} TIME.@I(13,I,{})@P
CHOOSE A VARIABLE TO REPRESENT THE DISTA
NCE BETWEEN THEM {}.@HUSE A VARIABLE TO
REPRESENT THE TOTAL DISTANCE.@HUSE A LET
TER, SUCH AS `T' TO REPRESENT THE TOTAL
DISTANCE.@I(19,I,&V)@RPARTS@PCALCULATE T
HE DISTANCE TRAVELLED BY EACH {}.@HRATE
* TIME = DISTANCE@HRATE \F05* \F07TIME
\F12= DISTANCE \N{}@I(17,I,{})@HRATE *
TIME = DISTANCE@HRATE \F05* \F07TIME \
F12= DISTANCE \N{}@I(18,I,{})&D(20, )@RW
HOLE@PSUBSTITUTE FOR D{}, D{} AND TOTAL
IN THE EQUATION: \N D{}+D{} = {}@HD{}=
{}, D{}={} AND TOTAL DISTANCE=&V.@H{} DI
ST. \F12+ {} DIST. \F25= {} \N{}.@I(20,I
,{})@S@RCOMPUTE@PSOLVE THE EQUATION FOR
"&V" (TOTAL DIST.) USE PAPER AND PENCIL,
OR USE THE CALCULATOR.@HISOLATE "&V" ON
ONE SIDE OF THE EQUATION.@HTHE CALCULAT
OR SOLVES EQUATIONS FOR YOU AND DISPLAYS
THE STEPS IN THE SOLUTION.@I(20,I,&V={}
)@PNOW ENTER YOUR ANSWER TO THE PROBLEM
IN THE GRID. REMEMBER THE QUESTION. {}&W
(20)@H{} EQUAL TO THE VALUE OF "&V".@H&V
={}, SO {}@I(19,I,{})@S@RCHECK&D(0,REREA
D THE PROBLEM. CHECK YOUR ANSWERS. GET R
EADY FOR A NEW PROBLEM.)@FAT 10 AM A BLU
E CAR LEAVES SPRINGFIELD HEADING WEST AT
48 MI/HR AND A GREEN CAR HEADS EAST AT
50 MI/HR. HOW FAR APART WILL THEY BE AT
1 PM?.BLUE.GREEN.TWO CARS DRIVE IN OPPOS
ITE DIRECTIONS AT DIFFERENT RATES FOR TH
E SAME LENGTH OF TIME..HOW FAR APART WIL
L THEY BE AT 1 PM?.B.BLUE'S.G.GREEN'S.B.
G.THE DISTANCE BETWEEN THEM WILL BE THE
SUM OF THE DISTANCES TRAVELLED BY EACH C
AR..`DB+DG=TOTAL'.BY EACH CAR.DB+DG=TOTA
L.THE DISTANCE BETWEEN THEM WILL BE THE
SUM OF THE DISTANCES TRAVELLED BY EACH C
AR..`DB+DG=TOTAL'.BY EACH CAR.MI/HR.4.MI
/HR.HOURS. (`HR').2.HR.MILES. (`MI').2.M
I.THE BLUE CAR GOES &H48 MI/HR&H.THE BLU
E CAR.`48' MI/HR.48.THE GREEN CAR GOES &
H50 MI/HR&H.THE GREEN CAR.`50' MI/HR.50.
THE CARS STARTED AT 10 AM AND STOPPED AT
1 PM..`3' HOURS.THE BLUE CAR'S.3.THE CA
RS STARTED AT 10 AM AND STOPPED AT 1 PM.
.`3' HOURS.THE GREEN CAR'S.3.AT 1 PM.CAR
. 48 \F05* \F07 3 \F12= `144'.144.
50 \F05* \F07 3 \F12= `150'.150
.B.G.B.G.TOTAL.B.144.G.150.BLUE.GREEN.TO
TAL. `144 \F12+ 150 \F
25= &V'.144+150=&V.294.&QHOW FAR APART
WILL THEY BE AT 1 PM?&Q.THE DISTANCE BET
WEEN THEM IS.294.THEY ARE `294' MILES AP
ART..294.@FSPARROWS FLY 40 MI/DA AND GEE
SE FLY 32 MI/DA. IF A FLOCK OF GEESE FLY
NORTH AND A FLOCK OF SPARROWS FLY SOUTH
FROM THE SAME SPOT AT THE SAME TIME, HO
W FAR APART ARE THEY AFTER 2.5 DAYS?.SPA
RROWS.GEESE.TWO FLOCKS FLY IN OPPOSITE D
IRECTIONS AT DIFFERENT RATES FOR THE SAM
E LENGTH OF TIME..HOW FAR APART ARE THEY
AFTER 2.5 DAYS?.S.SPA..G.GEESE'S.S.G.TH
E DISTANCE BETWEEN THEM WILL BE THE SUM
OF THE DISTANCES TRAVELLED BY EACH OF TH
E FLOCKS..`DS+DG = TOTAL'.BY EACH FLOCK.
DS+DG = TOTAL.THE DISTANCE BETWEEN THEM
WILL BE THE SUM OF THE DISTANCES TRAVELL
ED BY EACH OF THE FLOCKS..`DS+DG = TOTAL
'.BY EACH FLOCK.MI/DA.5.MI/DA.DAYS. (`DA
').2.DA.MILES. (`MI').2.MI.&HSPARROWS FL
Y 40 MI/DA&H.SPARROWS.`40' MI/DA.40.&HGE
ESE FLY 32 MI/DA&H.GEESE.`32' MI/DA.32.T
HE FLOCKS WILL BE FLYING FOR 2.5 DAYS..`
2.5' DAYS.SPARROWS'.2.5.THE GEESE WILL B
E FLYING FOR THE SAME AMOUNT OF TIME AS
THE SPARROWS..`2.5'.THE GEESE'S.2.5.AFTE
R 2.5 DAYS.OF THE FLOCKS.40 \F05* \F0
6 2.5 \F12= `100'.100. 32 \F05* \F07 2.
5 \F12=`80'.80.S.G.S.G.TOTAL.C.100.S.80
.SPARROW.GEESE.TOTAL DIST..` 100 \F12+
80 \F25= &V'.100+80=&V.180.&QHOW FAR
APART ARE THEY AFTER 2.5 DAYS?&Q.`180'
IS.180.ENTER `180' MILES..180.@FA FAST T
ROUT SWIMS 50 YARDS EACH MINUTE AND A SL
OWER TROUT SWIMS 40 YD/MIN. IF THE TWO T
ROUT START AT THE SAME POINT AND SWIM IN
OPPOSITE DIRECTIONS, HOW FAR APART ARE
THEY AFTER 45 SECONDS?.FAST.SLOW.TWO TRO
UT SWIM IN OPPOSITE DIRECTIONS AT DIFFER
ENT RATES FOR THE SAME LENGTH OF TIME..H
OW FAR APART ARE THEY AFTER 45 SECONDS?.
F.FAST.S.SLOW.F.S.THE DISTANCE BETWEEN T
HEM IS THE SUM OF THE DISTANCES THAT EAC
H OF THE TROUT SWIMS..`DF+DS=TOTAL'.BY E
ACH TROUT.DF+DS = TOTAL.THE DISTANCE BET
WEEN THEM IS THE SUM OF THE DISTANCES TH
AT EACH OF THE TROUT SWIMS..`DF+DS=TOTAL
'.BY EACH TROUT.YD/MIN.6.YD/MIN.MINUTES.
(`MIN').3.MIN.YARDS. (`YD').2.YD.&HA FA
ST TROUT SWIMS 50 YARDS EACH MINUTE&H.TH
E FAST TROUT.`50' YD/MIN.50.&HA SLOWER T
ROUT SWIMS 40 YD/MIN&H.THE SLOWER TROUT.
`40' YD/MIN.40.THEY SWIM FOR 45 SEC. (1
MIN.=60 SEC. AND SINCE TIME IS MEASURED
IN MIN., THE 45 SEC. MUST BE CONVERTED T
O MIN.).`.75' MINUTES.THE FAST TROUT'S..
75.THE SLOW TROUT SWIMS FOR THE SAME AMO
UNT OF TIME AS THE FAST TROUT..`.75' MIN
UTES.THE SLOW TROUT'S..75.AFTER 45 SECON
DS.OF THE TROUT.50 \F05* \F07 .75 \F1
2= `37.5' YARDS.37.5.40 \F05* \F07 .75
\F12= `30' YARDS.30.F.S.F.S.TOTAL.F.37
.5.S.30.FAST.SLOW.TOTAL.`37.5 \F12+
30 \F25= &V'.37.5+30=&V.67.5.&QHOW FAR
APART ARE THEY AFTER 45 SECONDS?&Q.THE
TOTAL DISTANCE IS.67.5.AFTER 45 SECONDS,
THEY WERE `67.5' YARDS APART..67.5.@FA
STATE TROOPER AND A DETECTIVE LEFT THE S
AME HEADQUARTERS AT 2 PM. THE TROOPER HE
ADED EAST AT 50 MI/HR AND THE DETECTIVE
HEADED WEST AT 58 MI/HR. HOW FAR APART W
ILL THEY BE AT 5:30 PM?.TROOPER.DETECTIV
E.THEY TRAVEL IN OPPOSITIVE DIRECTIONS F
OR THE SAME LENGTH OF TIME, BUT AT DIFFE
RENT RATES..HOW FAR APART WILL THEY BE A
T 5:30 PM?.T.TROOPER'S.D.DETECTIVE'S.T.D
.SINCE THEY ARE TRAVELLING IN OPPOSITIVE
DIRECTIONS, THE TOTAL DISTANCE IS THE S
UM OF EACH OF THEIR DISTANCES..`DT+DD =
TOTAL'.BY EACH OFFICER.`DT+DD = TOTAL'.S
INCE THEY ARE TRAVELLING IN OPPOSITIVE D
IRECTIONS, THE TOTAL DISTANCE IS THE SUM
OF EACH OF THEIR DISTANCES..`DT+DD = TO
TAL'.BY EACH OFFICER.MI/HR.4.MI/HR.HOURS
(`HR')..2.HR.MILES (`MI')..2.MI.&HTHE T
ROOPER HEADED EAST AT 50 MI/HR&H.THE TRO
OPER.`50' MI/HR.50.&HTHE DETECTIVE HEADE
D WEST AT 58 MI/HR&H.DETECTIVE.`58' MI/H
R.58.THEY DROVE FROM 2:30 TO 5:30, WHICH
IS 3.5 HOURS..`3.5'.THE TROOPER'S DRIVI
NG.3.5.THE DETECTIVE'S TIME IS THE SAME
AS THE TROOPER'S..`3.5'.THE DETECTIVE'S
DRIVING.3.5.AT 5:30.OFFICER.`50 \F05*
\F07 3.5' \F12= DISTANCE.50*3.5.`58 \F05
* \F07 3.5' \F12= DISTANCE.58*3.5.T.D.T
.D.&V.T.175.D.203.T'S.D'S.TOTAL. `175 \
F12+ 203 \F25= &V'.175+203=&V.378.&Q
HOW FAR APART WILL THEY BE AT 5:30 PM?&Q
.THE TOTAL DISTANCE IS.378.THE TOTAL DIS
TANCE IS `378' MILES..378.|.
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