,
, significant
Date : 23.08.21
-
numbers
✗ ample si Explanation :
5 →
1 ,
so →
7,500 →
7,5000--7 . . .
-
In a number , the zero is not significant
50 Once
becomes
put decimal the
-
.
→
2 , 50.0 → 3
, 50.00 →
4 . .
we a , number
significant . In the case of 50 .
,
we round to the nearest decimal . The
accuracy is . . .
[ 49.5
,
50 .
5)
With
•
a decimal ,
the zeros after the non
-
Zero numbers a re
significant .
5 1 The [ 0.045,0 )
-
→
0.5
→
7 0.05 7 of 0.05 is OSS
→
accuracy
. - -
. . .
.
, ,
The 0 050 [ 5)
'
0.050 →
2 ,
0 .
0500 → 3 .
.
.
accuracy
of .
. . . 0 .
0495,0 .
050
and 0.0500 . . -
[ 0.04995 , 0.05005 )
contexts :
-
The temperature outside is
28°C .
→
2 Significant
numbers .
1am km/h There
- -
travelling at 100 are 3
Sig .
fig because the accuracy is .
. .
→
3 numbers [ 99.5
significant , 100.5 )
There
>
30 students There
-
a re in a re infinite Sig fig because the number
my
-
.
classroom .
is exact . So . . . 30.000 - - -
Order of oppe ration :
Additions and Subtractions : In this case , we round off to the least
accurate number .
EXAMPLE .
-15.7+2.593-6.46=11.833 ( on calculator)
* Since the least accurate number is 15.913 Sig figs -
, the awnser is 11.8
Multiplications and divisions : You round off
your awnser to the least Sig .
figs given . .
EXAMPLE 16.78×0.0048=-1.39--0.051909352
:
* Since the number with the least Sig -
figs . is 0.004812 Sig figs) .
, the awnser is 0.052
EXAMPLE :
23.499-4.39×1.29=23.499 -
92.8=15.99 →
16.011 point after decimal )
, Date
temperature
-
.
24.08.21 .
Formulas :
conversion
T( F)
•
= IT c) +32
1- ( F)
•
1,8T / c)
☐
=
-132
-
Examples :
Solving steps :
I cook chicken at 400°F , T( F) 1,817°C ) -132
☐
=
a
-
what will be the temperature 400 =
1,817°C) -132
in Celsius ? T= 204,4°C
volumetric
- density
Formula :
Solving steps :
A) ① v=¥ñr3 ② f= F-
7- ✓ =
¥7120.033 m=f
f- v
-
✓ = 33 510cm
>
m =
9.87 ¥3 -
33510dm
'
m : mass in
grams
m= 2.675×1058×1100*099=11
✓ : volume in cm3( solids) 2451b¥
Or volume in m2 ( liquids) m= 581.5 lbs
Examples :
B) ① ② V= Ab -
h
-
A 40,0cm diameter 18in ✗
74k¥ V= 45.92cm ✗ 60.96cm -
Cannonball has a den =
45.92cm =
2989h cms
sity of
7.87g / cm3 .
"% ¥°"
✓
"
" ^^ " " " " be +" ^^ " "*
× ✗
"
in pounds (V =
¥1T r 3) =
60.96cm .
=
0.0606µm
③ 7-
g=
Aluminum density ¥4s
>
has a of ✓ =
. 6×10 -3g
2-
70g / cm ? If I cut out a
piece 2787h --
2.9¥
with a mass of
45.6mg and h =
6.06×10
-
•
Clm ✗
t¥m
dimensions
with the
following :
aetbyieoincnes
is the foil in
µm
,
?
now thick
CONVERSION TRICK !
•
632hm in m .
.
.
•
99.5MHz in H2
-
EXTRA . .
.
nano is ✗ 10-9 Mega is ✗ 10°
°
SO -
- -
632×10-9m So - . .
99.5×10 HE
A has fuel
consumption
car a
42.85¥ C) ① of - If 600km ✗ 0.6>214 F¥m×t£¥mi 98¥ .¥6pt× ✗ ×
the distance between 1-7×65-1
Mtl and Toronto is 600km ,
=
53.85$
how much would it cost to
travel to Toronto ?
, significant
Date : 23.08.21
-
numbers
✗ ample si Explanation :
5 →
1 ,
so →
7,500 →
7,5000--7 . . .
-
In a number , the zero is not significant
50 Once
becomes
put decimal the
-
.
→
2 , 50.0 → 3
, 50.00 →
4 . .
we a , number
significant . In the case of 50 .
,
we round to the nearest decimal . The
accuracy is . . .
[ 49.5
,
50 .
5)
With
•
a decimal ,
the zeros after the non
-
Zero numbers a re
significant .
5 1 The [ 0.045,0 )
-
→
0.5
→
7 0.05 7 of 0.05 is OSS
→
accuracy
. - -
. . .
.
, ,
The 0 050 [ 5)
'
0.050 →
2 ,
0 .
0500 → 3 .
.
.
accuracy
of .
. . . 0 .
0495,0 .
050
and 0.0500 . . -
[ 0.04995 , 0.05005 )
contexts :
-
The temperature outside is
28°C .
→
2 Significant
numbers .
1am km/h There
- -
travelling at 100 are 3
Sig .
fig because the accuracy is .
. .
→
3 numbers [ 99.5
significant , 100.5 )
There
>
30 students There
-
a re in a re infinite Sig fig because the number
my
-
.
classroom .
is exact . So . . . 30.000 - - -
Order of oppe ration :
Additions and Subtractions : In this case , we round off to the least
accurate number .
EXAMPLE .
-15.7+2.593-6.46=11.833 ( on calculator)
* Since the least accurate number is 15.913 Sig figs -
, the awnser is 11.8
Multiplications and divisions : You round off
your awnser to the least Sig .
figs given . .
EXAMPLE 16.78×0.0048=-1.39--0.051909352
:
* Since the number with the least Sig -
figs . is 0.004812 Sig figs) .
, the awnser is 0.052
EXAMPLE :
23.499-4.39×1.29=23.499 -
92.8=15.99 →
16.011 point after decimal )
, Date
temperature
-
.
24.08.21 .
Formulas :
conversion
T( F)
•
= IT c) +32
1- ( F)
•
1,8T / c)
☐
=
-132
-
Examples :
Solving steps :
I cook chicken at 400°F , T( F) 1,817°C ) -132
☐
=
a
-
what will be the temperature 400 =
1,817°C) -132
in Celsius ? T= 204,4°C
volumetric
- density
Formula :
Solving steps :
A) ① v=¥ñr3 ② f= F-
7- ✓ =
¥7120.033 m=f
f- v
-
✓ = 33 510cm
>
m =
9.87 ¥3 -
33510dm
'
m : mass in
grams
m= 2.675×1058×1100*099=11
✓ : volume in cm3( solids) 2451b¥
Or volume in m2 ( liquids) m= 581.5 lbs
Examples :
B) ① ② V= Ab -
h
-
A 40,0cm diameter 18in ✗
74k¥ V= 45.92cm ✗ 60.96cm -
Cannonball has a den =
45.92cm =
2989h cms
sity of
7.87g / cm3 .
"% ¥°"
✓
"
" ^^ " " " " be +" ^^ " "*
× ✗
"
in pounds (V =
¥1T r 3) =
60.96cm .
=
0.0606µm
③ 7-
g=
Aluminum density ¥4s
>
has a of ✓ =
. 6×10 -3g
2-
70g / cm ? If I cut out a
piece 2787h --
2.9¥
with a mass of
45.6mg and h =
6.06×10
-
•
Clm ✗
t¥m
dimensions
with the
following :
aetbyieoincnes
is the foil in
µm
,
?
now thick
CONVERSION TRICK !
•
632hm in m .
.
.
•
99.5MHz in H2
-
EXTRA . .
.
nano is ✗ 10-9 Mega is ✗ 10°
°
SO -
- -
632×10-9m So - . .
99.5×10 HE
A has fuel
consumption
car a
42.85¥ C) ① of - If 600km ✗ 0.6>214 F¥m×t£¥mi 98¥ .¥6pt× ✗ ×
the distance between 1-7×65-1
Mtl and Toronto is 600km ,
=
53.85$
how much would it cost to
travel to Toronto ?
