please select a team
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Please select a Team. Team 1 Team 2 Team 3 Team 4 Team 5 Team 6 Team 7 Team 8 Team 9 Team 10. Response Grid. 15. Countdown. Ch.8 Jeopardy. Response Grid. 15. Countdown. Response Grid. 15. Countdown. Response Grid. 15. Countdown. Response Grid. 15. Countdown. - PowerPoint PPT PresentationTRANSCRIPT
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Please select a Team.
a. b. c. d. e. f. g. h. i. j.
0 0 0 0 000000
a. Team 1
b. Team 2
c. Team 3
d. Team 4
e. Team 5
f. Team 6
g. Team 7
h. Team 8
i. Team 9
j. Team 10
Response Grid
Countdown
15
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Ch.8 JeopardyExponentialEquations
Do I solve for x?
Lincoln Logs and
Solving EQ’s
Common Logs
Like Oak
Natural Logs
Like Pine
Applications Of Logs
Like Cabins
10 10 10 10 10
20 20 20 20 20
30 30 30 30 30
40 40 40 40 40
50 50 50 50 50
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a. b. c. d.
0 000
.A
.B
.C
.D
CHECKPOINT
327 3
273 3
313
27
3 13
27
a.
b.
c.
d.
3
1 log 3 in exponential form.
27Write
Response Grid
Countdown
15
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a. b. c. d.
0 000
.A
.B
.C
.D
CHECKPOINT
1x
1x
7x
7x
a.
b.
c.
d.
4: 5 125xSolve
Response Grid
Countdown
15
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a. b. c. d.
0 000
.A
.B
.C
.D
CHECKPOINT
0x
1b
2
3x
5
3x
a.
b.
c.
d.
2 2 2: 256 4b bSolve
Response Grid
Countdown
15
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a. b. c. d.
0 000
.A
.B
.C
.D
CHECKPOINT
2
5x
14x
5
2x
1x
a.
b.
c.
d.
4 12 11
: 82
xxSolve
Response Grid
Countdown
15
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a. b. c. d.
0 000
.A
.B
.C
.D
CHECKPOINT
13
2x
5
8x
11
8x
3x
a.
b.
c.
d.
5 1 42 27
: 3 8
x x
Solve
Response Grid
Countdown
15
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a. b. c. d.
0 000
.A
.B
.C
.D
CHECKPOINT
a.
b.
c.
d.
6 7 117649 logarithmic from.Write
7log 117649 6
117649log 7 6
6log 117649 7
6log 7 117649Response Grid
Countdown
15
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a. b. c. d.
0 000
.A
.B
.C
.D
CHECKPOINT
.3333x
3.6123x
3x
8512x
a.
b.
c.
d.
8: log 512Evaluate
Response Grid
Countdown
15
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a. b. c. d.
0 000
.A
.B
.C
.D
CHECKPOINT
3125x
3.6239x
5x
8881784197x
a.
b.
c.
d.
25
5: log
2Solve x
Response Grid
Countdown
15
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a. b. c. d.
0 000
.A
.B
.C
.D
CHECKPOINT
.0117x
85.3333x
28.4444x
.0351x
a.
b.
c.
d.
3 3 3
1: 3log 8 log 36 log
2Solve x
Response Grid
Countdown
15
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a. b. c. d.
0 000
.A
.B
.C
.D
CHECKPOINT
10n
5n
1n
2n
a.
b.
c.
d.
10 10: log ( ) log ( 9) 1Solve n n
Response Grid
Countdown
15
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a. b. c. d.
0 000
.A
.B
.C
.D
CHECKPOINT
1
5
25
10
a.
b.
c.
d.
Common Logs have a base of ?
Response Grid
Countdown
15
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a. b. c. d.
0 000
.A
.B
.C
.D
CHECKPOINT
.4565x
726.1060x
5.6563x
2.4565x
a.
b.
c.
d.
your calculator: log 286.1= xUse
Response Grid
Countdown
15
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a. b. c. d.
0 000
.A
.B
.C
.D
CHECKPOINT
.0380
.2417
.1523
a.
b.
c.
d.
your calculator: antilog (-1.42)Use
Response Grid
Countdown
15
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a. b. c. d.
0 000
.A
.B
.C
.D
CHECKPOINT
.5439x
.0535x
.0453x
1.5424x
a.
b.
c.
d.
3 1: 6 8xSolve
Response Grid
Countdown
15
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a. b. c. d.
0 000
.A
.B
.C
.D
CHECKPOINT
2.4527x
25.9164x
3.3215x
.2634x
a.
b.
c.
d.
2 3: 7 5x xSolve
Response Grid
Countdown
15
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a. b. c. d.
0 000
.A
.B
.C
.D
CHECKPOINT
10
7
e
a.
b.
c.
d.
logs (ln) are really logs with base ___?Natural
Response Grid
Countdown
15
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a. b. c. d.
0 000
.A
.B
.C
.D
CHECKPOINT
6.1123
7.4295
3.2266
1.6850
a.
b.
c.
d.
your calculator: ln 1685Use
Response Grid
Countdown
15
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a. b. c. d.
0 000
.A
.B
.C
.D
CHECKPOINT
2980.96x
.9031x
2x
2.0794x
a.
b.
c.
d.
: e 8xSolve
Response Grid
Countdown
15
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a. b. c. d.
0 000
.A
.B
.C
.D
CHECKPOINT
5.9896x
.2518x
.1094x
2.1758x
a.
b.
c.
d.
: ln x = .7774Solve
Response Grid
Countdown
15
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a. b. c. d.
0 000
.A
.B
.C
.D
CHECKPOINT
5.9915x
133.1440x
.0075x
3.1011x
a.
b.
c.
d.
.045: 2000 = 5e xSolve
Response Grid
Countdown
15
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a. b. c. d.
0 000
.A
.B
.C
.D
CHECKPOINT
/Growth Decay
Compounded InterestContinous
Appreciation
Depreciation
a.
b.
c.
d.
following equation is used for _______ ?
y kt
The
ae
Response Grid
Countdown
15
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a. b. c. d.
0 000
.A
.B
.C
.D
CHECKPOINT
AmountEnding
Time
AmountInitial
RateInterest
a.
b.
c.
d.
the fomrula , what does the P represent?rtIn A Pe
Response Grid
Countdown
15
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a. b. c. d.
0 000
.A
.B
.C
.D
CHECKPOINT
V and P switchn
Take away the n
IDK
, switch
a.
b.
c.
d.
In the formulas for Appreciation/Depreciation,
what is the only difference between the two formulas?
Response Grid
Countdown
15
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a. b. c. d.
0 000
.A
.B
.C
.D
CHECKPOINT
17.67years
.56years
12years
3.6years
a.
b.
c.
d.
Suppose $500 is invested at 8% annual interest compounded twice a year. When will the investment be worth $2000?
Response Grid
Countdown
15
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a. b. c. d.
0 000
.A
.B
.C
.D
CHECKPOINT
3r
.0580r
.0617r
.0168r
a.
b.
c.
d.
Dave bought a new car 30 years ago for $5400. To buy a new car comparably equipped now would cost $32,500. Assuming a steady rate of increase, what was the yearly rate of inflation in car prices over the 8-year period?
Response Grid
Countdown
15
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Team Scores0 Team 1
0 Team 2
0 Team 3
0 Team 4
0 Team 5
0 Team 6
0 Team 7
0 Team 8
0 Team 9
0 Team 10