algebra i concept test # 11 – exponents practice test © 2007-09 by s-squared, inc. all rights...

Algebra I Concept Test # 11 – Exponents Practice Test

© 2007-09 by S-Squared, Inc. All Rights Reserved.

1. a) Expand 3.49 x 10−6

.00000349

Note: To expand, shift the decimal point.

b) Write 7132 in scientific notation.

Note: Shift the decimal point to create a number between 1 and 10.

7.132 x 103

2. Multiply and leave the result in scientific notation.

(4.2 x 108)(3.0 x 10−3)

12.6


x 105

1.26 x 106

Note: Multiply the decimal form of numbers together and like base exponent terms

together.

Note: When a number is in scientific notation,the first number must be greater than orequal to 1 and less than 10.

49

Power to a Power Multiply exponents

=

Simplify:

3. (7−1y5)−2

72 y−10

Negative Exponent property y10

72

y10


4. (− 3x5 y)2 • 6x2(y7)4

Power to a Power Multiply exponents

(− 3)2x10 y2 • 6x2y28

9x10 y2 • 6x2y28

Simplify

54x10 y2 • x2y28

Multiply

54x12 y30 Product of Powers Property

Quotient of Powers Subtract exponents

Zero Exponent property

Simplify:

5. d−14 • d8 d−6

Product of Powers property

d−6 d−6 d−6 – (−6)

d0 =

1


6. − 52m6n−7

Product of Powers property

Quotient of Powers Subtract exponents

Negative Exponent property

m2n•

m4

m−7n − 52m−1n−6 m6n

− 25m−1n−6 m6n

− 25m−1– 6n−6 – 1

− 25m−7n−7

− 25 m7n7


7. You bought a new car for $10,000. It depreciates at a rate of 4% annually.

a) Write the decay model that represents the valueof your car over t , number of years.

Decay Model: y = C(1 – r)t , where C = initial amount r = rate t = time r = 4% or .04

C = 10000

t = t

y = 10000(1 – .04)t OR y = 10000(.96) t

Exponential Decay Table Decay Rates

Years 1% 2% 3% 4% 5% 6% 7% 8%

1 0.990000 0.980000 0.970000 0.960000 0.950000 0.940000 0.930000 0.920000

2 0.980100 0.960400 0.940900 0.921600 0.902500 0.883600 0.864900 0.846400

3 0.970299 0.941192 0.912673 0.884736 0.857375 0.830584 0.804357 0.778688

4 0.960596 0.922368 0.885293 0.849347 0.814506 0.780749 0.748052 0.716393

5 0.950990 0.903921 0.858734 0.815373 0.773781 0.733904 0.695688 0.659082



Years 1% 2% 3% 4% 5% 6% 7% 8%

1 0.990000 0.980000 0.970000 0.960000 0.950000 0.940000 0.930000 0.920000

2 0.980100 0.960400 0.940900 0.921600 0.902500 0.883600 0.864900 0.846400

3 0.970299 0.941192 0.912673 0.884736 0.857375 0.830584 0.804357 0.778688

4 0.960596 0.922368 0.885293 0.849347 0.814506 0.780749 0.748052 0.716393

5 0.950990 0.903921 0.858734 0.815373 0.773781 0.733904 0.695688 0.659082

Note: The number that you extract from the table is the (1 – r)t value.

$ 9,216.00

Growth Model: y = 10000(.96)2 y = 10000(.921600)


b) Using the “Exponential Decay” table below,determine the value of the car after 2 years.



Years 1% 2% 3% 4% 5% 6% 7% 8%

1 0.990000 0.980000 0.970000 0.960000 0.950000 0.940000 0.930000 0.920000

2 0.980100 0.960400 0.940900 0.921600 0.902500 0.883600 0.864900 0.846400

3 0.970299 0.941192 0.912673 0.884736 0.857375 0.830584 0.804357 0.778688

4 0.960596 0.922368 0.885293 0.849347 0.814506 0.780749 0.748052 0.716393

5 0.950990 0.903921 0.858734 0.815373 0.773781 0.733904 0.695688 0.659082

Note: Use the table knowing the rate of 4% and the amount $8493.47 to find the number of years t.

4 years


c) How long will it take the car to depreciate to $8493.47?

3 4 5 6-4 -3 -2 -1 71-7 -6 -5 2

1234

67

-1-2

8

121110

9


Note: Substitute each of the given x-coordinates into the given exponential equation to find the f (x)-coordinate.

b) Graph the function using the ordered pairs from part (a).

8. Given the exponential function, f (x) = 2x + 3 complete the following:

a) Complete the table:

x f (x)

0 4

− 1

1

2

− 2

5

7/2

7

13/4

5

8. Given the exponential function, f (x) = 2x + 3 complete the following:


c) Identify, if it exists, the y-intercept.

d) Identify, if it exists, the x-intercept.

(0, 4)

Does not exist

f) State the range.

y > 3

e) State the domain.

x is any real number

3 4 5 6-4 -3 -2 -1 71-7 -6 -5 2

1234567

-1-2

8

121110

9

9. A scientist recorded the growth of a certain bacteria. Theexperiment began with three bacteria (time = 0.) The growthof the number of bacteria is given in the table below:


Hours Number of bacteria

2 11

3

1

0

4

5

29

3

83

Using the information and the exponentialmodel, A(x) = bx + c , predict the number of bacteria that will exist after 5 hours.

Note: x = number hours.A(x) = number of bacteria

after x hours.

A(x) = bx + cEquation

3 = b0 + cSubstitute

**Substitute A(x) = 3 and x = 0 into the equation to find c.

3 = 1 + cSimplify **Anything to the zero power is 1.

2 = c **Subtract 1 to solve for c.


9. A scientist recorded the growth of a certain bacteria. Theexperiment began with three bacteria (time = 0). The growthof the number of bacteria is given in the table below:

Using the information and the exponentialmodel, A(x) = bx + c , predict the number of bacteria that will exist after 5 hours.

Hours Number of bacteria

2 11

3

1

0

4

5

29

3

83

5 = b1 + 2Substitute

**Substitute A(x) = 5, c = 2 and x = 1 into the equation to find b.

3 = b1 Solve

3 = b

**Substitute b = 3, c = 2 and x = 5 into the equation to find A(5).

A(5) = 35 + 2Substitute

A(5) = 243 + 2 Simplify

A(5) = 245

245 bacteria

A(x) = 3x + 2

Nice Work

algebra i concept test # 11 – exponents practice test © 2007-09 by s-squared, inc. all rights...

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