Ch. 8 Test Exponent Properties And Scientific Notation

To evaluate the expression for x = -3 and y = 2, we substitute these values into the expression. So,

= |x^2 - y^2|
= |-3^2 - 2^2|
= |-9 - 4|
= |-13|
= 13

Therefore, the answer is 13, not 20.25.

The given expression can be simplified by dividing the exponents of the same base. In this case, the base is "d" and the exponents are 12 and 7. When dividing exponents with the same base, we subtract the exponents. Therefore, d^12 / d^7 simplifies to d^(12-7) which is d^5. Similarly, when dividing exponents with the same base "e", we subtract the exponents. Therefore, e^2 / e^6 simplifies to e^(2-6) which is e^(-4). Combining the simplified expressions, we get d^5 / e^(-4). To make the exponent positive, we can rewrite e^(-4) as 1/e^4. Therefore, the simplified expression is d^5 / (1/e^4) which can be further simplified to d^5 * e^4.

To simplify (–k^2)^3, we need to raise the negative k^2 to the power of 3. This means we need to multiply the exponent of k^2, which is 2, by 3. So, (-k^2)^3 becomes -k^6. Therefore, the correct answer is –k^6.

To convert a number written in scientific notation to decimal form, we move the decimal point to the left or right depending on the exponent. In this case, the exponent is -3, which means we move the decimal point 3 places to the left. Therefore, 2 • 10–3 as a decimal is 0.002.

To convert 10–5 into a decimal, we need to move the decimal point five places to the left. Since there are no digits before the decimal point, we add zeros as placeholders. Therefore, the correct answer is 0.00001.

The expression (j5)6 simplifies to j^30. This is because when we raise a number to a power, we multiply the exponent of the base by the exponent of the power. In this case, the base is j^5 and the power is 6. So, we have (j^5)^6, which is equal to j^(5*6) = j^30. Therefore, the correct answer is j^30.

The correct answer is 1/49 because when we simplify 7–2, we subtract 2 from 7 which gives us 5. Then, we express the answer as a fraction with 1 as the numerator and 49 as the denominator, resulting in 1/49.

The expression (11p4)–2 simplifies to 1 / (121p^8).

The given expression represents the product of two terms: (2cd^2)^3 and (cd)^6. In the first term, we raise 2cd^2 to the power of 3, which means we multiply the exponents of each variable by 3. This gives us 2^3 * c^3 * (d^2)^3, which simplifies to 8c^3 * d^6. In the second term, we raise cd to the power of 6, which means we multiply the exponents of each variable by 6. This gives us c^6 * d^6. Finally, we multiply the two simplified terms together: 8c^3 * d^6 * c^6 * d^6 = 8c^9 * d^12. Therefore, the correct answer is 8c^9 d^12.

The given expression, d2(d4)5, can be simplified by first evaluating the exponent inside the parentheses, which is d^4. Then, we multiply this result by 5, resulting in d^20. Finally, we simplify d2 to d^2. Therefore, the simplified expression is d^22.

To simplify the expression –3x^2(6x^3 + 2x), we need to distribute the –3x^2 to both terms inside the parentheses. This gives us –18x^5 – 6x^3.

When evaluating b–3 for b = 3, we substitute the value of b into the expression. So, 3–3 becomes 3 raised to the power of -3. This can be rewritten as 1/3^3, which simplifies to 1/27. Therefore, the correct answer is 1/27.

To simplify the given expression, we multiply the coefficients (0.4 and 0.7) and add the exponents (-6 and -2).

0.4 × 0.7 = 0.28

10^(-6) × 10^(-2) = 10^(-6-2) = 10^(-8)

Therefore, the simplified expression in scientific notation is 0.28 × 10^(-8), which is equivalent to 2.8 × 10^(-9).

The expression 4–6 • 40 can be simplified by performing the multiplication before the subtraction. First, we multiply 6 and 40 to get 240. Then, we subtract 240 from 4, resulting in -236. Therefore, the answer is -236.

To solve the given system of equations, we can use the method of elimination. By multiplying the second equation by 5 and the first equation by 7, we can eliminate the variable x when we subtract the two equations. This gives us 35x + 56y = -469 and 35x + 10y = -335. By subtracting these equations, we get 46y = -134, which simplifies to y = -134/46 = -67/23. Substituting this value of y into the second equation, we can solve for x: -9x - 2(-67/23) = -67, which simplifies to -9x + 134/23 = -67. Solving for x gives us x = 2. Therefore, the solution to the system of equations is (-9, 2).

To simplify –6–3, we can combine the two negative signs and subtract the numbers. When we subtract 3 from -6, we get -9. Therefore, the simplified form of –6–3 is -9. However, none of the given options match this answer, so the correct answer is not available.

To simplify the expression (–)–3, we need to evaluate the negative sign first. The negative sign in front of the parentheses indicates that we need to change the sign of the expression inside the parentheses. So, the expression becomes -(-3). When we multiply two negative signs, it results in a positive sign. Therefore, -(-3) simplifies to 3. Now, we can divide 3 by 27/64. Dividing 3 by a fraction is the same as multiplying by its reciprocal. So, 3 divided by 27/64 is equal to 3 multiplied by 64/27. Multiplying 3 by 64 and dividing by 27 gives us -64/27. Hence, the answer is -64/27.

To simplify 0.7(2.8 × 10^6), we multiply 0.7 by 2.8 and multiply 10^6 by 10^6, which gives us 1.96 × 10^6.

The number 5.3 × 10^–7 is written in scientific notation because it is written as a decimal number between 1 and 10 (5.3) multiplied by a power of 10 (10^–7). This format is commonly used to represent very large or very small numbers in a concise and standardized way.

The given expression involves simplifying the product of several terms. To simplify, we can combine the coefficients and the variables separately. In this case, the coefficient is -4 * 2 * 5 = -40. The variables x and y are multiplied together, so we can add their exponents. The exponent of x is 3 + (-8) = -5, and the exponent of y is (-2) + 5 = 3. Therefore, the simplified expression is -40y^3 / x^5.

The given expression is a multiplication of two numbers in scientific notation. To simplify this expression, we multiply the two decimal numbers (2.3 and 1.4) to get 3.22. Then, we add the exponents (-5 and -6) to get -11. Therefore, the simplified expression is 3.22 × 10^–11.

The given system of equations can be solved by substituting the values of x and y from the ordered pairs into the equations and checking if they satisfy both equations. By substituting x = 5 and y = -3 into the equations, we get -2(5) - 4(-3) = 26 and 2(5) - 6(-3) = 48, which are both true. Therefore, (5, -3) is a solution to the system of equations.

The given system of equations is 3g + 2h = 7 and h = -3g + 11. We can solve this system using substitution by substituting the value of h from the second equation into the first equation. Substituting -3g + 11 for h in the first equation, we get 3g + 2(-3g + 11) = 7. Simplifying this equation, we get 3g - 6g + 22 = 7. Combining like terms, we get -3g + 22 = 7. Subtracting 22 from both sides, we get -3g = -15. Dividing both sides by -3, we get g = 5. Substituting g = 5 into the second equation, we get h = -3(5) + 11 = -4. Therefore, the solution to the system is (5, -4).

The given list has the numbers in order from least to greatest because 5.4 × 10^3 is the smallest number, followed by 4.5 × 10^4, and then 5.4 × 10^4, which is the largest number.

To simplify the expression 5(2.7 × 10–4), we multiply 5 by 2.7 and multiply 10–4 by 10. This gives us 13.5 × 10–3. Since 13.5 is between 10 and 100, we can write it as 1.35 and move the decimal point three places to the left to get 1.35 × 10–3. Therefore, the correct answer is 1.35 × 10^–3.

The given system of equations has infinitely many solutions because the two equations are linearly dependent, meaning one equation is a multiple of the other. In this case, the second equation can be obtained by multiplying the first equation by -1. Therefore, any values of x and y that satisfy the first equation will also satisfy the second equation, resulting in infinitely many solutions.

The given expression is 6 × 10^2. In scientific notation, numbers are written in the form of a × 10^n, where "a" is a number between 1 and 10, and "n" is an integer. In this case, the number 6 is between 1 and 10, and the exponent 2 indicates that the decimal point should be moved two places to the right. Therefore, the answer is 6 × 10^2.

Chase scored 12 points on Monday and doubled his score each day thereafter. To find out how many points he scored on Sunday, we can work backwards. On Saturday, he would have scored 384 points (12 * 2^5). And on Friday, he would have scored 192 points (12 * 2^4). Continuing this pattern, on Thursday he would have scored 96 points, on Wednesday 48 points, and on Tuesday 24 points. Finally, on Sunday, he would have scored 768 points (12 * 2^6).