Two Precalculus Tests

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Do all the questions in tests 7&8 (attached)

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WebAssign 131-lab#7(5.1-5.3) (Homework) Current Score : – / 25 Sarah Saad MTH131-SP2018, section 01, Spring 2018 Instructor: Mojtaba Sirjani Due : Monday, April 30 2018 12:00 PM EDT 1. –/3 pointsHarMathAp11 5.1.005.MI.SA. This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise Graph the function. y = 5x 2. –/1 pointsHarMathAp11 5.1.021. Given that y = 5 x , write an equivalent equation in the form y = b−x, with b > 1. 6 y= 3. –/1 pointsHarMathAp11 5.1.029.MI. If $2000 is invested for x years at 8%, compounded quarterly, the future value that will result is represented by the following equation. S = 2000(1.02)4x What amount will result in 4 years? (Round your answer to the nearest cent.) $ 4. –/1 pointsHarMathAp11 5.1.031. We will show in the next chapter that if $P is invested for n years at 5% compounded continuously, the future value of the investment is given by S = Pe0.05n. Use P = 1000 and graph this function for 0 ≤ n ≤ 20. 5. –/1 pointsHarMathAp11 5.2.039. Write the expression as the sum or difference of two logarithmic functions containing no exponents. log3(x 3 x + 8) 6. –/1 pointsHarMathAp11 5.2.041. Use the properties of logarithms to write the expression as a single logarithm. ln(3x) − ln(3y) 7. –/1 pointsHarMathAp11 5.2.043. Use the properties of logarithms to write the expression as a single logarithm. log5(x + 6) + 1 log5(x) 2 8. –/2 pointsHarMathAp11 5.2.051.MI. Use a change-of-base formula to evaluate each logarithm. (Round your answers to four decimal places.) (a) log6(17) (b) log3(0.52) 9. –/2 pointsHarMathAp11 5.2.053. Use a change-of-base formula to rewrite the logarithm. (Write your answer using base e logarithms.) y = log3(x) y= Use a graphing utility to graph the function. 10.–/1 pointsHarMathAp11 5.2.063. Use the fact that the loudness of sound (in decibels) perceived by the human ear depends on intensity levels according to L = 10 log(I/I0) where I0 is the threshold of hearing for the average human ear. Find the loudness when I is 10,000 times I0. This is the intensity level of the average voice. L= dB 11.–/3 pointsHarMathAp11 5.2.068. Chemists use the pH (hydrogen potential) of a solution to measure its acidity or basicity. The pH is given by the formula shown below, where [H+] is the concentration of hydrogen ions in moles per liter. pH = −log[H+] Find the approximate pH of each of the following. (Round your answers to one decimal place.) (a) blood: [H+] = 3.98 (b) beer: [H+] = 6.31 × 10−8 = 0.0000000398 × 10−5 = 0.0000631 (c) vinegar: [H+] = 6.3 × 10−3 = 0.0063 12.–/2 pointsHarMathAp11 5.3.001.MI.SA. This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise Solve the equation. 96x = 4,782,969 13.–/1 pointsHarMathAp11 5.3.009. Solve the equation. Give your answer correct to 3 decimal places. 79 = 100 − 100e−0.04x x= 14.–/1 pointsHarMathAp11 5.3.015. Solve the equation. log4(9x + 10) = 3 x= 15.–/1 pointsHarMathAp11 5.3.017. Solve the equation. Give your answer correct to 3 decimal places. ln(x) − ln(7) = 10 x= 16.–/3 pointsHarMathAp11 5.3.021.MI.SA. This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise Solve the equation. ln(x + 2) + ln(x) = ln(x + 72) WebAssign 131-lab#8(5.3,6.1-6.2) (Homework) Current Score : – / 18 Sarah Saad MTH131-SP2018, section 01, Spring 2018 Instructor: Mojtaba Sirjani Due : Monday, April 30 2018 12:00 PM EDT 1. –/1 pointsHarMathAp11 5.3.029. An initial amount of 100 g of the radioactive isotope thorium-234 decays according to Q(t) = 100e−0.02828t where t is in years. How long before half of the initial amount has disintegrated? This time is called the half-life of this isotope. (Round your answer to one decimal place.) t= yr 2. –/1 pointsHarMathAp11 5.3.033. For the years from 2006 and projected to 2021, the national health care expenditures H, in billions of dollars, can be modeled by H = 2009e0.05194t where t is the number of years past 2005.† In what year are national health care expenditures expected to reach $2.7 trillion (that is, $2700 billion)? 3. –/2 pointsHarMathAp11 5.3.035. The demand function for a certain commodity is given by p = 100e−q/2. (p is the price per unit and q is the number of units.) (a) At what price per unit will the quantity demanded equal 6 units? (Round your answer to the nearest cent.) $ (b) If the price is $2.73 per unit, how many units will be demanded, to the nearest unit? units 4. –/2 pointsHarMathAp11 5.3.043. If $7500 is invested at 11.6% compounded continuously, the future value S at any time t (in years) is given by the following formula. (Round your answers to two decimal places.) S = 7500e0.116t (a) What is the amount after 18 months? S=$ (b) How long before the investment doubles? t= yr 5. –/1 pointsHarMathAp11 5.3.051. Suppose the supply of x units of a product at price p dollars per unit is given by p = 12 + 6 ln(4x + 1). How many units would be supplied when the price is $48 each? (Round your answer to the nearest whole number.) units 6. –/2 pointsHarMathAp11 6.1.005. $17,000 is invested for 2 years at an annual simple interest rate of 15%. (a) How much interest will be earned? $ (b) What is the future value of the investment at the end of the 2 years? $ 7. –/1 pointsHarMathAp11 6.1.009. If you borrow $700 for 6 months at 18% annual simple interest, how much must you repay at the end of the 6 months? $ 8. –/1 pointsHarMathAp11 6.1.021. If $5000 is invested at 8% annual simple interest, how long does it take to be worth $9000? yr 9. –/2 pointsHarMathAp11 6.2.005. For the investment situation below, identify the annual interest rate, the length of the investment in years, the periodic interest rate, and the number of periods of the investment. 12% compounded quarterly for 8 years (a) the annual interest rate % (b) the length of the investment in years yr (c) the periodic interest rate % (d) the number of periods of the investment periods 10.–/2 pointsHarMathAp11 6.2.011. What are the future value and the interest earned if $4000 is invested for 4 years at 8% compounded quarterly? (Round your answers to the nearest cent.) future value $ interest earned $ 11.–/1 pointsHarMathAp11 6.2.013. What lump sum do parents need to deposit in an account earning 12%, compounded monthly, so that it will grow to $50,000 for their son's college fund in 15 years? (Round your answer to the nearest cent.) $ 12.–/1 pointsHarMathAp11 6.2.023. Which investment will earn more money, a $4000 investment for 8 years at 8% compounded annually or a $4000 investment for 8 years compounded continuously at 7%? 8% compounded annually 7% compounded continuously 13.–/1 pointsHarMathAp11 6.2.035. How long (in years) would $700 have to be invested at 11.9%, compounded continuously, to earn $500 interest? (Round your answer to the nearest whole number.) yr
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TEST 7

1. I need the parts of this question.
This graph increases rapidly, always positive and has 1 at 𝑥 = 0.

5 𝑥

𝟔 −𝒙

2. 𝑦 = (6) = (𝟓) .

3. 𝑆(4) = 2000(1.02)16 ≈ 𝟐𝟕𝟒𝟓. 𝟓𝟕.

4. 𝑆(0) = 𝑃 = 1000, 𝑆(20) = 1000𝑒 1 ≈ 2700, so the upper right graph.

𝟏

3

5. log 3 (𝑥 √𝑥 + 8) = 𝐥𝐨𝐠 𝟑 (𝒙) + 𝟑 𝐥𝐨𝐠 𝟑 (𝒙 + 𝟖).

𝟑𝒙

6. ln(3𝑥) − ln(3𝑦) = 𝐥𝐧 (𝟑𝒚).

1

7. log 5 (𝑥 + 6) + log 5 (𝑥) = 𝐥𝐨𝐠 𝟓 ((𝒙 + 𝟔)√𝒙).
2

8. (a) log 6 17 =

ln 17
ln 6

≈ 𝟏. 𝟓𝟖𝟏𝟐. (b) log 3 0.52 =

ln 0.52
ln 3

≈ −𝟎. 𝟓𝟗𝟓𝟐.

𝐥𝐧 𝒙

9. 𝑦 = log 3 𝑥 = 𝐥𝐧 𝟑, this function increases and has zero at 𝑥 = 1, so the upper right
graph.

𝐼

10. 𝐿 = 10 log (𝐼 ) = 10 log 10000 = 10 log10 104 = 𝟒𝟎.
0

11.
(a) − log(3.98 ∙ 10−8 ) = 8 log(3.98) ≈ 𝟒. 𝟖.
(b) − log(6.31 ∙ 10−5 ) = 5 log(6.31) ≈ 𝟒. 𝟎.
(c) − log(6.30 ∙ 10−3 ) = 3 log(6.30) ≈ 𝟐. 𝟒.

12. 9(6𝑥) = 4782969, ln: ln 9 ∙ 6𝑥 = ln(4782969) , 𝑥 =

ln(4782969)
6 ln 9

7

= 6 ≈ 1.1666.

100−79

13. 79 = 100 − 100𝑒 −0.04𝑥 , 𝑒 −0.04𝑥 = 100 = 0.21, ln: − 0.04𝑥 = ln(0.21),
− ln(0.21)
𝑥=
≈ 𝟑𝟗. 𝟎𝟏𝟔.
0.04
14. log 4 (9𝑥 + 10) = 3. Apply 4𝑦 to both parts and get 9𝑥 + 10 = 43 = 64,
so 9𝑥 = 54, 𝒙 = 𝟔.

𝑥

15. ...


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