1 Choice C is correct
The correct answer is \(7\) Adding \(6\) to each side of the inequality \(4\ n\ \ 3\ ≥\ 1\) yields the inequality \(4\ n\ +\ 3\ ≥\ 7\). Therefore, the least possible value of \(4\ n\ +\ 3\) is \(7\).

2 Choice A is correct
The correct answer is \(4\ x^4\ +\ 4x^3\ \ 12\ x^2\) Simplify and combine like terms. \((6\ x^3\ \ 8\ x^2\ +\ 2\ x^4 )\ \ (4\ x^2\ \ 2\ x^4\ +\ 2\ x^3 )\) ⇒ \( (6\ x^3\ \ 8\ x^2\ +\ 2\ x^4 )\ \ 4\ x^2\ +\ 2\ x^4\ \ 2\ x^3\ ⇒\ 4\ x^4\ +\ 4\ x^3\ \ 12\ x^2\)

3 Choice C is correct
The correct answer is \(38\%\) The population is increased by \(15\%\) and \(20\%\). \(15\%\) increase changes the population to \(115\%\) of original population. For the second increase, multiply the result by \(120\%\). \((1.15)\ ×\ (1.20)=1.38=138\%\) \(38\) percent of the population is increased after two years.

4 Choice D is correct
Solve for \(x\). \(\ 2\ ≤\ 2\ x\ \ 4\ <\ 8\ ⇒\) (add \(4\) all sides) \(\ 2\ +\ 4\ ≤\ 2\ x\ \ 4\ + \ 4\ <\ 8\ +\ 4 ⇒\) \(2\ ≤\ 2\ x\ <\ 12 ⇒\) (divide all sides by \(2\)) \(1\ ≤\ x\ <\ 6\) \(x\) is between \(1\) and \(6\). Choice \(D\) represent this inequality.

5 Choice D is correct
The correct answer is \(120\) cm\(^3\) Volume of a box \(=\) length \(×\) width \(×\) height \(=4\ ×\ 5\ ×\ 6=120\)

6 Choice C is correct
The correct answer is \(60\) To find the number of possible outfit combinations, multiply number of options for each factor: \(3\ ×\ 5\ ×\ 4=60\)

7 Choice A is correct
The correct answer is \(12,324\) In the stadium the ratio of home fans to visiting fans in a crowd is \(5:7\). Therefore, total number of fans must be divisible by \(12: 5\ +\ 7 = 12\). Let’s review the choices: A. \(12,324:\) \(12,324\ ÷\ 12=1,027\) B. \(42,326\) \(42,326\ ÷\ 12=3,527.166\) C. \(44,566\) \(44,566\ ÷\ 12=3,713.833\) D. \(66,812\) \(66,812\ ÷\ 12=5,567.666\) Only choice \(A\) when divided by \(12\) results a whole number.

8 Choice C is correct
The correct answer is \(60,000\) Three times of \(24,000\) is \(72,000\). One sixth of them cancelled their tickets. One sixth of \(72,000\) equals \(12,000\ (\frac{1}{6}\ ×\ 72000 = 12000)\). \(60,000\ (72,000\ –\ 12,000=60,000)\) fans are attending this week

9 Choice A is correct
The correct answer is \(97.6\) The area of the square is \(595.36\). Therefore, the side of the square is square root of the area. \(\sqrt{595.36}=24.4\) Four times the side of the square is the perimeter: \(4\ ×\ 24.4=97.6\)

10 Choice A is correct
The correct answer is \((\ 2,\ 3)\) \(x\ +\ 2\ y=4\). Plug in the values of \(x\) and \(y\) from choices provided. Then: A. \((\ 2,\ 3)\) \(x\ +\ 2\ y=4\ →\ \ 2\ +\ 2\ (3)=4\ →\ \ 2\ +\ 6=4\) This is true! B. \((1,\ 2)\) \(x\ +\ 2\ y=4\ →\ 1\ +\ 2\ (2)=4\ →\ 1\ +\ 4 =4\) This is NOT true! C. \((\ 1,\ 3)\) \(x\ +\ 2\ y = 4\ →\ \ 1\ +\ 2\ (3)=4\ →\ \ 1\ +\ 6=4\) This is NOT true! D. \((\ 3,\ 4)\) \(x\ +\ 2\ y=4\ →\ \ 3\ +\ 2\ (4)=4\ →\ \ 3\ +\ 8=4\) This is NOT true!

11 Choice A is correct
The correct answer is \(10\) meters The width of the rectangle is twice its length. Let \(x\) be the length. Then, width\(=2\ x\) Perimeter of the rectangle is \(2\) (width \(+\) length) \(= 2\ (2\ x\ +\ x)=60\ ⇒\ 6\ x=60\ ⇒\ x=10\) Length of the rectangle is \(10\) meters.

12 Choice D is correct
The correct answer is \(\frac{2}{3} ,\ 67\%,\ 0.68,\ \frac{4}{5}\) Change the numbers to decimal and then compare. \(\frac{2}{3}=0.666\)… \(67\%=0.67\) \(\frac{4}{5}=0.80\) Then: \(\frac{2}{3}\ <\ 67\%\ <\ 0.68\ <\ \frac{4}{5}\)

13 Choice C is correct
The correct answer is \(87.5\) average (mean)\(= \frac{sum\ of\ terms}{number\ of\ terms}\ ⇒\ 88= \frac{sum\ of\ terms}{50}\ ⇒\) sum\(=88\ ×\ 50=4400\) The difference of \(94\) and \(69\) is \(25\). Therefore, \(25\) should be subtracted from the sum. \(4400\ –\ 25=4375\), mean\(= \frac{sum\ of\ terms}{number\ of\ terms}\ ⇒\) mean\(=\frac{4375}{50}=87.5\)

14 Choice B is correct
The correct answer is \(\frac{1}{4}\) To get a sum of \(6\) for two dice, we can get \(5\) different options: \((5,\ 1),\ (4,\ 2),\ (3,\ 3),\ (2,\ 4),\ (1,\ 5)\) To get a sum of \(9\) for two dice, we can get \(4\) different options: \((6,\ 3),\ (5,\ 4),\ (4,\ 5),\ (3,\ 6)\) Therefore, there are \(9\) options to get the sum of 6 or \(9\). Since, we have \(6\ ×\ 6 = 36\) total options, the probability of getting a sum of \(6\) and \(9\) is \(9\) out of \(36\) or \(\frac{1}{4}\).

15 Choice D is correct
The correct answer is \(8\) Use formula of rectangle prism volume. \(V=\)(length) (width) (height) \(⇒ 2000=(25)\ (10)\) (height) \(⇒\) height\(=2000\ ÷\ 250=8\)

16 Choice B is correct
The correct answer is \(32\) The diagonal of the square is \(8\). Let \(x\) be the side. Use Pythagorean Theorem: \(a^2\ +\ b^2=c^2\) \(x^2\ +\ x^2=8^2\ ⇒\ 2\ x^2 = 82\ ⇒\ 2\ x^2 = 64\ ⇒\ x^2 = 32\ ⇒\ x= \sqrt{32}\) The area of the square is: \(\sqrt{32}\ ×\ \sqrt{32}=32\)

17 Choice B is correct
The correct answer is \(\frac{1}{4}\) Probability\(=\frac{number\ of\ desired\ outcomes}{number\ of\ total\ outcomes} = \frac{18}{12\ +\ 18\ +\ 18\ +\ 24} = \frac{18}{72} =\frac{1}{4}\)

18 Choice D is correct
The correct answer is \(16\) average \(=\frac{sum\ of\ terms}{number\ of\ terms}\ ⇒\) (average of \(6\) numbers) \(12 = \frac{sum\ of\ numbers}{6}\ ⇒\) sum of \(6\) numbers is \(12\ ×\ 6 = 72\) (average of \(4\) numbers) \(10 = \frac{sum\ of\ numbers}{4}\ ⇒\) sum of \(4\) numbers is \(10\ ×\ 4 = 40\) sum of \(6\) numbers \(–\) sum of \(4\) numbers \(=\) sum of \(2\) numbers \(72\ –\ 40 = 32\) average of \(2\) numbers \(= \frac{32}{2}=16\)

19 Choice C is correct
The correct answer is \(\ 2\) Solving systems of Equations by Elimination Multiply the first equation by (\(\ 2\)), then add it to the second equation \(\underline{\ 2\ (2\ x\ +\ 5\ y=11)\\ 4\ x\ \ 2\ y =\ 14}\) \(\Rightarrow\) \( \ 4\ x\ \ 10\ y= \ 22\\ 4\ x\ \ 2\ y= \ 14\) \(\Rightarrow\) \(\ 12\ y=\ 36 \Rightarrow \ y=3\) Plug in the value of \(y\) into one of the equations and solve for \(x\). \(2\ x\ +\ 5\ (3)= 11\ ⇒ \ 2\ x\ +\ 15= 11\ ⇒\ 2\ x= \ 4\ ⇒\ x= \ 2\)

20 Choice B is correct
The correct answer is \(70\) cm\(^2\) The perimeter of the trapezoid is \(36\) cm. Therefore, the missing side (height) is \(= 36\ –\ 8\ –\ 12\ –\ 6=10\). Area of a trapezoid: \(A=\frac{1}{2}\ h\ (b_1\ +\ b_2)= \frac{1}{2}\ (10)\ (6\ +\ 8)=70\)

21 Choice A is correct
The correct answer is \(45\) First, find the number. Let \(x\) be the number. Write the equation and solve for \(x\). \(150\%\) of a number is \(75\), then: \(1.5\ ×\ x=75\ ⇒\ x=75\ ÷\ 1.5=50\) \(90\%\) of \(50\) is: \(0.9\ ×\ 50=45\)

22 Choice A is correct
The correct answer is \(\frac{3\ x\ \ 1}{x^2\ \ x}\) \((\frac{f}{g})\ (x) = \frac{f\ (x)}{g\ (x)} = \frac{3\ x\ –\ 1}{x^2\ \ x}\)

23 Choice B is correct
The correct answer is \(\frac{1}{4}\) The probability of choosing a Hearts is \(\frac{13}{52}=\frac{1}{4}\)

24 Choice D is correct
The correct answer is \(27\) First, find the sum of five numbers. average \(=\frac{sum\ of\ terms}{number\ of\ terms}\ ⇒\ 24 = \frac{sum\ of\ 5\ numbers}{5}\ ⇒\) sum of \(5\) numbers \(= 24\ ×\ 5 = 120\) The sum of \(5\) numbers is \(120\). If a sixth number that is greater than \(42\) is added to these numbers, then the sum of \(6\) numbers must be greater than \(162\). \(120\ +\ 42 = 162\) If the number was \(42\), then the average of the numbers is: average \(=\frac{sum\ of\ terms}{number\ of\ terms}=\frac{162}{6}=27\) Since the number is bigger than \(42\). Then, the average of six numbers must be greater than \(27\). Choice \(D\) is greater than \(27\).

25 Choice B is correct
The correct answer is \(45\) Let \(L\) be the length of the rectangular and \(W\) be the with of the rectangular. Then, \(L=4\ W\ +\ 3\) The perimeter of the rectangle is \(36\) meters. Therefore: \(2\ L\ +\ 2\ W=36\) \(L\ +\ W=18\) Replace the value of \(L\) from the first equation into the second equation and solve for \(W\): \((4\ W\ +\ 3)\ +\ W=18\ →\ 5\ W\ +\ 3=18\ →\ 5\ W=15\ →\ W=3\) The width of the rectangle is \(3\) meters and its length is: \(L=4\ W\ +\ 3=4\ (3)\ +\ 3=15\) The area of the rectangle is: length \(×\) width \(= 3\ ×\ 15 = 45\)
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26 Choice C is correct
The correct answer is \(\sqrt[5]{b^3}\) \(b^{\frac{m}{n}}=\sqrt[n]{b^m}\) For any positive integers \(m\) and \(n\). Thus, \(b^{\frac{3}{5}}=\sqrt[5]{b^3}\).

27 Choice A is correct
The correct answer is \(\frac{1}{22}\) \(2,500\) out of \(55,000\) equals to \(\frac{2500}{55000}=\frac{25}{550}=\frac{1}{22}\)

28 Choice D is correct
The correct answer is \(60\) Jason needs an \(75\%\) average to pass for five exams. Therefore, the sum of \(5\) exams must be at lease \(5\ ×\ 75 = 375\). The sum of \(4\) exams is: \(68\ +\ 72\ +\ 85\ +\ 90=315\) The minimum score Jason can earn on his fifth and final test to pass is: \(375\ –\ 315=60\)

29 Choice D is correct
The correct answer is \(\frac{1}{4}\) Isolate and solve for \(x\). \(\frac{2}{3}\ x\ +\ \frac{1}{6}= \frac{1}{3}\ ⇒\ \frac{2}{3}\ x= \frac{1}{3}\ \ \frac{1}{6} = \frac{1}{6}\ ⇒\ \frac{2}{3}\ x= \frac{1}{6}\) Multiply both sides by the reciprocal of the coefficient of \(x\). \((\frac{3}{2})\ \frac{2}{3}\ x=\frac{1}{6}\) \((\frac{3}{2})\ ⇒\ x= \frac{3}{12}=\frac{1}{4}\)

30 Choice C is correct
The correct answer is \(66\ π\) in\(^2\) Surface Area of a cylinder \(= 2\ π\ r\ (r\ +\ h)\), The radius of the cylinder is \(3\ (6\ ÷\ 2)\) inches and its height is \(8\) inches. Therefore, Surface Area of a cylinder \(= 2\ π\ (3)\ (3\ +\ 8) = 66\ π\)

31 Choice C is correct
The correct answer is \(\frac{125}{512}\) The square of a number is \(\frac{25}{64}\), then the number is the square root of \(\frac{25}{64}\) \(\sqrt{\frac{25}{64}}= \frac{5}{8}\) The cube of the number is: \((\frac{5}{8})^3 = \frac{125}{512}\)

32 Choice B is correct
The correct answer is \(28\) Write the numbers in order: \(2,\ 19,\ 27,\ 28,\ 35,\ 44,\ 67\) Median is the number in the middle. So, the median is \(28\).

33 Choice D is correct
The correct answer is \(10\) Use Pythagorean Theorem: \(a^2\ +\ b^2= c^2\) \(6^2\ +\ 8^2= c^2\ ⇒\ 100= c^2 \ ⇒\ c=10\)

34 Choice B is correct
The correct answer is \(40\) Plug in \(104\) for \(F\) and then solve for \(C\). \(C= \frac{5}{9}\ (F\ –\ 32)\ ⇒\ C= \frac{5}{9}\ (104\ –\ 32)\ ⇒ C= \frac{5}{9}\ (72)=40\)

35 Choice C is correct
The correct answer is \(10\) Let \(x\) be the number. Write the equation and solve for \(x\). \(40\%\) of \(x=4\ ⇒\ 0.40\ x=4 \ ⇒\ x=4\ ÷\ 0.40=10\)

36 Choice D is correct
The correct answer is \($810\) Let \(x\) be all expenses, then \(\frac{22}{100}\ x=$660\ →\ x=\frac{100\ ×\ $660}{22}=$3,000\) He spent for his rent: \(\frac{27}{100}\ ×\ $3,000=$810\)

37 Choice C is correct
The correct answer is \(6\) hours The distance between Jason and Joe is \(9\) miles. Jason running at \(5.5\) miles per hour and Joe is running at the speed of \(7\) miles per hour. Therefore, every hour the distance is \(1.5\) miles less. \(9\ ÷\ 1.5=6\)

38 Choice D is correct
The correct answer is \(80\%\) The failing rate is \(11\) out of \(55 = \frac{11}{55}\). Change the fraction to percent: \(\frac{11}{55}\ ×\ 100\%=20\%\) \(20\) percent of students failed. Therefore, \(80\) percent of students passed the exam.

39 Choice B is correct
The correct answer is \($840\) Use simple interest formula: \(I=\) prt (\(I =\) interest, \(p =\) principal, \(r =\) rate, \(t =\) time) \(I=(12000)\ (0.035)\ (2)=840\)

40 Choice D is correct
The correct answer is \(48\ x^8\ y^6\) Simplify. \(6\ x^2\ y^3\ (2\ x^2\ y)^3= 6\ x^2\ y^3\ (8\ x^6\ y^3 ) = 48\ x^8\ y^6\)

41 Choice C is correct
The correct answer is \(g(x)=\ 2\ x\ +\ 1\) Plugin the values of \(x\) in the choices provided. The points are \((1,\ \ 1),\ (2,\ \ 3)\),and \((3,\  \ 5)\) For \((1,\ \ 1)\) check the options provided: A. \(g(x)=2\ x\ +\ 1\ →\ \ 1=2\ (1)\ +\ 1\ →\ \ 1=3\) This is NOT true. B. \(g(x)=2\ x\ \ 1\ →\ \ 1=2\ (1)\ \ 1=1\) This is NOT true. C. \(g(x)=\ 2\ x\ +\ 1\ →\ \ 1=2\ (\ 1)\ +\ 1\ →\ \ 1=\ 1\) This is true. D. \(g(x)=x\ +\ 2\ →\ \ 1=1\ +\ 2\ →\ \ 1=3\) This is NOT true. From the choices provided, only choice C is correct.

42 Choice B is correct
The correct answer is \(\frac{3\ x}{4}\) \(\sqrt{\frac{x^2}{2}\ +\ \frac{x^2}{16}}=\sqrt{\frac{8\ x^2}{16}\ +\ \frac{x^2}{16}}=\sqrt{\frac{9\ x^2}{16}}=\sqrt{\frac{9}{16}\ x^2} =\sqrt{\frac{9}{16}}\ ×\ \sqrt{x^2}=\frac{3}{4}\ ×\ x=\frac{3\ x}{4}\)

43 Choice D is correct
The correct answer is \(\ 4\) To find the yintercept of a line from its equation, put the equation in slopeintercept form: \(x\ \ 3\ y=12\), \(\ 3\ y=\ x\ +\ 12\), \(3\ y=x\ \ 12\), \(y=\frac{1}{3}\ x\ \ 4\) The \(y\ \)intercept is what comes after the \(x\). Thus, the \(y\ \)intercept of the line is \(\ 4\).

44 Choice C is correct
The correct answer is \(30\) Adding both side of \(4\ a\ \ 3=17\) by \(3\) gives \(4\ a=20\) Divide both side of \(4\ a=20\) by \(4\) gives \(a=5\), then \(6\ a=6\ (5)=30\)

45 Choice B is correct
The correct answer is \(1\) The easiest way to solve this one is to plug the answers into the equation. When you do this, you will see the only time \(x=x^{\ 6}\) is when \(x=1\) or \(x=0\). Only x=1 is provided in the choices.

46 Choice C is correct
The correct answer is \(\frac{12\ x\ +\ 1}{x^3}\) First find a common denominator for both of the fractions in the expression \(\frac{5}{x^2}\ +\frac{7\ x\ \ 3}{x^3}\) . of \(x^3\), we can combine like terms into a single numerator over the denominator: \(\frac{5\ x\ +\ 4}{x^3}\ +\ \frac{7\ x\ \ 3}{x^3} =\frac{(5\ x\ +\ 4)\ +\ (7\ x\ \ 3)}{x^3} =\frac{12\ x\ +\ 1}{x^3}\)

47 Choice D is correct
The correct answer is \(y=4\ (x\ \ 3)^2\ \ 3\) Let’s find the vertex of each choice provided: A. \(y=3\ x^2\ \ 3\) The vertex is: \((0,\ \ 3)\) B. \(y=\ 3\ x^2\ +\ 3\) The vertex is: \((0,\ 3)\) C. \(y=x^2\ +\ 3\ x\ \ 3\) The value of \(x\) of the vertex in the equation of a quadratic in standard form is: \(x=\frac{\ b}{2\ a}=\frac{\ 3}{2}\) (The standard equation of a quadratic is: \(a\ x^2\ +\ b\ x\ +\ c=0)\) The value of \(x\) in the vertex is \(3\) not \(\frac{\ 3}{2}\). D. \(y=4\ (x\ \ 3)^2\ \ 3\) Vertex form of a parabola equation is in form of \(y=a\ (x\ \ h)^2\ +\ k\), where \((h,\ k)\) is the vertex. Then \(h=3\) and \(k=\ 3\). (This is the answer)

48 Choice D is correct
The correct answer is \(2\ x\ \ \frac{1}{3}\) To find the average of three numbers even if they’re algebraic expressions, add them up and divide by \(3\). Thus, the average equals: \(\frac{(4\ x\ +\ 2)\ +\ (\ 6\ x\ \ 5)\ +\ (8\ x\ +\ 2)}{3}=\frac{6\ x\ \ 1}{3}=2\ x\ \frac{1}{3}\)

49 Choice B is correct
The correct answer is \(\frac{1}{2}\) The equation of a line in slope intercept form is: \(y=m\ x\ +\ b\). Solve for \(y\). \(4\ x\ \ 2\ y=12\ ⇒\ \ 2\ y=12\ \ 4\ x\ ⇒\ y=(12\ \ 4\ x)\ ÷\ (\ 2)\ ⇒\ y=2\ x\ \ 6\). The slope is \(2\). The slope of the line perpendicular to this line is: \(m\ 1\ ×\ m\ 2 = \ 1\ ⇒\ 2\ ×\ m\ 2 = \ 1\ ⇒\ m\ 2=\ \frac{1}{2}\)

50 Choice D is correct
The correct answer is \((1,\ 6),\ (2,\ 5),\ (−\ 5,\ 8)\) Since the triangle \(A\ B\ C\) is reflected over the \(y\ \) axis, then all values of y’s of the points don’t change and the sign of all x’s change. (remember that when a point is reflected over the \(y\ \)axis, the value of \(y\) does not change and when a point is reflected over the \(x\ \)axis, the value of \(x\) does not change). Therefore: \((−\ 1,\ 6)\) changes to \((1,\ 6)\). \((−\ 2,\ 5)\) changes to \((2,\ 5)\). \((5,\ 8)\) changes to \((−\ 5,\ 8)\)

51 Choice B is correct
The correct answer is \(39\) The area of rectangle is:\(9\ ×\ 4=36\) cm\(^2\). The area of circle is: \(π\ r^2=π\ ×(\frac{10}{2})^2=3\ ×\ 25=75\) cm\(^2\). Difference of areas is: \(75\ \ 36=39\)

52 Choice D is correct
The correct answer is \(\frac{2}{x^3}\ +\ 2\) \(f(g(x) )=2\ ×(\frac{1}{x})^3\ +\ 2=\frac{2}{x^3}\ +\ 2\)

53 Choice B is correct
The correct answer is \(4\) \(1269=6^4\) \(→\ 6^x=6^4\ →\ x=4\)

54 Choice D is correct
The correct answer is \(170\) miles Use the information provided in the question to draw the shape. Use Pythagorean Theorem: \(a^2\ +\ b^2 = c^2\) \(80^2\ +\ 150^2 = c^2\ ⇒\ 6400\ +\ 22500 = c^2\ ⇒\ 28900 = c^2\ ⇒ c = 170\)

55 Choice D is correct
The correct answer is \(45\) m\(^2\) Let \(L\) be the length of the rectangular and \(W\) be the with of the rectangular. Then, \(L=4\ W\ +\ 3\) The perimeter of the rectangle is \(36\) meters. Therefore: \(2\ L\ +\ 2\ W=36\) \(L\ +\ W=18\) Replace the value of \(L\) from the first equation into the second equation and solve for \(W\): \((4\ W\ +\ 3)\ +\ W=18\ →\ 5\ W\ +\ 3=18\ →\ 5\ W=15\ →\ W=3\) The width of the rectangle is \(3\) meters and its length is: \(L=4\ W\ +\ 3=4\ (3)\ +\ 3=15\) The area of the rectangle is: length \(×\) width \(= 3\ ×\ 15 = 45\)

56 Choice B is correct
The correct answer is \(5\) Let \(x\) be the number of adult tickets and \(y\) be the number of student tickets. Then: \(x\ +\ y=12\), \(12.50\ x\ +\ 7.50\ y=125\) Use elimination method to solve this system of equation. Multiply the first equation by \(\ 7\).\(5\) and add it to the second equation. \(\ 7.5\ (x\ +\ y=12)\), \(\ 7.5\ x\ \ 7.5\ y=\ 90\), \(12.50\ x\ +\ 7.50\ y=125\). \(5\ x=35\), \(x=7\) There are \(7\) adult tickets and \(5\) student tickets.

57 Choice C is correct
The correct answer is \(\frac{1}{25}\) Write the ratio of \(5\ a\) to \(2\ b\). \(\frac{5\ a}{2\ b}=\frac{1}{10}\) Use cross multiplication and then simplify. \(5\ a\ ×\ 10=2\ b\ ×\ 1\ →\ 50\ a=2\ b\ →\ a=\frac{2\ b}{50}=\frac{b}{25}\) Now, find the ratio of \(a\) to \(b\). \(\frac{a}{b}=\frac{\frac{b}{25}}{b}\ →\ \frac{b}{25}\ ÷\ b=\frac{b}{25}\ ×\ \frac{1}{b}=\frac{b}{25\ b}=\frac{1}{25}\)

58 Choice A is correct
The correct answer is \(30\) Plug in the value of \(x\) in the equation and solve for \(y\). \(2\ y=\frac{ 2\ x^2}{3}\ +\ 6\ →\ 2\ y = \frac{2\ (9)^2}{3}\ +\ 6\ →\ 2\ y=\frac{2\ (81)}{3}\ +\ 6\ →\ 2\ y= 54\ +\ 6=60\) \(2\ y = 60\ →\ y=30\)

59 Choice A is correct
The correct answer is \(1.085\ (3\ p)\ +\ 6\) Since a box of pen costs \($3\), then \(3\ p\) Represents the cost of \(p\) boxes of pen. Multiplying this number times \(1.085\) will increase the cost by the \(8.5\%\) for tax. Then add the \($6\) shipping fee for the total: \(1.085\ (3\ p)\ +\ 6\)

60 Choice D is correct
The correct answer is \(y(x)=8\ x\) Rate of change (growth or \(x\)) is \(8\) per week. \(40\ ÷\ 5=8\) Since the plant grows at a linear rate, then the relationship between the height \((y)\) of the plant and number of weeks of growth \((x)\) can be written as: \(y(x)=8\ x\)
