Math.

\( x = 3 \) Step 1: Apply the distributive law The first step is to expand the terms \( 5\left(4x + 3 \right) \) and \( 4 \left(2x + 1 \right) \) by distribution so \( 5\left(4x \: + \: 3 \right) \:, \: 4 \left(2x \: + \: 1 \right) = 47 \) becomes \( 20x + 15 \:, \: 8x \:, \: 4 = 47 \) Step 2: Combine like terms We can then collect like terms in: \( 20x \:, \: 8x + 15 \:, \:4 = 47 \) \( 12x + 11 = 47 \) Step 3: Subtract 11 on both sides We then have to subtract 11 on both sides to isolate the term with \( x \). \( 12x + 11 \:, 11 = 47 \:, 11 \) \( 12x = 36 \) Step 4: Divide by 12 on both sides We then divide both sides of the equation by 12 to isolate \( x \). \( \large \frac{12x}{12} = \frac{36}{12} \) \( 1x = 3 \) We can drop the 1 on the left-hand side to \( x = 3 \) \( x = 3\) is the solution of the equation.

\( x = 7 \) or \( -7 \) Step 1: Add 61 on both sides The first step in solving this problem is to add 61 from both sides of the equation. \( 4x^2 \:, 61 + 61 = 135 + 61 \) You should then carry out the addition of the numbers on both sides of the equation. \( 4x^2 = 196 \) Step 2: Divide by 4 on both sides We then divide both sides of the equation by 4. \( \large \frac{4x^2}{4} = \frac{196}{4} \) \( x^2 = 49 \) Step 3: Take the square root on both sides The final step is to take the square root on both sides to solve for \( x \). \( x^2 = 49 \) \( \sqrt{x^2} = \sqrt{49} \) \( x = 7 \) or \(, 7 \) If you are asking WHY the -7 gets included as well, it’s because any negative number squared, is positive. \( 7 \times 7 = 49 \) But so does: \( -7 \times -7 = 49 \) Since both options work, we have to include both.

$24.50 This is a proportion problem and the given information allows us to formulate the equation below: \( \large \frac{5 \: \text{tickets}}{$17.50} = \frac{7 \: \text{tickets}}{x} \) where \( x \) is the cost of 7 lottery tickets. We can solve this equation by first cross-multiplying. \( 5 \: \text{tickets} \times x = 7 \: \text{tickets} \times 17.50 \:$ \) Carrying out the product on the right-hand side, we get; \( 5 \: \text{tickets} \times x = 122.5 \: \text{tickets} \bullet $ \) We can solve for \( x \) by dividing both sides of the equation by 5 tickets. \( \large \frac{5 \: \text{tickets} \: \times \: x}{5 \: \text{tickets}} = \frac{122.5 \: \text{tickets} \bullet $}{5 \: \text{tickets}} \) Carrying out the division, we get a 1 on the left-hand side and a 24.5 on the right-hand side. \( 1x = 24.5 \:$ \) We can drop the 1 next to \( x \), to get \( x = 24.5 \: $ \). It would cost Jordy $24.50 to buy 7 lottery tickets.

28 trains This is a proportion problem and the given information allows us to formulate the equation below: \( \large \frac{42 \: \text{trains}}{18 \: \text{hours}} = \frac{x}{12 \: \text{hours}} \) where \( x \) is the number of trains that pass near Emma’s house in 12 hours. We can solve this equation by first cross-multiplying. \( 18 \: \text{hours} \times x = 42 \: \text{trains} \times 12 \: \text{hours} \) Carrying out the product on the right-hand side, we get: \( 18 \: \text{hours} \times x = 504 \: \text{trains•hours} \) We can solve for \( x \) by dividing both sides of the equation by 18 hours. \( \large \frac{18 \:\text{hours} \times x}{18 \: \text{hours}} = \frac{504 \: \text{trains•hours}}{18 \: \text{hours}} \) Carrying out the division, we get a 1 on the left-hand side and a 28 on the right-hand side. \( 1x = 28 \: \text{trains} \) We can drop the 1 next to \( x \), to get \( x = 28\) trains. In 12 hours, 28 trains pass near Emma’s house.

\( x = 15 \) To find the unknown value in a ratio, we rewrite the ratios as fractions and use some conversions or cross-multiply. Step 1: Rewrite the ratios as fractions \( 5:12 = x:36 \) is equivalent to \( \large \frac{5}{12} = \frac{x}{36} \) Step 2: Get the same denominator on both sides Note that \( 36 = 12 \times 3 \), so we can just multiply the numerator and denominator of the left-hand side fraction by 3 to get the same denominator. \( \large \frac{\left(5 \: \times \: 3\right)}{\left(12 \: \times \: 3\right)} = \frac{x}{36}\) \( \large \frac{15}{36} = \frac{x}{36} \) Step 3: Get rid of the denominator. Multiply both of the sides by 36. \(\left(\large \frac{15}{36}\right) \: \normalsize{ \times \:36 \:=} \left(\large \frac{x}{36}\right) \: \normalsize{\times \: 36} \) \( x = 15 \)

\( x = 27 \) To find the unknown value in a ratio, we rewrite the ratios as fractions and use some conversions or cross-multiply. Step 1: Rewrite the ratios as fractions \( 30:45 = 18:x \) is equivalent to \( \left(\large \frac{30}{45}\right) = \left(\large\frac{18}{x}\right) \) Step 2: Cross-multiply the fractions \( 30 \times x = 18 \times 45 \) \( 30x = 810 \) Step 3: Divide both of the sides by 30 to get \( x \) \( \left(\large \frac{30x}{30}\right) = \left(\large\frac{810}{30}\right) \) \( 810 \div 30 = 27 \) \( x = 27 \)

\( x = 51 \) To find the unknown value in a ratio, we rewrite the ratios as fractions and use some conversions or cross-multiply. Step 1: Rewrite the ratios as fractions. \( 0.75:x = 3:204 \) is equivalent to \( \left(\large \frac{0.75}{x} \right) = \left(\large\frac{3}{204}\right)\) Step 2: Cross-multiply the fractions. \( 0.75 \times 204 = 3 \times x \) \( 153 = 3x \) Step 3: Divide both of the sides by 3 to get \( x \). \( \left(\large \frac{3x}{3} \right) = \left(\large \frac{153}{3}\right)\) \( 153 \div 3 = 51 \) \( x = 51 \)

0.55 There are two methods we can use to solve this problem. Method 1: Using Equivalent Fractions Notice that the denominator of the fraction is a factor of 100. We can multiply both the numerator and the denominator by 5 to get an equivalent fraction with a denominator of 100: \( \large \frac{11}{20} = \frac{11\: \times \: 5}{20 \: \times \:5} \) \( \large \frac{55}{100} \) \(  \frac{55}{100} \), or fifty-five hundredths, is written as 0.55 as a decimal.   Method 2: Use a calculator to divide the numerator and denominator. The simplest way to work out the fraction is by just dividing 11 by 20 using a calculator: \( \large \frac{11}{20} \: \normalsize{ = 11 \div 20} \) \( = 0.55 \)

\(262 \: boxes\) Step 1: Establish a match The expression “\(29 \%\) less than on the second day” means that \(100 \% \), \(29 \%\) = \(71 \%\) of the number of boxes collected on the second day were collected on the first day. Now, we know that \(186\) is \(71 \% \) of the number that we are looking for. Let this number be \(x\). We will setup an equation for \(x\) and solve it.   Step 2: Set up equalities First of all, we will define the relationship between values and percentages: \(100 \% = x \) and \(71 \% = 186\)   Step 3: Set up an equation We will divide the second equation by the first equation: \( \large \frac{100}{71} = \frac{x}{186} \) Remember that you can perform the division only if the corresponding sides of the equations have the same units (e.g. percents must be divided by percents).   Step 4: Solve the equation for \(x\) \( \large \frac{100}{71} = \large \frac{x}{186} \) Finally, we will multiply both sides of the equation by 186: \( \large \frac{x}{186}  \normalsize \times 186= \large \frac{100}{71} \normalsize \times 186\) \( \large \frac{186x}{186} = \large \frac{18,600}{71} \) \(x = 18,600 \div 71 \) \( x = 261.97 \) Since we are rounding to the nearest whole number, the answer is \(262\). Therefore, \(262 \: boxes\) were collected on the second day.

48  pints For this problem, we are going from bigger units, i.e. gallons, to smaller units, i.e. pints. We will first convert the gallon into quarts, and then convert the quarts into pints. To convert gallons into quarts, we multiply the number of gallons by 4 and to convert quarts into pints, we multiply the number of quarts by 2.  \( 6 \: \text{gallons} \times \left(4 \: \text{quarts/gallon}\right) \times \left(2 \: \text{pints/quart}\right) = 48 \: \text{pints} \) Therefore, there are 48 pints in 6 gallons.

0.17235 Remember that kilo means “one thousand” so there are 1,000 grams in 1 kilogram. To convert grams (g) into kilograms (kg), we divide the number of grams (g) by 1,000. \( \large \frac{172.35 \: \text{g}}{1,000} = 0.17235 \: \text{kg} \) 172.35 grams (g) are equal to 0.17235 kilograms (kg).

262.78 Make sure the numbers are written vertically with decimal points aligned:  \(\begin{array}{cccccccccc}\require{enclose}& \small{\color{blue}{1}} & \small{\color{blue}{1}} & \small{\color{blue}{1}} & \\ && 6 & 7 & . & 9 & 8 & \\ + & 1 & 9 & 4 & . & 8 & \color{red}{0} & \\ \hline & 2 & 6 & 2 & . & 7 & 8 & \end{array}\)

16 days When Shawn takes 8 ounces of his medication, he decreases the contents of the bottle by 8 ounces. Then, he repeats it for the second time, then for the third, he will decrease the contents of the bottle by 8 ounces every time. We need to know, how many times 8 goes into 128 so we use division: \( \require{enclose} \begin {array}{r} 016 \\ 8 \enclose {longdiv}{128} \\ -\underline{08}\phantom{.0} \\ 48 \\ -\underline{48} \\ 0 \end{array} \) \( 128 \div 8 = 16 \) The answer is 16 days, with no medication remaining.

71.60 To answer this question, we must divide the total amount collected by 3 (the number of servers that worked that evening): Each person received $71.60 after the dinner shift.

40 miles Step 1: This month Jane walked a fraction of the distance she walked last month, 36 miles is part of a whole. To get the whole,  you first have to turn the percentage into a decimal.  \(90 \% =  \large \frac{90}{100} \) \( = 0.9 \) Step 2: We then divide the part by the decimal in the last step to get the whole.  \( \large \frac{36}{0.9} = 40 \) As a way of checking, you can divide 36 by 40 and take note of the decimal you get.  \( \frac{36}{40} = 0.9 \), which tells us that our answer is correct.   Jane walked 40 miles last month.