Question

29.) a city park has an area of \\( \frac{2}{3} \\) square miles. the park is shaped like a rectangle. the width of the park is \\( \frac{3}{5} \\) miles. select the equation and solution that represents the length of the park.
a rectangle: \\( a = \ell \cdot w \\), where \\( a = \frac{2}{3} \text{mi}^2 \\), \\( w = \frac{3}{5} \text{mi} \\), \\( \ell = \underline{\quad\quad} \text{mi} \\).
options for equation and solution:

• \\( \frac{2}{3} \ell = \frac{3}{5} \\); \\( \ell = \frac{9}{10} \\) miles
• \\( \frac{3}{5} \ell = \frac{2}{3} \\); \\( \ell = \frac{9}{10} \\) miles
• \\( \frac{2}{3} \ell = \frac{3}{5} \\); \\( \frac{2}{5} \\) miles
• \\( \frac{3}{5} \ell = \frac{2}{3} \\); \\( \ell = \frac{10}{9} \\) miles

30.) the coldest temperature recorded in siberia was \\( -90^{\circ} \text{f} \\). this was \\( 10^{\circ} \text{f} \\) colder than the coldest temperature recorded in alaska. which equation can be used to find the coldest temperature recorded in alaska? what is the coldest temperature recorded in alaska?
options:

• \\( t + 10 = -90 \\); \\( -100^{\circ} \text{f} \\)
• \\( t - 10 = -90 \\); \\( -80^{\circ} \text{f} \\)
• \\( t - 10 = -90 \\); \\( -100^{\circ} \text{f} \\)
• \\( t + 10 = -90 \\); \\( -80^{\circ} \text{f} \\)

31.) solve the absolute value equation \\( |2 - x| = 3 \\).
options for solution:

• \\( x = 5 \\)
• \\( x = -1 \\) and \\( x = -5 \\)
• \\( x = -1 \\)

Explanation:

Problem 29:

Step1: Recall the area formula for a rectangle

The area \( A \) of a rectangle is given by \( A = l \cdot w \), where \( l \) is the length and \( w \) is the width. We know \( A=\frac{2}{3}\) square miles and \( w = \frac{3}{5}\) miles. We need to solve for \( l \). So we start with the formula \( \frac{2}{3}=l\times\frac{3}{5}\).

Step2: Solve for \( l \)

To solve for \( l \), we can divide both sides of the equation by \( \frac{3}{5} \), which is the same as multiplying both sides by \( \frac{5}{3} \). So \( l=\frac{2}{3}\div\frac{3}{5}=\frac{2}{3}\times\frac{5}{3}=\frac{10}{9}\)? Wait, no, wait. Wait, \( A = l\times w \), so \( l=\frac{A}{w}\). Substituting the values, \( A = \frac{2}{3}\), \( w=\frac{3}{5}\), so \( l=\frac{\frac{2}{3}}{\frac{3}{5}}=\frac{2}{3}\times\frac{5}{3}=\frac{10}{9}\)? Wait, no, the options are given. Wait, let's check the options again. The options are:

• \( \frac{2}{3}l=\frac{3}{5} \); \( l = \frac{9}{10}\) miles
• \( \frac{3}{5}l=\frac{2}{3} \); \( l=\frac{9}{10}\) miles
• \( \frac{2}{3}l=\frac{3}{5} \); \( l=\frac{10}{9}\) miles

Wait, no, let's do it correctly. \( A=l\times w \), so \( \frac{2}{3}=l\times\frac{3}{5} \). To solve for \( l \), we can write the equation as \( \frac{3}{5}l=\frac{2}{3} \)? No, wait, \( l\times\frac{3}{5}=\frac{2}{3} \), so \( l=\frac{2}{3}\div\frac{3}{5}=\frac{2}{3}\times\frac{5}{3}=\frac{10}{9} \)? Wait, no, that's not matching. Wait, maybe I mixed up \( A \) and the values. Wait, the area \( A=\frac{2}{3}\) mi², width \( w = \frac{3}{5}\) mi. So \( l=\frac{A}{w}=\frac{\frac{2}{3}}{\frac{3}{5}}=\frac{2}{3}\times\frac{5}{3}=\frac{10}{9} \)? But the options have \( \frac{3}{5}l=\frac{2}{3} \); \( l=\frac{10}{9}\) miles? Wait, no, the third option is \( \frac{3}{5}l=\frac{2}{3} \); \( l=\frac{9}{10}\) miles? Wait, no, let's re - express the equation. From \( A = l\times w \), so \( \frac{2}{3}=l\times\frac{3}{5} \), which can be rewritten as \( \frac{3}{5}l=\frac{2}{3} \). Then, solving for \( l \), we multiply both sides by \( \frac{5}{3} \), so \( l=\frac{2}{3}\times\frac{5}{3}=\frac{10}{9} \)? Wait, the fourth option is \( \frac{3}{5}l=\frac{2}{3} \); \( l=\frac{10}{9}\) miles? Wait, the options are:

1. \( \frac{2}{3}l=\frac{3}{5} \); \( l=\frac{9}{10}\) miles
2. \( \frac{3}{5}l=\frac{2}{3} \); \( l=\frac{9}{10}\) miles
3. \( \frac{3}{5}l=\frac{2}{3} \); \( l=\frac{10}{9}\) miles
4. \( \frac{2}{3}l=\frac{3}{5} \); \( l=\frac{10}{9}\) miles

Wait, let's do the algebra again. \( A = l\times w \), so \( \frac{2}{3}=l\times\frac{3}{5} \). Let's solve for \( l \):

\( l=\frac{2}{3}\div\frac{3}{5}=\frac{2}{3}\times\frac{5}{3}=\frac{10}{9} \). So the equation is \( \frac{3}{5}l=\frac{2}{3} \) (because \( l\times\frac{3}{5}=\frac{2}{3} \)) and \( l = \frac{10}{9}\) miles. So the correct option is the one where the equation is \( \frac{3}{5}l=\frac{2}{3} \) and \( l=\frac{10}{9}\) miles.

Step1: Define the variables

Let \( T \) be the coldest temperature in Alaska. The coldest temperature in Siberia is \( - 90^{\circ}\text{F} \), and it is \( 10^{\circ}\text{F} \) colder than the coldest temperature in Alaska. "Colder" means that the temperature in Siberia is lower than in Alaska, so \( T-10=-90 \) (because Alaska's temperature minus 10 degrees is Siberia's temperature).

Step2: Solve the equation

To solve \( T - 10=-90 \), we add 10 to both sides of the equation. So \( T=-90 + 10=-80 \). So the equation is \( T - 10=-90 \) and the solution is \( T=-80^{\circ}\text{F} \).

Step1: Recall the absolute value equation property

For an absolute value equation \( |a|=b \) (where \( b\geq0 \)), we have \( a = b \) or \( a=-b \). So for \( |2 - x|=3 \), we can split it into two cases:
Case 1: \( 2 - x = 3 \)
Case 2: \( 2 - x=-3 \)

Step2: Solve Case 1

For \( 2 - x = 3 \), subtract 2 from both sides: \( -x=3 - 2 = 1 \), then multiply both sides by - 1: \( x=-1 \)

Step3: Solve Case 2

For \( 2 - x=-3 \), subtract 2 from both sides: \( -x=-3 - 2=-5 \), then multiply both sides by - 1: \( x = 5 \)

Answer:

The equation is \( \frac{3}{5}l=\frac{2}{3} \) and the solution is \( l = \frac{10}{9}\) miles (the option with \( \frac{3}{5}l=\frac{2}{3} \); \( l=\frac{10}{9}\) miles)

Problem 30: