Let's solve this step by step using the **midpoint formula**, which assumes a linear change between the surrounding years. ### Step 1: Identify the surrounding years and their cutoffs - 1980: cutoff = 8063 - 1990: cutoff = 13484 1987 lies between these two years. --- ### Step 2: Find the number of years between 1980 and 1990 1990 - 1980 = 10 years --- ### Step 3: Find the number of years from 1980 to 1987 1987 - 1980 = 7 years --- ### Step 4: Find the increase in the cutoff between 1980 and 1990 13484 - 8063 = 5421 --- ### Step 5: Find the increase per year 5421 ÷ 10 = 542.1 --- ### Step 6: Find the increase from 1980 to 1987 542.1 × 7 = 3794.7 --- ### Step 7: Add this increase to the 1980 cutoff 8063 + 3794.7 = 11857.7 --- ### Step 8: Round to the nearest dollar The poverty level cutoff in 1987 was **$11,858** (rounded). --- **Final Answer:** >The poverty level cutoff in 1987 was **$11,858**.

Let's define the variables: - \( t \) = number of minutes talked - \( C_1 \) = cost of plan 1 - \( C_2 \) = cost of plan 2 ### Step 1: Write the equations for each plan **Plan 1:** Charges 26 cents per minute, which is \$0.26 per minute. \[ C_1 = 0.26 \times t \] **Plan 2:** Charges a fixed fee of \$39.95 plus 11 cents per minute (\$0.11 per minute). \[ C_2 = 39.95 + 0.11 \times t \] --- ### Step 2: Find the number of minutes where both plans cost the same Set \( C_1 = C_2 \): \[ 0.26 t = 39.95 + 0.11 t \] ### Step 3: Solve for \( t \) Subtract \( 0.11 t \) from both sides: \[ 0.26 t - 0.11 t = 39.95 \] Simplify: \[ 0.15 t = 39.95 \] Divide both sides by 0.15: \[ t = \frac{39.95}{0.15} \] Calculate: \[ t \approx 266.33 \] ### **Final answer:** The number of minutes that makes both plans cost the same is approximately **266.3 minutes** (rounded to one decimal place).

Let's analyze this problem step by step. --- ### Given: - **Annual growth rate of emissions:** 4% per year - **Time span:** from 2010 to 2040 (which is 30 years) ### Goal: - **Approximate increase factor:** Using the approximate rule of 70 - **Exact increase factor:** Using the exponential growth formula --- ### Step 1: Approximate increase factor using the rule of 70 The rule of 70 estimates how many years it takes for a quantity to double given a certain percentage growth rate: \[ \text{Doubling time} \approx \frac{70}{\text{growth rate in percent}} \] For a 4% growth rate: \[ \frac{70}{4} = 17.5 \text{ years} \] - Total years from 2010 to 2040: 30 years Number of doubling periods in 30 years: \[ \frac{30}{17.5} \approx 1.714 \] - **Approximate increase factor:** Since each doubling roughly doubles emissions, the total increase factor is approximately: \[ 2^{1.714} \] Calculating: \[ 2^{1.714} \approx e^{1.714 \times \ln 2} \approx e^{1.714 \times 0.693} \approx e^{1.188} \approx 3.28 \] **So,** **Approximate increase factor ≈ 3.28** --- ### Step 2: Exact increase factor using exponential growth The exact formula for exponential growth over \( t \) years: \[ \text{Increase factor} = (1 + r)^t \] Where: - \( r = 0.04 \) (4% as a decimal) - \( t = 30 \) years Calculating: \[ (1 + 0.04)^{30} = 1.04^{30} \] Using logarithms: \[ \ln(1.04^{30}) = 30 \times \ln(1.04) \] Calculate \( \ln(1.04) \): \[ \ln(1.04) \approx 0.03922 \] Then: \[ 30 \times 0.03922 \approx 1.1766 \] Exponentiate: \[ e^{1.1766} \approx 3.242 \] **So,** **Exact increase factor ≈ 3.24** --- ### **Final answers:** - **Using the approximate formula:** **3.28** - **Using the exact formula:** **3.24** *(both rounded to two decimal places)*

One plus one equals two.

Let's break down this problem step-by-step: ### Given Data: - Monthly take-home pay: \$2,750 - Recommended housing expense: 1/4 of take-home pay - Down payment: \$15,600 - Annual Percentage Rate (APR): 3.85% - Loan term: 15 years - Compounding: Monthly --- ### Step 1: Calculate the maximum monthly mortgage payment \[ \text{Monthly payment} = \frac{1}{4} \times 2750 \] \[ \text{Monthly payment} = 687.50 \] --- ### Step 2: Convert APR to monthly interest rate \[ \text{Monthly interest rate} = \frac{3.85\%}{12} = \frac{0.0385}{12} \] \[ \text{Monthly interest rate} \approx 0.0032083 \] --- ### Step 3: Determine the total number of payments \[ n = 15 \text{ years} \times 12 \text{ months/year} = 180 \text{ months} \] --- ### Step 4: Use the mortgage formula to find the maximum loan amount (principal \( P \)) The formula for the monthly payment \( M \): \[ M = P \times \frac{r (1 + r)^n}{(1 + r)^n - 1} \] Rearranged to solve for \( P \): \[ P = M \times \frac{(1 + r)^n - 1}{r (1 + r)^n} \] Where: - \( M = 687.50 \) - \( r = 0.0032083 \) - \( n = 180 \) --- ### Step 5: Calculate \( (1 + r)^n \) \[ (1 + r)^n = (1 + 0.0032083)^{180} \] Using a calculator: \[ (1.0032083)^{180} \approx e^{180 \times \ln(1.0032083)} \] Calculate \( \ln(1.0032083) \): \[ \ln(1.0032083) \approx 0.0032 \] Now multiply: \[ 180 \times 0.0032 \approx 0.576 \] Exponentiate: \[ e^{0.576} \approx 1.779 \] --- ### Step 6: Calculate numerator and denominator for \( P \) Numerator: \[ (1 + r)^n - 1 = 1.779 - 1 = 0.779 \] Denominator: \[ r \times (1 + r)^n = 0.0032083 \times 1.779 \approx 0.0032083 \times 1.779 \approx 0.00571 \] --- ### Step 7: Calculate the maximum loan amount \( P \) \[ P = 687.50 \times \frac{0.779}{0.00571} \] Calculate numerator: \[ 687.50 \times 0.779 \approx 535.94 \] Now divide: \[ \frac{535.94}{0.00571} \approx 93,911.59 \] ### Step 8: Calculate the total house price Total house price = Loan amount + Down payment \[ \text{Total cost} = 93,911.59 + 15,600 \] \[ \text{Total cost} \approx 109,511.59 \] ### **Final answer:** The total cost of the house you can afford is approximately **\$109,511.59**. --- **Note:** Rounding to the nearest cent, the total house price is **\$109,511.59**.

Let's analyze this step by step: --- ### Given: - Cost of a set of clubs: \$1,400 - Cost per round with clubs: \$85 - Cost per round with own clubs: \$60 --- ### Part (a): Find the number of rounds needed to recover the cost **Step 1:** Define variables: Let \( R \) = number of rounds played. **Step 2:** Set up the inequality: The total cost for playing \( R \) rounds with the clubs must be **greater than or equal to** the cost of buying the clubs (since playing enough rounds will recover the cost): \[ 85 R \geq 1400 + 60 R \] **Step 3:** Solve the inequality: Subtract \( 60 R \) from both sides: \[ 85 R - 60 R \geq 1400 \] Simplify: \[ 25 R \geq 1400 \] Divide both sides by 25: \[ R \geq \frac{1400}{25} \] Calculate: \[ R \geq 56 \] **Step 4:** Round to the nearest whole number (since rounds are whole): \[ R = 56 \] --- ### **Part (a) Final answer:** **You must play at least 56 rounds** with the clubs to recover the cost and start saving money. --- ### Part (b): Interpret the answer in a complete sentence: **You need to play at least 56 rounds using the golf course's clubs for the total cost to be equal to or greater than the expense of purchasing a set of clubs, after which you will start saving money by using your own clubs.**

Let's analyze the function: \[ f(x) = \frac{10x^2 + 10x + 19}{9 - 6x^2} \] ### Step 1: Understand the behavior for large \(|x|\) When \( x \) becomes very large (positive or negative), the highest degree terms dominate. - Numerator leading term: \( 10x^2 \) - Denominator leading term: \( -6x^2 \) ### Step 2: Find the horizontal asymptote The horizontal asymptote of a rational function where numerator and denominator are polynomials is given by the ratio of the leading coefficients, provided the degrees are equal. Since both numerator and denominator are degree 2: \[ \text{Asymptote } y = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}} \] which is: \[ y = \frac{10}{-6} = -\frac{5}{3} \] ### Step 3: Convert to decimal and round to two decimal places \[ -\frac{5}{3} \approx -1.6667 \] Rounded to two decimal places: \[ -1.67 \] --- ### **Final answer:** \(\boxed{-1.67}\)

Let's solve each equation step-by-step. --- ### Part (a): \(\tan \alpha = \cot (\alpha - 20^\circ)\) **Step 1:** Recall that \(\cot x = \tan (90^\circ - x)\). So, rewrite the right side: \[ \cot (\alpha - 20^\circ) = \tan (90^\circ - (\alpha - 20^\circ)) \] which simplifies to: \[ \tan (90^\circ - \alpha + 20^\circ) = \tan (110^\circ - \alpha) \] **Step 2:** Set equal: \[ \tan \alpha = \tan (110^\circ - \alpha) \] **Step 3:** Use properties of tangent: \[ \text{Since } \tan A = \tan B, \text{ then } A = B + k \times 180^\circ, \text{ for integer } k \] Given the problem asks for *acute* angles (\(0^\circ < \alpha < 90^\circ\)), the principal solution occurs when: \[ \alpha = 110^\circ - \alpha + 180^\circ k \] For the main solution (\(k=0\)): \[ 2\alpha = 110^\circ \] \[ \alpha = 55^\circ \] This is within the acute range. **Answer:** \[ \boxed{\alpha = 55^\circ} \] --- ### Part (b): \(\sin(3\beta + 50^\circ) = \cos \beta\) **Step 1:** Use the identity \(\cos \beta = \sin (90^\circ - \beta)\): \[ \sin(3\beta + 50^\circ) = \sin(90^\circ - \beta) \] **Step 2:** For \(\sin A = \sin B\): \[ A = B + 360^\circ k \quad \text{or} \quad A = 180^\circ - B + 360^\circ k \] Given \(\beta\) is acute, we'll consider \(k=0\). Set: 1. \(3\beta + 50^\circ = 90^\circ - \beta\) \[ 3\beta + 50^\circ = 90^\circ - \beta \] \[ 3\beta + \beta = 90^\circ - 50^\circ \] \[ 4\beta = 40^\circ \] \[ \beta = 10^\circ \] 2. \(3\beta + 50^\circ = 180^\circ - (90^\circ - \beta)\) \[ 3\beta + 50^\circ = 180^\circ - 90^\circ + \beta \] \[ 3\beta + 50^\circ = 90^\circ + \beta \] \[ 3\beta - \beta = 90^\circ - 50^\circ \] \[ 2\beta = 40^\circ \] \[ \beta = 20^\circ \] Both \(\beta = 10^\circ\) and \(\beta = 20^\circ\) are within the acute range. **Answers:** \[ \boxed{\beta = 10^\circ \text{ or } 20^\circ} \] --- ### Part (c): \(\sec(2\theta + 20^\circ) = \csc (\theta + 10^\circ)\) Recall: \[ \sec x = \frac{1}{\cos x}, \quad \csc x = \frac{1}{\sin x} \] Thus: \[ \frac{1}{\cos (2\theta + 20^\circ)} = \frac{1}{\sin (\theta + 10^\circ)} \] Cross-multiplied: \[ \sin (\theta + 10^\circ) = \cos (2\theta + 20^\circ) \] **Step 1:** Use identity: \[ \cos A = \sin (90^\circ - A) \] So, \[ \sin (\theta + 10^\circ) = \sin [90^\circ - (2\theta + 20^\circ)] \] \[ \sin (\theta + 10^\circ) = \sin (70^\circ - 2\theta) \] **Step 2:** Set equal: \[ \theta + 10^\circ = 70^\circ - 2\theta + 360^\circ k \] or \[ \theta + 10^\circ = 180^\circ - (70^\circ - 2\theta) + 360^\circ k \] Calculate each: **First:** \[ \theta + 10^\circ = 70^\circ - 2\theta \] \[ \theta + 2\theta = 70^\circ - 10^\circ \] \[ 3\theta = 60^\circ \] \[ \theta = 20^\circ \] **Second:** \[ \theta + 10^\circ = 180^\circ - 70^\circ + 2\theta \] \[ \theta + 10^\circ = 110^\circ + 2\theta \] \[ \theta - 2\theta = 110^\circ - 10^\circ \] \[ - \theta = 100^\circ \] \[ \theta = -100^\circ \] Given the problem probably expects a positive acute solution, the relevant solution is: \[ \boxed{\theta = 20^\circ} \] --- ### **Final answers:** - **(a):** \(\boxed{55^\circ}\) - **(b):** \(\boxed{\beta = 10^\circ \text{ or } 20^\circ}\) - **(c):** \(\boxed{\theta = 20^\circ}\)