Translating Context to Equations - Math 3

The time $T$ required for a pendulum to complete one full swing is given by $T = 2\pi\sqrt{\frac{L}{g}}$ , where $L$ is the length of the pendulum and $g$ is gravitational acceleration. If a pendulum must complete a swing in exactly 1.5 seconds when $g = 9.8$ m/s², which equation determines the required length $L$ ?
Answer: $1.5 = 2\pi\sqrt{\frac{L}{9.8}}$ Substituting T = 1.5 and g = 9.8 into T = 2π√(L/g) gives: 1.5 = 2π√(L/9.8). Choice B incorrectly places L outside the square root and puts 1.5 inside. Choice C moves L outside the square root entirely. Choice D attempts to isolate √L but uses incorrect algebra.
An investment account earns interest compounded continuously at a rate of 4.5% per year. If the initial deposit is $2500, which equation models the account balance $A$ after $t$ years?
Answer: $A = 2500e^{0.045t}$ When you encounter continuous compound interest problems, you need to recognize the specific formula that applies to this type of growth. Unlike regular compound interest that occurs at discrete intervals, continuous compounding uses the mathematical constant $e$ and follows the formula $A = Pe^{rt}$ , where $P$ is the principal, $r$ is the annual interest rate as a decimal, and $t$ is time in years. For this problem, you have $P = 2500$ , $r = 0.045$ (4.5% converted to decimal), so the equation becomes $A = 2500e^{0.045t}$ , which is choice D. Choice A, $A = 2500(e^{0.045})^t$ , incorrectly applies the exponent rules. This would calculate $e^{0.045}$ first (approximately 1.046), then raise that to the power $t$ , which doesn't represent continuous compounding. Choice B, $A = 2500(1.045)^t$ , uses the discrete compound interest formula where interest is compounded once per year. While 1.045 correctly represents the growth factor for annual compounding, this isn't the right model for continuous compounding. Choice C, $A = 2500e^{4.5t}$ , makes the common error of using the percentage form (4.5) instead of the decimal form (0.045) in the exponent. This would result in extremely unrealistic growth. Remember: continuous compound interest always uses the form $A = Pe^{rt}$ where the rate must be in decimal form in the exponent. If you see "compounded continuously," look for the answer with $e$ raised to the power of (decimal rate × time).
The resistance $R$ of two resistors connected in parallel, with individual resistances $R_1$ and $R_2$ , is given by $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$ . If one resistor has resistance $r$ ohms and the other has resistance $(r + 6)$ ohms, and the total resistance is 4 ohms, which equation can be solved to find $r$ ?
Answer: $\frac{1}{4} = \frac{1}{r} + \frac{1}{r + 6}$ When you encounter parallel resistance problems, remember that the key formula is $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$ , where $R$ is the total resistance and $R_1$ , $R_2$ are the individual resistances. To find the correct equation, substitute the given values directly into this formula. You know the total resistance $R = 4$ ohms, one resistor has resistance $r$ ohms, and the other has resistance $(r + 6)$ ohms. Plugging these into the parallel resistance formula gives: $\frac{1}{4} = \frac{1}{r} + \frac{1}{r + 6}$ This matches answer choice D exactly. Now let's examine why the other options are incorrect: Choice A sets $4 = r + (r + 6)$ , which incorrectly uses the formula for resistors in series, not parallel. Series resistance is simply the sum of individual resistances, but that's not what this problem describes. Choice B writes $4 = \frac{1}{r} + \frac{1}{r + 6}$ , which puts the total resistance $R$ on the wrong side of the equation. The parallel resistance formula has $\frac{1}{R}$ on the left, not $R$ . Choice C shows $\frac{1}{4} = \frac{r + (r + 6)}{r(r + 6)}$ , which appears to be an attempt to combine the fractions from the parallel formula, but this form isn't useful for solving and represents an unnecessary intermediate step. Study tip: Always substitute given values directly into the original formula first. Don't try to manipulate or simplify the formula before substituting—this often leads to errors and makes the problem more complicated than necessary.
The speed $v$ (in km/h) of a car is related to its stopping distance $d$ (in meters) by the equation $d = 0.05v^2 + 1.2v$ . If a car's stopping distance is 85 meters, which equation can be used to find the car's speed?
Answer: $85 = 0.05v^2 + 1.2v$ When you're given a formula that relates two variables and asked to find one variable given the other, you need to substitute the known value and set up an equation to solve for the unknown. You're given the relationship $d = 0.05v^2 + 1.2v$ where $d$ is stopping distance and $v$ is speed. Since you know the stopping distance is 85 meters and need to find the speed, you substitute 85 for $d$ in the original equation. This gives you $85 = 0.05v^2 + 1.2v$ , which is answer choice B. This equation can then be rearranged into standard quadratic form and solved for $v$ . Choice A incorrectly substitutes 85 for $v$ instead of $d$ , which would calculate a stopping distance when you already know the distance is 85 meters. Choice C sets the expression equal to zero rather than 85, which would find when the stopping distance is zero (not useful here). Choice D makes the same substitution error as A, but keeps $d$ as a variable instead of setting it equal to 85. The key insight is recognizing which variable is known and which is unknown, then substituting correctly. When you know the output of a function and need to find the input, substitute the known output value and solve for the input variable. Don't get confused about which variable goes where – always check that your substitution matches what the problem is telling you.
A rectangular garden has length $\ell$ meters and width $w$ meters. The garden is surrounded by a uniform walkway that is 2 meters wide. If the total area (garden plus walkway) is 528 square meters and the garden itself has area 360 square meters, which system of equations models this situation?
Answer: $\ell w = 360$ and $(\ell + 4)(w + 4) = 528$ When you encounter word problems involving shapes with borders or frames, you need to carefully track how the border affects the overall dimensions. Here, you have a rectangular garden with dimensions $\ell$ by $w$ , surrounded by a 2-meter-wide walkway on all sides. Start with what you know directly: the garden itself has area 360 square meters, so $\ell w = 360$ . This is your first equation. For the second equation, visualize how the walkway affects the total dimensions. Since the walkway is 2 meters wide and goes around the entire garden, it adds 2 meters on each side of both length and width. This means the total length becomes $\ell + 2 + 2 = \ell + 4$ , and the total width becomes $w + 2 + 2 = w + 4$ . The total area is 528 square meters, so $(\ell + 4)(w + 4) = 528$ . Looking at the wrong answers: Choice A incorrectly assigns the areas to the wrong expressions—it says the garden alone has area 528 and the total has area 360, which contradicts the problem. Choice B only adds 2 to each dimension instead of 4, forgetting that the walkway extends 2 meters on both sides of each dimension. Choice D attempts to calculate the walkway area separately but uses an incorrect formula that doesn't account for the corner overlap regions. Remember this key insight: when something surrounds a rectangle completely with uniform width $d$ , the new dimensions are the original dimensions plus $2d$ , not just $d$ .
A radioactive substance decays so that after $t$ years, the amount remaining is $A(t) = A_0 e^{-0.052t}$ , where $A_0$ is the initial amount. If 75 grams remain after 8 years, which equation can be used to find the initial amount $A_0$ ?
Answer: $75 = A_0 e^{-0.052 \cdot 8}$ Substituting t = 8 and A(8) = 75 into A(t) = A₀ $e^{-0.052t}$ gives: 75 = A₀ $e^{-0.052×8}$ . Choice B incorrectly solves for A₀ by multiplying instead of dividing. Choice C uses positive exponent, representing growth instead of decay. Choice D combines errors from both B and C.
A water tank is being drained. The height $h$ of water (in feet) after $t$ minutes is modeled by $h = 12 - \sqrt{4t + 1}$ . Which equation represents the time when the water height reaches 8 feet?
Answer: $8 = 12 - \sqrt{4t + 1}$ When you encounter a function modeling a real-world situation, you need to substitute the given value for the output variable and set up an equation to solve for the input variable. Here, you're given the function $h = 12 - \sqrt{4t + 1}$ where $h$ represents the water height and $t$ represents time. To find when the height reaches 8 feet, you substitute $h = 8$ into the equation: $8 = 12 - \sqrt{4t + 1}$ . This directly matches answer choice D. Let's examine why the other options are incorrect: Choice A rearranges the equation incorrectly, moving the 12 to the wrong side and making it negative: $8 = \sqrt{4t + 1} - 12$ . This changes the mathematical relationship entirely. Choice B confuses the variables by substituting 8 for $t$ instead of $h$ : $t = 12 - \sqrt{4 \cdot 8 + 1}$ . This treats time as the output and tries to solve for some unknown input, which doesn't match the question. Choice C represents an intermediate step you might reach while solving the equation ( $\sqrt{4t + 1} = 4$ ), but it's not the equation that represents the original problem. The question asks for the equation to set up, not a simplified version. The key strategy here is to identify which variable you're given (the height of 8 feet) and which you're solving for (the time $t$ ). Always substitute the known value for its corresponding variable in the original function, then solve from there.
The velocity $v$ of water flowing from a tank through a hole is given by Torricelli's law: $v = \sqrt{2gh}$ , where $g = 9.8$ m/s² and $h$ is the height of water above the hole. If water must flow at 14 m/s, which equation determines the required height?
Answer: $14 = \sqrt{2 \cdot 9.8 \cdot h}$ When you encounter physics equations with square roots, the key is carefully substituting known values while preserving the equation's structure. Torricelli's law relates velocity to height through the square root relationship $v = \sqrt{2gh}$ . To find the required height when water must flow at 14 m/s, you substitute the known values directly into the original equation. With $v = 14$ m/s and $g = 9.8$ m/s², the equation becomes: $14 = \sqrt{2 \cdot 9.8 \cdot h}$ . This is answer choice C. Let's examine why the other options are incorrect: Choice A ( $14 = 2 \cdot 9.8 \cdot \sqrt{h}$ ) incorrectly separates the 2 and $g$ from under the square root. The original formula has $\sqrt{2gh}$ , not $2g\sqrt{h}$ . Choice B ( $\sqrt{14} = 2 \cdot 9.8 \cdot h$ ) makes two errors: it puts the square root on the velocity (14) instead of on the right side, and it removes the square root from the $2gh$ term entirely. Choice D ( $14^2 = 2 \cdot 9.8 \cdot h$ ) represents what you'd get after squaring both sides of the correct equation, but the question asks for "which equation determines the required height," meaning the direct substitution step, not the solved form. Study tip: When substituting into equations with radicals, always substitute into the original form first before manipulating. Don't rearrange the equation structure during the substitution step—that comes after you've set up the problem correctly.
A water tank is being filled by two pipes simultaneously. Pipe A can fill the tank in $h$ hours when working alone, while Pipe B can fill the same tank in $(h+3)$ hours when working alone. When both pipes work together, they can fill the tank in 2 hours. Which equation correctly models this situation?
Answer: $\frac{1}{h} + \frac{1}{h+3} = \frac{1}{2}$ When working together, the rates of work add up. Pipe A fills 1/h of the tank per hour, Pipe B fills 1/(h+3) of the tank per hour, and together they fill 1/2 of the tank per hour. Therefore: 1/h + 1/(h+3) = 1/2. Choice B incorrectly uses 2 instead of 1/2 for the combined rate. Choice C adds times instead of rates. Choice D uses times instead of rates and sets equal to 2.
The concentration $C$ of a medication in the bloodstream $t$ hours after injection follows the model $C(t) = \frac{12t}{t^2 + 4}$ mg/L. A doctor needs to determine when the concentration will be exactly 2.4 mg/L. Which equation should be solved?
Answer: $2.4 = \frac{12t}{t^2 + 4}$ When you're given a function that models a real-world situation and asked to find when that function equals a specific value, you need to set up an equation where the function equals the target value. Here, you have the concentration function $C(t) = \frac{12t}{t^2 + 4}$ and you want to find when the concentration equals 2.4 mg/L. This means you need to solve for when $C(t) = 2.4$ , which gives you the equation $2.4 = \frac{12t}{t^2 + 4}$ . This is exactly what option B shows. Looking at the wrong answers: Option A incorrectly places 2.4 in the numerator with $t$ , which would change the original function completely. Option C shows the result after multiplying both sides by $(t^2 + 4)$ , but the question asks which equation should be solved initially, not what you get after algebraic manipulation. Option D makes an error in the cross-multiplication process—when you multiply $2.4$ by $(t^2 + 4)$ , you should get $2.4t^2 + 9.6$ , not $2.4t^2 + 4$ . Remember this key strategy: when a problem asks "find when [function] equals [value]," always set up your equation as "[value] = [function]" or "[function] = [value]." Don't be tempted to choose an answer that shows intermediate algebraic steps—stick with the direct setup unless specifically asked for a simplified form.
A projectile is launched vertically upward from a building. The height $h(t)$ in feet above ground level after $t$ seconds is given by $h(t) = -16t^2 + v_0t + h_0$ , where $v_0$ is the initial velocity and $h_0$ is the initial height. If the projectile reaches its maximum height of 180 feet at $t = 2.5$ seconds and was launched from a height of 20 feet, which equation can be used to find $v_0$ ?
Answer: $-16(2.5)^2 + v_0(2.5) + 20 = 180$ At maximum height (t = 2.5), h(2.5) = 180. Substituting into h(t) = -16t² + v₀t + h₀: -16(2.5)² + v₀(2.5) + 20 = 180. Choice B uses the derivative condition but isn't complete for finding v₀ directly. Choice C omits the initial height h₀ = 20. Choice D incorrectly uses +16 instead of -16 for the gravity term.
The time $t$ (in seconds) for a pendulum to complete one full swing is given by $t = 2\pi\sqrt{\frac{L}{9.8}}$ , where $L$ is the length in meters. If a pendulum takes 3.2 seconds for one swing, which equation can be used to find its length?
Answer: $3.2 = 2\pi\sqrt{\frac{L}{9.8}}$ Substituting $t = 3.2$ into the given formula gives $3.2 = 2\pi\sqrt{\frac{L}{9.8}}$ . Choice B incorrectly swaps variables. Choice C moves the $2\pi$ inside the square root. Choice D attempts to solve for $\sqrt{L}$ but uses incorrect algebra.
A company's revenue $R$ (in thousands of dollars) is related to advertising expenditure $x$ (in thousands of dollars) by the equation $R = \frac{120x}{x + 15}$ . If the company wants to achieve a revenue of 90 thousand dollars, which equation must be solved to find the required advertising expenditure?
Answer: $90 = \frac{120x}{x + 15}$ When you encounter a problem where you're given a function and asked to find the input that produces a specific output, you're working with equation setup and solving. Here, you have the revenue function $R = \frac{120x}{x + 15}$ and want to find what advertising expenditure $x$ will produce a revenue of 90 thousand dollars. To set up this equation, you substitute the desired revenue (90) for $R$ in the original function. This gives you $90 = \frac{120x}{x + 15}$ , which is exactly what answer choice B shows. Let's examine why the other options are incorrect: Choice A reverses the setup by putting 90 in place of the coefficient 120, creating $\frac{90x}{x + 15} = 120$ . This would be solving for a different scenario entirely. Choice C shows $90(x + 15) = 120x$ , which is actually the result you'd get after cross-multiplying the correct equation from choice B. While this equation would give the same solution, it's not the direct substitution the question asks for. Choice D places 90 in the wrong position, creating $90x = \frac{120}{x + 15}$ , which completely changes the mathematical relationship. Key strategy: When a problem asks "which equation must be solved," look for the direct substitution of given values into the original function. Don't overthink it by jumping ahead to algebraically manipulated forms – stick with the immediate substitution step.
The pH of a solution is given by $pH = -\log_{10}[H^+]$ , where $[H^+]$ is the hydrogen ion concentration in moles per liter. If a solution has a pH of 6.5, which equation can be used to find the hydrogen ion concentration?
Answer: $6.5 = -\log_{10}[H^+]$ When you encounter pH problems, remember that you're working with a logarithmic relationship between pH and hydrogen ion concentration. The key is understanding how to substitute known values into the given formula. You're given that $pH = -\log_{10}[H^+]$ and told the solution has a pH of 6.5. To find which equation helps determine the hydrogen ion concentration, you simply substitute the known pH value into the original formula. When you substitute 6.5 for pH, you get: $6.5 = -\log_{10}[H^+]$ This matches answer choice C exactly. Let's examine why the other options are incorrect. Choice A gives $6.5 = \log_{10}[H^+]$ , which omits the crucial negative sign from the original formula. Choice B shows $-6.5 = \log_{10}[H^+]$ , which incorrectly makes the pH negative and removes the negative sign from the logarithm term. Choice D presents $[H^+] = -\log_{10}(6.5)$ , which confuses the relationship by putting the hydrogen ion concentration on the left side and treating 6.5 as the input to the logarithm rather than as the pH value. When working with logarithmic formulas like the pH equation, always substitute your known values directly into the original formula without rearranging first. This prevents sign errors and variable mix-ups. Focus on maintaining the exact structure of the given equation while replacing the appropriate variable with its numerical value.
The resistance $R$ (in ohms) of two resistors connected in parallel with individual resistances $R_1$ and $R_2$ is given by $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$ . If one resistor has resistance 6 ohms and the combined resistance is 2 ohms, which equation can be used to find the resistance $R_2$ of the second resistor?
Answer: $\frac{1}{2} = \frac{1}{6} + \frac{1}{R_2}$ When you encounter problems involving resistors in parallel, you're working with reciprocal relationships. The key formula $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$ tells us that the reciprocal of the total resistance equals the sum of the reciprocals of individual resistances. To find the correct equation, substitute the known values directly into the original formula. You know the combined resistance $R = 2$ ohms and one individual resistance $R_1 = 6$ ohms. Substituting these values: $\frac{1}{2} = \frac{1}{6} + \frac{1}{R_2}$ . This matches answer choice C exactly. Let's examine why the other options are incorrect. Choice A gives $\frac{1}{R_2} = \frac{1}{2} + \frac{1}{6}$ , which incorrectly isolates $\frac{1}{R_2}$ by adding the other terms instead of subtracting. Choice B shows $2 = 6 + R_2$ , which completely ignores the reciprocal nature of the parallel resistance formula and treats it like a simple addition problem. Choice D states $R_2 = \frac{1}{2} - \frac{1}{6}$ , which confuses $R_2$ with $\frac{1}{R_2}$ and jumps ahead to a calculation step rather than setting up the substitution equation. Remember this strategy: when working with parallel resistance problems, always start by substituting known values directly into the given formula before attempting any algebraic manipulation. This prevents confusion between the resistance values and their reciprocals, and ensures you set up the equation correctly from the start.
The population of a bacterial culture triples every 4 hours. If the initial population is $P_0$ bacteria, which equation models the population $P(t)$ after $t$ hours?
Answer: $P(t) = P_0 \cdot 3^{\frac{t}{4}}$ Since the population triples every 4 hours, after t hours, there have been t/4 tripling periods. Each tripling multiplies by 3, so P(t) = P₀ · $3^{t/4}$ . Choice B incorrectly uses 4t in the exponent, making growth too rapid. Choice C confuses the tripling factor with the time period. Choice D incorrectly multiplies P₀ by 3 initially and uses base 4.
The pH of a solution is related to its hydrogen ion concentration $[H^+]$ (in moles per liter) by the equation $\text{pH} = -\log_{10}[H^+]$ . If a solution has a pH of 5.2, which equation can be used to find its hydrogen ion concentration?
Answer: $5.2 = -\log_{10}[H^+]$ When you encounter pH problems, remember that you're working with a logarithmic relationship between pH and hydrogen ion concentration. The key is understanding how to manipulate the given formula when you know one variable and need to find the other. Given that pH = 5.2 and the formula $\text{pH} = -\log_{10}[H^+]$ , you can directly substitute the known pH value into the equation. This gives you $5.2 = -\log_{10}[H^+]$ , which is exactly what choice B shows. This equation is now set up perfectly to solve for $[H^+]$ using inverse logarithmic operations. Choice A reverses the roles incorrectly, treating the pH value as if it were the hydrogen ion concentration in the logarithm. Choice C drops the negative sign entirely, which fundamentally changes the relationship—this would give you a positive logarithm rather than the negative one required by the pH definition. Choice D also omits the crucial negative sign and would result in an incorrect hydrogen ion concentration that's far too large for a pH of 5.2. The negative sign in the pH formula is critical because hydrogen ion concentrations are typically very small decimal numbers (like 0.0001), and the negative logarithm converts these into positive, manageable pH values. Study tip: When working with logarithmic formulas, always substitute known values directly into the original equation first, then solve for the unknown. Don't try to rearrange the formula in your head—write out the substitution step to avoid sign errors and variable mix-ups.
A bacteria culture grows exponentially, tripling every 4 hours. If there are initially 200 bacteria, which equation models the population $P$ after $h$ hours?
Answer: $P = 200 \cdot 3^{\frac{h}{4}}$ The population triples every 4 hours, so after $h$ hours it multiplies by $3^{\frac{h}{4}}$ . Choice B has the time relationship inverted. Choice C confuses the growth factor and time period. Choice D incorrectly adds 1 to the growth factor.
The half-life of a certain isotope is 12 years. If a sample initially contains $N_0$ atoms, which equation models the number of atoms $N(t)$ remaining after $t$ years?
Answer: $N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{12}}$ When you encounter radioactive decay problems, you're dealing with exponential decay where the key insight is understanding what "half-life" means: the time it takes for half the substance to decay. The correct model is $N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{12}}$ . Here's why this works: after 12 years (one half-life), we get $N(12) = N_0 \left(\frac{1}{2}\right)^{\frac{12}{12}} = N_0 \left(\frac{1}{2}\right)^1 = \frac{N_0}{2}$ , which is exactly half the original amount. After 24 years (two half-lives), we get $N_0 \left(\frac{1}{2}\right)^2 = \frac{N_0}{4}$ , confirming the pattern. Choice A is wrong because $N(t) = N_0 \left(\frac{1}{2}\right)^{t-12}$ gives you the full amount $N_0$ at $t = 12$ years, not half. This incorrectly shifts the decay timeline. Choice B incorrectly uses $N(t) = N_0 \left(\frac{1}{2}\right)^{12t}$ , making the decay far too rapid. At just $t = 1$ year, you'd have $N_0 \left(\frac{1}{2}\right)^{12}$ , which is virtually zero. Choice C, $N(t) = \frac{N_0}{2^{12t}}$ , is mathematically equivalent to choice B (since $\frac{1}{2^{12t}} = \left(\frac{1}{2}\right)^{12t}$ ) and has the same problem of excessively rapid decay. The key pattern to remember: in half-life problems, the exponent should always be $\frac{\text{time}}{\text{half-life period}}$ . This ensures that when the time equals one half-life period, the exponent equals 1, giving you exactly half the original amount.
The intensity of light from a source varies inversely with the square of the distance from the source. If the intensity is 400 lumens per square meter at a distance of 3 meters, which equation models the intensity $I$ at distance $d$ meters?
Answer: $I = \frac{3600}{d^2}$ For inverse square variation, I = k/d² where k is constant. Using I = 400 when d = 3: 400 = k/3², so k = 400 × 9 = 3600. Therefore I = 3600/d². Choice B omits the distance factor from the constant. Choice C uses 400 × 3 instead of 400 × 9. Choice D shows the calculation but doesn't simplify 400 × 9 = 3600.

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