Algebra Flashcards: Recognize Constant Rate Changes

What this deck covers

This deck focuses on Recognize Constant Rate Changes, giving you a quick way to review the definitions, rules, and examples that matter most for Algebra.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: Identify whether the relationship is constant rate: points $(0,1)$, $(1,2)$, $(2,4)$.

Answer: No; slopes change ($1$ then $2$). Slopes differ: $(2-1)/(1-0) = 1$, $(4-2)/(2-1) = 2$.

Flashcard 2: Find the unit rate if y=-12 when x=3.

Answer: $-4$. Dividing change: $\frac{-12}{3} = -4$ units per unit.

Flashcard 3: What is the name of the constant ratio y/x for a linear relationship?

Answer: Slope (constant rate of change). In linear equations, the coefficient of $x$ represents the constant rate.

Flashcard 4: Find the constant rate if $y$ changes from $30$ to $18$ while $x$ changes from $2$ to $8$.

Answer: $-2$. Rate equals $\frac{18-30}{8-2} = \frac{-12}{6} = -2$.

Flashcard 5: Identify the constant rate of change from the points $(3,2)$ and $(9,14)$.

Answer: $2$. Using slope formula: $\frac{14-2}{9-3} = \frac{12}{6} = 2$.

Flashcard 6: Which expression represents a constant rate of change: $y=4x+9$ or $y=4^x+9$?

Answer: $y=4x+9$. Linear has constant rate; exponential has variable rate.

Flashcard 7: Identify the constant rate if a taxi charges $3 plus $2 per mile.

Answer: $2$ dollars per mile. The coefficient of miles gives the per-mile rate.

Flashcard 8: Identify the constant rate of change from the points $(-2,7)$ and $(4,-5)$.

Answer: $-2$. Using slope formula: $\frac{-5-7}{4-(-2)} = \frac{-12}{6} = -2$.

Flashcard 9: Find the unit rate (slope) if $y=18$ when $x=6$.

Answer: $3$. Dividing total change: $\frac{18}{6} = 3$ units per unit.

Flashcard 10: A table has equal $x$ steps of $1$ and $y$ values $1,4,9,16$. Is the rate constant?

Answer: No; the changes are $3,5,7$. These are perfect squares; differences increase: $3, 5, 7$.

Flashcard 11: Identify if the rate is constant: $(0,0)$, $(2,6)$, $(5,15)$.

Answer: Yes; each slope is $3$. All slopes equal $3$: $\frac{6-0}{2-0} = 3$, $\frac{15-6}{5-2} = 3$.

Flashcard 12: Identify the slope (rate) of the line through $(2,-1)$ and $(2,6)$.

Answer: Undefined (division by $0$). Both points have same $x$-value, so $\Delta x = 0$.

Flashcard 13: Choose the word that signals constant rate: "per" or "percent"?

Answer: "Per" (as in "per unit"). "Per" indicates rate per unit; "percent" often suggests proportional change.

Flashcard 14: Which model matches constant rate: $y=y_0+rt$ or $y=y_0(1+r)^t$?

Answer: $y=y_0+rt$. Linear form $y_0 + rt$ has constant rate; exponential has variable rate.

Flashcard 15: What does it mean for $y$ to change at a constant rate per unit of $x$?

Answer: The ratio $\Delta y / \Delta x$ is constant for all intervals. This defines linear relationships where change in $y$ per unit $x$ stays the same.

Flashcard 16: Identify the initial value in $y=mx+b$ that pairs with constant rate $m$.

Answer: $b$ (the value of $y$ when $x=0$). The $y$-intercept $b$ is the starting value when $x = 0$.

Flashcard 17: Identify if the rate is constant: $(0,2)$, $(2,5)$, $(5,11)$.

Answer: No; slopes are $\frac{3}{2}$ and $2$. Slopes differ: $\frac{5-2}{2-0} = 1.5$, $\frac{11-5}{5-2} = 2$.

Flashcard 18: Identify the slope (rate) of the line through $(0,4)$ and $(5,4)$.

Answer: $0$. Both points have same $y$-value, so $\Delta y = 0$.

Flashcard 19: Identify the slope between $(2,5)$ and $(6,13)$ to test if the rate is constant.

Answer: $2$. Using slope formula: $\frac{13-5}{6-2} = \frac{8}{4} = 2$.

Flashcard 20: What is the constant rate of change in the equation $C=2.5t+40$?

Answer: $2.5$. The coefficient $2.5$ represents the rate per unit of $t$.

Flashcard 21: Which situation shows constant rate: "$y$ increases by $8$ every $2$ minutes" or "$y$ increases by $8$ every minute"?

Answer: Both (each states a constant per-time change). Both describe fixed amounts per time interval (constant rates).

Flashcard 22: What does a negative constant rate of change indicate about the relationship between $x$ and $y$?

Answer: As $x$ increases, $y$ decreases at a constant rate. Negative slope means $y$ decreases as $x$ increases.

Flashcard 23: Identify the constant rate of change from the points $(3,2)$ and $(9,14)$.

Answer: $2$. Using slope formula: $\frac{14-2}{9-3} = \frac{12}{6} = 2$.

Flashcard 24: What does a vertical line mean about a rate per unit of $x$?

Answer: Undefined rate; it is not a function of $x$. Vertical lines have zero $\Delta x$, making slope undefined.

Flashcard 25: What does a positive constant rate of change indicate about the relationship between $x$ and $y$?

Answer: As $x$ increases, $y$ increases at a constant rate. Positive slope means $y$ increases as $x$ increases.

Flashcard 26: A quantity follows $y=50+6t$. What is the constant rate and in what units?

Answer: $6$ units of $y$ per $1$ unit of $t$. The coefficient $6$ gives the rate: $6$ $y$-units per $t$-unit.

Flashcard 27: A table has equal $x$ steps of $1$ and $y$ values $2,6,10,14$. Is the rate constant?

Answer: Yes; the change is $+4$ each step. Each $y$-value increases by exactly $4$ for each unit $x$ increase.

Flashcard 28: Identify the slope (rate) of the line through $(0,4)$ and $(5,4)$.

Answer: $0$. Both points have same $y$-value, so $\Delta y = 0$.

Flashcard 29: Which phrase indicates constant rate: "$+12$ each year" or "multiplied by $1.12$ each year"?

Answer: "$+12$ each year.". Addition shows constant rate; multiplication shows exponential growth.

Flashcard 30: A plant grows $15$ cm in $3$ weeks at a steady pace. What is the constant rate per week?

Answer: $5$ cm per week. Divide total growth by time: $\frac{15}{3} = 5$ cm per week.

Flashcard 31: Which model matches constant rate: $y=y_0+rt$ or $y=y_0(1+r)^t$?

Answer: $y=y_0+rt$. Linear form $y_0 + rt$ has constant rate; exponential has variable rate.

Flashcard 32: Which phrase indicates constant rate: "$+12$ each year" or "multiplied by $1.12$ each year"?

Answer: "$+12$ each year.". Addition shows constant rate; multiplication shows exponential growth.

Flashcard 33: Choose the word that signals constant rate: "per" or "percent"?

Answer: "Per" (as in "per unit"). "Per" indicates rate per unit; "percent" often suggests proportional change.

Flashcard 34: If $x$ increases by $4$ and $y$ decreases by $12$, what is the constant rate of change?

Answer: $-3$. Rate equals $\frac{-12}{4} = -3$ (negative because $y$ decreases).

Flashcard 35: Identify the initial value in $y=mx+b$ that pairs with constant rate $m$.

Answer: $b$ (the value of $y$ when $x=0$). The $y$-intercept $b$ is the starting value when $x = 0$.

Flashcard 36: Identify whether the relationship is constant rate: points $(0,2)$, $(1,5)$, $(2,8)$.

Answer: Yes; slope is $3$ each step. Each consecutive slope equals $3$: $(5-2)/(1-0) = 3$, $(8-5)/(2-1) = 3$.

Flashcard 37: In $y=-5x+12$, what is the constant rate and what does it mean per $1$ unit of $x$?

Answer: Rate $-5$; $y$ decreases by $5$ for each $+1$ in $x$. The coefficient $-5$ means $y$ drops $5$ units per $x$ unit.

Flashcard 38: Find the unit rate if $y=-12$ when $x=3$.

Answer: $-4$. Dividing change: $\frac{-12}{3} = -4$ units per unit.

Flashcard 39: A car travels $180$ miles in $3$ hours at constant speed. What is the constant rate?

Answer: $60$ miles per hour. Divide distance by time: $\frac{180}{3} = 60$ miles per hour.

Flashcard 40: Which equation form shows a constant rate of change: $y=mx+b$ or $y=ax^2+bx+c$?

Answer: $y=mx+b$. Linear form has constant slope $m$; quadratic has variable rate.

Flashcard 41: What does a horizontal line on a graph mean about the constant rate of change?

Answer: Rate is $0$ (no change in $y$ as $x$ changes). Horizontal lines have zero slope since $y$ never changes.

Flashcard 42: A table has $x$ steps of $2$ and $y$ values $1,7,13,19$. What is the constant rate?

Answer: $3$. Changes are $6, 6, 6$ for steps of $2$, so rate is $\frac{6}{2} = 3$.

Flashcard 43: What is the name of the constant ratio y/x for a linear relationship?

Answer: Slope (constant rate of change). In linear equations, the coefficient of $x$ represents the constant rate.

Flashcard 44: Identify the slope (rate) of the line through $(2,-1)$ and $(2,6)$.

Answer: Undefined (division by $0$). Both points have same $x$-value, so $\Delta x = 0$.

Flashcard 45: Identify if the rate is constant: $(0,2)$, $(2,5)$, $(5,11)$.

Answer: No; slopes are $\frac{3}{2}$ and $2$. Slopes differ: $\frac{5-2}{2-0} = 1.5$, $\frac{11-5}{5-2} = 2$.

Flashcard 46: Which situation is constant rate: $y$ increases by $3$ when $x$ increases by $1$, always; or increases by $3,6,12$?

Answer: The first situation (always $+3$ per $+1$). Consistent change shows constant rate; varying increases indicate non-linear.

Flashcard 47: Convert "increases by $8$ every $2$ minutes" to a per-$1$-minute constant rate.

Answer: $4$ per minute. Divide the change by time: $\frac{8}{2} = 4$ per minute.

Flashcard 48: Which situation is constant rate: $y$ increases by $3$ when $x$ increases by $1$, always; or increases by $3,6,12$?

Answer: The first situation (always $+3$ per $+1$). Consistent change shows constant rate; varying increases indicate non-linear.

Flashcard 49: Identify the constant rate if the line passes through $(0,-3)$ and $(4,5)$.

Answer: $2$. Using slope formula: $\frac{5-(-3)}{4-0} = \frac{8}{4} = 2$.

Flashcard 50: A car travels $180$ miles in $3$ hours at constant speed. What is the constant rate?

Answer: $60$ miles per hour. Divide distance by time: $\frac{180}{3} = 60$ miles per hour.

Flashcard 51: What is the constant rate of change in the equation $C=2.5t+40$?

Answer: $2.5$. The coefficient $2.5$ represents the rate per unit of $t$.

Flashcard 52: A table has equal $x$ steps of $1$ and $y$ values $2,6,10,14$. Is the rate constant?

Answer: Yes; the change is $+4$ each step. Each $y$-value increases by exactly $4$ for each unit $x$ increase.

Flashcard 53: Which equation form shows a constant rate of change: $y=mx+b$ or $y=ax^2+bx+c$?

Answer: $y=mx+b$. Linear form has constant slope $m$; quadratic has variable rate.

Flashcard 54: What does a vertical line mean about a rate per unit of $x$?

Answer: Undefined rate; it is not a function of $x$. Vertical lines have zero $\Delta x$, making slope undefined.

Flashcard 55: What does a positive constant rate of change indicate about the relationship between $x$ and $y$?

Answer: As $x$ increases, $y$ increases at a constant rate. Positive slope means $y$ increases as $x$ increases.

Flashcard 56: In $y=-5x+12$, what is the constant rate and what does it mean per $1$ unit of $x$?

Answer: Rate $-5$; $y$ decreases by $5$ for each $+1$ in $x$. The coefficient $-5$ means $y$ drops $5$ units per $x$ unit.

Flashcard 57: If $x$ increases by $2$ each step and $y$ increases by $10$ each step, what is the constant rate $\Delta y / \Delta x$?

Answer: $5$. Rate equals $\frac{\Delta y}{\Delta x} = \frac{10}{2} = 5$.

Flashcard 58: Identify the slope between $(1,9)$ and $(4,3)$.

Answer: $-2$. Using slope formula: $\frac{3-9}{4-1} = \frac{-6}{3} = -2$.

Flashcard 59: Identify the slope between $(2,5)$ and $(6,13)$ to test if the rate is constant.

Answer: $2$. Using slope formula: $\frac{13-5}{6-2} = \frac{8}{4} = 2$.

Flashcard 60: Identify if the rate is constant: $(0,0)$, $(2,6)$, $(5,15)$.

Answer: Yes; each slope is $3$. All slopes equal $3$: $\frac{6-0}{2-0} = 3$, $\frac{15-6}{5-2} = 3$.

Flashcard 61: What must be true about slopes between any two pairs of points on a line?

Answer: All computed slopes are equal (constant). Linear relationships maintain identical slope between any point pairs.

Flashcard 62: If $x$ increases by $2$ each step and $y$ increases by $10$ each step, what is the constant rate y/x?

Answer: $5$. Rate equals $\frac{\Delta y}{\Delta x} = \frac{10}{2} = 5$.

Flashcard 63: Find the unit rate (slope) if y=18 when x=6.

Answer: $3$. Dividing total change: $\frac{18}{6} = 3$ units per unit.

Flashcard 64: Which graph shape indicates a constant rate of change: straight line or curved line?

Answer: Straight line. Constant slope produces straight lines; variable slope creates curves.

Flashcard 65: What is the constant rate of change in the equation y=-x+10?

Answer: -. The coefficient of $x$ gives the constant rate of change.

Flashcard 66: What does a negative constant rate of change indicate about the relationship between $x$ and $y$?

Answer: As $x$ increases, $y$ decreases at a constant rate. Negative slope means $y$ decreases as $x$ increases.

Flashcard 67: What is the constant rate of change in the equation y=-x+10?

Answer: -. The coefficient of $x$ gives the constant rate of change.

Flashcard 68: Which description indicates a constant rate: "adds $5$ each hour" or "doubles each hour"?

Answer: "Adds $5$ each hour.". Fixed addition indicates constant rate; doubling shows exponential growth.

Flashcard 69: What does a horizontal line on a graph mean about the constant rate of change?

Answer: Rate is $0$ (no change in $y$ as $x$ changes). Horizontal lines have zero slope since $y$ never changes.

Flashcard 70: Identify the slope between $(1,9)$ and $(4,3)$.

Answer: $-2$. Using slope formula: $\frac{3-9}{4-1} = \frac{-6}{3} = -2$.

Flashcard 71: Identify whether the relationship is constant rate: points $(0,1)$, $(1,2)$, $(2,4)$.

Answer: No; slopes change ($1$ then $2$). Slopes differ: $(2-1)/(1-0) = 1$, $(4-2)/(2-1) = 2$.

Flashcard 72: What is the constant rate of change in the equation $y=7x-3$?

Answer: $7$. The coefficient of $x$ is the slope (constant rate of change).

Flashcard 73: A quantity follows $y=50+6t$. What is the constant rate and in what units?

Answer: $6$ units of $y$ per $1$ unit of $t$. The coefficient $6$ gives the rate: $6$ $y$-units per $t$-unit.

Flashcard 74: What does it mean for $y$ to change at a constant rate per unit of $x$?

Answer: The ratio y/x is constant for all intervals. This defines linear relationships where change in $y$ per unit $x$ stays the same.

Flashcard 75: What is the slope formula for the rate of change from $(x_1,y_1)$ to $(x_2,y_2)$?

Answer: $m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}$. This formula calculates the constant rate between two points.

Flashcard 76: Identify whether the statement implies constant rate: "The balance decreases by $25$ dollars each month."

Answer: Yes; it is a constant decrease of $25$ per month. Fixed decrease per month indicates linear (constant rate) change.

Flashcard 77: A plant grows $15$ cm in $3$ weeks at a steady pace. What is the constant rate per week?

Answer: $5$ cm per week. Divide total growth by time: $\frac{15}{3} = 5$ cm per week.

Flashcard 78: Identify whether the relationship is constant rate: points $(0,2)$, $(1,5)$, $(2,8)$.

Answer: Yes; slope is $3$ each step. Each consecutive slope equals $3$: $(5-2)/(1-0) = 3$, $(8-5)/(2-1) = 3$.

Flashcard 79: A table has equal $x$ steps of $1$ and $y$ values $1,4,9,16$. Is the rate constant?

Answer: No; the changes are $3,5,7$. These are perfect squares; differences increase: $3, 5, 7$.

Flashcard 80: Identify the constant rate of change from the points $(-2,7)$ and $(4,-5)$.

Answer: $-2$. Using slope formula: $\frac{-5-7}{4-(-2)} = \frac{-12}{6} = -2$.

Flashcard 81: Which table pattern indicates constant rate: constant first differences in $y$ or constant second differences?

Answer: Constant first differences in $y$ (for equal $x$ steps). Linear relationships show equal $y$-differences for equal $x$-steps.

Flashcard 82: Identify whether the statement implies constant rate: "The balance decreases by $10\%$ each month."

Answer: No; percent decrease suggests exponential change, not constant difference. Percentage decreases create exponential, not linear relationships.

Flashcard 83: Identify the constant rate if a taxi charges $3 plus $2 per mile.

Answer: $2$ dollars per mile. The coefficient of miles gives the per-mile rate.

Flashcard 84: What is the slope formula for the rate of change from $(x_1,y_1)$ to $(x_2,y_2)$?

Answer: $m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}$. This formula calculates the constant rate between two points.