A transport service models the total cost by C = 35+8n. A second student writes C−35 = 8n. Which comparison is accurate?

The first equation is $C = 35 + 8n$.
The second equation is $C - 35 = 8n$.

Starting with $C = 35 + 8n$, subtract 35 from both sides of the equation:

This shows that the two equations are mathematically equivalent.

In the context of a transport service cost model:

• $C$: Represents the total cost of the transport service.
• 35: Represents a fixed fee or base charge, which is incurred regardless of the number of units ($n$).
• 8: Represents the variable cost per unit, meaning the cost for each additional unit of service.
• $n$: Represents the number of units of the transport service (e.g., number of trips, miles, hours).
• $8n$: Represents the total variable cost, which depends on the number of units.

Option C states that the equations are equivalent, and subtracting the fixed fee (35) isolates the variable charge ($8n$), while units remain consistent. This aligns with our analysis.

The correct comparison is $\boxed{\text{C. The equations are equivalent; subtracting the fixed fee isolates the variable charge, and the units remain consistent.}}$.

Zanele needs to adapt an equation task for a diverse group of Grade 6 learners, aiming for the "fairest conceptual access." This means choosing a method that minimizes barriers to understanding and maximizes engagement for all students.

• A. Long verbal description with irrelevant facts and algebraic algorithm first: This is likely to confuse Grade 6 learners, especially a diverse group, as it introduces abstraction before conceptual understanding.
• B. Unfamiliar context (snow-ski rental) without explanation: Unfamiliar contexts increase cognitive load and can create an access barrier for learners who lack background knowledge.
• C. Smartphone shopping application: While digital, not all learners may have equal access to or familiarity with smartphones or specific apps, potentially creating a digital divide barrier.
• D. Familiar transport-fare context, simple table or counters, and ask learners to name quantities: This option uses a familiar context (transport fare), which reduces cognitive load. It provides concrete tools (table/counters) for hands-on exploration, which is excellent for Grade 6. Asking learners to name quantities before forming an equation helps them build conceptual understanding of variables.

• Access barrier avoided: Unfamiliar context or abstract representation without concrete grounding. By using a familiar context and concrete manipulatives (table/counters), learners can connect the problem to their real-world experiences, reducing the cognitive demand associated with abstract concepts.
• Mathematical idea retained: The concept of a variable representing an unknown quantity, and the idea of modeling a real-world situation with an equation. Learners identify the quantities (e.g., fixed cost, cost per unit, total cost, number of units) and understand how they relate to form an equation.

The choice that gives the fairest conceptual access is $\boxed{\text{D. Use a familiar transport-fare context, offer a simple table or counters, and ask learners to name the quantities before forming an equation.}}$.

The sequence is $7, 11, 15, 19, \dots$
Calculate the difference between consecutive terms:
$11 - 7 = 4$
$15 - 11 = 4$
$19 - 15 = 4$
The common difference ($d$) is 4. This is an arithmetic sequence.

The terms are:
$T_1 = 7$
$T_2 = 11$
$T_3 = 15$
$T_4 = 19$
To find $T_5$, add the common difference to $T_4$:
$T_5 = T_4 + d = 19 + 4 = 23$.

The general formula for the $n$-th term of an arithmetic sequence is $T_n = T_1 + (n-1)d$.
For $T_5$:
$T_5 = T_1 + (5-1)d$
$T_5 = T_1 + 4d$
Substituting the values $T_1 = 7$ and $d = 4$:
$T_5 = 7 + 4(4)$
$T_5 = 7 + 16$
$T_5 = 23$
This shows that to get to the 5th term, you add the common difference 4 times to the first term.

Option C states $T_5 = 23$, using four equal increases. This matches our calculation and understanding of the relationship between the term number and the common difference.

The correct answer is $\boxed{\text{C. } T_5 = 23, \text{ using four equal increases}}$.