In an analytical chemistry laboratory, a student models the total titrant volume $V_{\text{total}}(t)$ in milliliters delivered by two automated syringe pumps during an acid-base neutralization experiment. Pump A dispenses volume modeled by $V_A(t) = 0.45(t^2 - 2.4t + 1.80)$, and Pump B dispenses volume modeled by $V_B(t) = 0.30(2.5t^2 + 1.6t - 0.90)$, where $t$ is the elapsed time in minutes. Combine all like terms to write a single simplified quadratic function for $V_{\text{total}}(t)$ with coefficients rounded to two decimal places, and compute $V

In an analytical chemistry laboratory, a student models the total titrant volume $V_{\text{total}}(t)$ in milliliters delivered by two automated syringe pumps during an acid-base neutralization experiment. Pump A dispenses volume modeled by $V_A(t) = 0.45(t^2 - 2.4t + 1.80)$ , and Pump B dispenses volume modeled by $V_B(t) = 0.30(2.5t^2 + 1.6t - 0.90)$ , where $t$ is the elapsed time in minutes. Combine all like terms to write a single simplified quadratic function for $V_{\text{total}}(t)$ with coefficients rounded to two decimal places, and compute $V_{\text{total}}(4.00)$ .

Final answer

$V_{\text{total}}(t) = 1.20t^2 - 0.60t + 0.54\text{ mL}$ ; $V_{\text{total}}(4.00) = 17.34\text{ mL}$

Steps

Expand each expression using the distributive property

To find the total delivered titrant volume $V_{\text{total}}(t)$ , add the volumes delivered by Pump A and Pump B, expand the expressions, and group like terms together.

For Pump A: $V_A(t) = 0.45(t^2 - 2.4t + 1.80)$ $V_A(t) = (0.45 \cdot 1)t^2 - (0.45 \cdot 2.4)t + (0.45 \cdot 1.80)$ $V_A(t) = 0.45t^2 - 1.08t + 0.81$

For Pump B: $V_B(t) = 0.30(2.5t^2 + 1.6t - 0.90)$ $V_B(t) = (0.30 \cdot 2.5)t^2 + (0.30 \cdot 1.6)t - (0.30 \cdot 0.90)$ $V_B(t) = 0.75t^2 + 0.48t - 0.27$

Combine like terms

The total volume is the sum of both functions: $V_{\text{total}}(t) = V_A(t) + V_B(t)$ $V_{\text{total}}(t) = (0.45t^2 - 1.08t + 0.81) + (0.75t^2 + 0.48t - 0.27)$

Group the terms by powers of $t$ :

Quadratic term ( $t^2$ ): $(0.45 + 0.75)t^2 = 1.20t^2$
Linear term ( $t$ ): $(-1.08 + 0.48)t = -0.60t$
Constant term: $0.81 - 0.27 = 0.54$

Putting them together gives the simplified quadratic equation: $V_{\text{total}}(t) = 1.20t^2 - 0.60t + 0.54\text{ mL}$

Common Pitfall: When expanding negative signs inside parentheses, ensure the sign is distributed correctly. For instance, $0.30 \cdot (-0.90) = -0.27$ , not $+0.27$ .

Evaluate $V_{\text{total}}(4.00)$

Substitute $t = 4.00$ into the combined equation: $V_{\text{total}}(4.00) = 1.20(4.00)^2 - 0.60(4.00) + 0.54$ $V_{\text{total}}(4.00) = 1.20(16.00) - 2.40 + 0.54$ $V_{\text{total}}(4.00) = 19.20 - 2.40 + 0.54$ $V_{\text{total}}(4.00) = 16.80 + 0.54 = 17.34\text{ mL}$

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets to distribute the leading scalar 0.45 to the remaining terms inside the first parenthesized expression, resulting in an incorrect function and evaluated volume.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$V_A(t) = 0.45t^2 - 2.4t + 1.80$
Correct
$V_A(t) = 0.45t^2 - 1.08t + 0.81$
You multiplied $0.45$ by the first term $t^2$ , but did not distribute it to $-2.4t$ and $1.80$ . Expanding fully gives $0.45(t^2) - 0.45(2.4t) + 0.45(1.80) = 0.45t^2 - 1.08t + 0.81$ .
Step 2
$V_B(t) = 0.75t^2 + 0.48t - 0.27$
Step 3
$V_{\text{total}}(t) = 1.20t^2 - 1.92t + 1.53$
Correct
$V_{\text{total}}(t) = 1.20t^2 - 0.60t + 0.54$
Your combining of like terms was executed correctly based on your previous lines, but the linear and constant terms inherited the expansion error from Step 1.
Step 4
$V_{\text{total}}(4.00) = 1.20(4.00)^2 - 1.92(4.00) + 1.53 \implies V_{\text{total}}(4.00) = 19.20 - 7.68 + 1.53 = 13.05\text{ mL}$
Correct
$V_{\text{total}}(4.00) = 1.20(4.00)^2 - 0.60(4.00) + 0.54 = 17.34\text{ mL}$
Your substitution and arithmetic were carried out accurately, but the final numerical value is affected by the earlier expansion mistake.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly distributed the scalar across all three terms of $V_B(t)$ .

You grouped and combined matching powers of $t$ systematically.

You substituted $t = 4.00$ and evaluated the resulting arithmetic accurately.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Partial Distribution

You distributed $0.45$ only to the leading term $t^2$ , leaving $-2.4t$ and $+1.80$ unmultiplied.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You treated the leading coefficient as attached only to the first term rather than applying the distributive property $a(b + c + d) = ab + ac + ad$ across every term inside the parentheses.

REASONING PATH

Your ACEpath teacher’s suggested approach

Count the Terms Inside Parentheses: An expression $k(p + q + r)$ contains three distinct terms inside the grouping symbols.
Draw distribution arcs from the outside factor to each term before multiplying to ensure no term is skipped.
Verify Coefficient Changes: Every non-zero coefficient inside the parentheses must be scaled by the outer constant.
Check that the coefficients in the expanded polynomial differ from the original parenthesized coefficients whenever the multiplier is not $1$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Expand $V_A(t)$ completely: $V_A(t) = 0.45(t^2) - 0.45(2.4t) + 0.45(1.80) = 0.45t^2 - 1.08t + 0.81$ .
Add the correctly expanded $V_B(t) = 0.75t^2 + 0.48t - 0.27$ to $V_A(t)$ .
Combine like terms: $(0.45 + 0.75)t^2 + (-1.08 + 0.48)t + (0.81 - 0.27) = 1.20t^2 - 0.60t + 0.54$ .
Substitute $t = 4.00$ : $V_{\text{total}}(4.00) = 1.20(16.00) - 0.60(4.00) + 0.54 = 19.20 - 2.40 + 0.54 = 17.34\text{ mL}$ .

Final answer

$V_{\text{total}}(t) = 1.20t^2 - 0.60t + 0.54\text{ mL}$ ; $V_{\text{total}}(4.00) = 17.34\text{ mL}$

NEXT STEPS

Final Takeaways from Your Teacher

Practice drawing multiplication arrows to every term inside a polynomial before writing down the expanded result.
Perform a sanity check on every expanded term to ensure the multiplier has scaled all coefficients, not just the first one.
Work through targeted drills that involve distributing decimal coefficients across trinomials.

OverviewACEpath Teacher's Diagnosis

02 The student mismanages the negative sign of the linear term when adding the linear coefficients -1.08 and 0.48, adding their absolute values instead of combining them with correct signs.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$V_A(t) = 0.45t^2 - 1.08t + 0.81 \implies V_B(t) = 0.75t^2 + 0.48t - 0.27$
Step 2
$V_{\text{total}}(t) = 1.20t^2 + 1.56t + 0.54$
Correct
$V_{\text{total}}(t) = 1.20t^2 - 0.60t + 0.54$
You combined the $t^2$ and constant terms correctly, but for the linear term you added $|-1.08| + 0.48 = 1.56$ instead of combining signed coefficients: $-1.08 + 0.48 = -0.60$ .
Step 3
$V_{\text{total}}(4.00) = 1.20(4.00)^2 + 1.56(4.00) + 0.54 \implies V_{\text{total}}(4.00) = 19.20 + 6.24 + 0.54 = 25.98\text{ mL}$
Correct
$V_{\text{total}}(4.00) = 1.20(4.00)^2 - 0.60(4.00) + 0.54 = 17.34\text{ mL}$
Your substitution and arithmetic were carried out correctly according to your polynomial, but the result is affected by the incorrect linear coefficient from the previous step.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly multiplied each term inside the parentheses by its outer constant factor.

You combined the quadratic coefficients and constant terms accurately.

You substituted $t = 4.00$ and evaluated the resulting arithmetic with proper order of operations.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Signed coefficient addition error

You dropped the negative sign on $-1.08$ and added $1.08 + 0.48$ instead of evaluating $-1.08 + 0.48$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Treating subtraction as a detached separator rather than attaching the negative sign directly to the coefficient when collecting like terms.

REASONING PATH

Your ACEpath teacher’s suggested approach

Attach Signs to Coefficients: Every term takes the sign directly preceding it, so $a - bt$ means the coefficient of $t$ is $-b$ .
Always write out $(-1.08 + 0.48)t$ before combining so that opposite signs are subtracted rather than added.

CORRECTED PATH

ACEpath Teacher's Demonstration

Combine the linear terms: $(-1.08 + 0.48)t = -0.60t$ .
Write the corrected function: $V_{\text{total}}(t) = 1.20t^2 - 0.60t + 0.54$ .
Substitute $t = 4.00$ : $V_{\text{total}}(4.00) = 1.20(16.00) - 0.60(4.00) + 0.54 = 19.20 - 2.40 + 0.54 = 17.34\text{ mL}$ .

Final answer

$V_{\text{total}}(t) = 1.20t^2 - 0.60t + 0.54\text{ mL}; V_{\text{total}}(4.00) = 17.34\text{ mL}$

NEXT STEPS

Final Takeaways from Your Teacher

Circle each term along with its preceding sign before combining.
Practice combining pairs of positive and negative decimal coefficients on a number line.
Re-evaluate the polynomial after correcting the sign of the linear coefficient.

OverviewACEpath Teacher's Diagnosis

03 The student mistakenly squares the product 1.20(4) instead of squaring only the time t = 4 in the quadratic term during evaluation.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$V_A(t) = 0.45t^2 - 1.08t + 0.81 \implies V_B(t) = 0.75t^2 + 0.48t - 0.27$
Step 2
$V_{\text{total}}(t) = 1.20t^2 - 0.60t + 0.54$
Step 3
$V_{\text{total}}(4.00) = (1.20 \times 4.00)^2 - 0.60(4.00) + 0.54$
Correct
$V_{\text{total}}(4.00) = 1.20(4.00)^2 - 0.60(4.00) + 0.54$
In the term $a t^2$ , the exponent applies only to the base $t$ , not to the coefficient $a$ . The term should be evaluated as $1.20 \times (4.00)^2$ , not $(1.20 \times 4.00)^2$ .
Step 4
$V_{\text{total}}(4.00) = (4.80)^2 - 2.40 + 0.54 \implies V_{\text{total}}(4.00) = 23.04 - 2.40 + 0.54 = 21.18\text{ mL}$
Correct
$V_{\text{total}}(4.00) = 1.20(16.00) - 2.40 + 0.54 = 17.34\text{ mL}$
Your arithmetic following the substitution was carried out consistently, but the final numerical value was thrown off by the earlier order-of-operations error.

WHAT YOU DID WELL

Compliments from Your Teacher

You expanded both polynomial expressions using the distributive property accurately.

You correctly grouped and combined like terms to construct $V_{\text{total}}(t)$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Order of Operations: Exponent Scope

You grouped the leading coefficient $1.20$ inside parentheses with $4.00$ before squaring.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing $a t^2$ with $(at)^2$ when substituting a numerical value into a monomial term, which violates the standard order of operations where exponents take precedence over multiplication.

REASONING PATH

Your ACEpath teacher’s suggested approach

Exponent Precedence: Unless grouping parentheses explicitly dictate otherwise, $a x^n = a \cdot (x^n)$ , never $(a \cdot x)^n$ .
Always evaluate $t^2$ first before multiplying the result by the coefficient.

CORRECTED PATH

ACEpath Teacher's Demonstration

Substitute $t = 4.00$ preserving the coefficient outside the exponent: $V_{\text{total}}(4.00) = 1.20(4.00)^2 - 0.60(4.00) + 0.54$ .
Compute the exponent first: $4.00^2 = 16.00$ .
Multiply the coefficient by the squared value: $1.20 \times 16.00 = 19.20$ .
Combine the remaining terms: $19.20 - 2.40 + 0.54 = 17.34\text{ mL}$ .

Final answer

$V_{\text{total}}(4.00) = 17.34\text{ mL}$

NEXT STEPS

Final Takeaways from Your Teacher

Review the distinction between $c \cdot x^2$ and $(c \cdot x)^2$ during function evaluation.
Practice inserting parentheses strictly around the substituted variable, such as $1.20(4.00)^2$ , before performing any arithmetic.

Lock it in

Question

What is the general distributive law for a scalar multiplied by a trinomial $a(x + y + z)$ ?

Answer

$ax + ay + az$ . Every single term inside the parentheses must be multiplied by the outside factor $a$ .

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student forgets to distribute the leading scalar 0.45 to the remaining terms inside the first parenthesized expression, resulting in an incorrect function and evaluated volume.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

02 The student mismanages the negative sign of the linear term when adding the linear coefficients -1.08 and 0.48, adding their absolute values instead of combining them with correct signs.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

03 The student mistakenly squares the product 1.20(4) instead of squaring only the time t = 4 in the quadratic term during evaluation.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

Lock it in

Similar question