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4 years ago

7

A truck stops at a gas station where diesel fuel costs $3.99 per gallon. The truck needs 8.8 gallons of diesel for the driver to

complete all his deliveries. Estimate whether $36.00 will be enough to pay for the fuel. Justify your answer.

2 answers:

Rudik [331]4 years ago

7 0

$36 will be enough because 3.99 multiplied by 8.8 is around $35.11 which is less than $36.

Dmitry [639]4 years ago

6 0

36.00 will be enough to pay for fuel because if you multiply the 8.8 gallons of diesel needed by the fuel cost per gallon you get 35.112

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General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

<u>Algebra I</u>

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<u>Calculus</u>

Derivatives

Derivative Notation

Solving Differentials - Integrals

Integration Constant C

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Step-by-step explanation:

*Note:

Ignore the Integration Constant C on the left hand side of the differential equation when integrating.

<u>Step 1: Define</u>

$\displaystyle \frac{ds}{dt} = 5t^3 + \frac{9}{t^2}$

t = 1

s = 8

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[Right Integral] Rewrite [Integration Property - Addition]: $\displaystyle s = \int {5t^3} \, dt + \int {\frac{9}{t^2}} \, dt$
[Right Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle s = 5\int {t^3} \, dt + 9\int {\frac{1}{t^2}} \, dt$
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Rewrite: $\displaystyle C = \frac{-9}{4}$

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Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Differentials Equations and Slope Fields

Book: College Calculus 10e

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