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3 years ago

14

After 6 months the simple interest earned annually on an investment of $3000 was $811. Find the interest rate to the nearest ten

th of a percent.
52.1%

54.1%

0.541%

0.5%

1 answer:

Talja [164]3 years ago

7 0

Answer:

52.1% percent using principal multiply by rate divided by 100

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Point A is located at (4, 8) and point B is located at (14, 10) . What point partitions the directed line segment AB⎯⎯⎯⎯⎯

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The 1:3 ratio means that the distance from A to the point is 1/4 of the distance from A to B.

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3 years ago

Find the function y = f(t) passing through the point (0, 18) with the given first derivative.

monitta

Answer:

$\displaystyle y = \frac{t^2}{16} + 18$

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>

Functions
Function Notation
Coordinates (x, y)

<u>Calculus</u>

Derivatives

Derivative Notation

Antiderivatives - 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]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

Point (0, 18)

$\displaystyle \frac{dy}{dt} = \frac{1}{8} t$

<u>Step 2: Find General Solution</u>

<em>Use integration</em>

[Derivative] Rewrite: $\displaystyle dy = \frac{1}{8} t\ dt$
[Equality Property] Integrate both sides: $\displaystyle \int dy = \int {\frac{1}{8} t} \, dt$
[Left Integral] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle y = \int {\frac{1}{8} t} \, dt$
[Right Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle y = \frac{1}{8}\int {t} \, dt$
[Right Integral] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle y = \frac{1}{8}(\frac{t^2}{2}) + C$
Multiply: $\displaystyle y = \frac{t^2}{16} + C$

<u>Step 3: Find Particular Solution</u>

Substitute in point [Function]: $\displaystyle 18 = \frac{0^2}{16} + C$
Simplify: $\displaystyle 18 = 0 + C$
Add: $\displaystyle 18 = C$
Rewrite: $\displaystyle C = 18$
Substitute in <em>C</em> [Function]: $\displaystyle y = \frac{t^2}{16} + 18$

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

Unit: Integration

Book: College Calculus 10e

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natta225 [31]

Answer:

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Step-by-step explanation:

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2) 16/4=4

for this sequence the common ratio

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