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3 years ago

What is (4ab^0c^9)^3 ?

Mathematics

1 answer:

fomenos3 years ago

3 0

Answer:

64a³c²⁷

Step-by-step explanation:

The expression inside parentheses can be simplified by noting that b^0 = 1. Then you have ...

(4ac^9)^3

The rules of exponents say the exponent outside applies to each of the inside factors. For c^9, the exponents multiply.

Here are the applicable rules:

(ab)^c = (a^c)(b^c)

(a^b)^c = a^(b·c)

(4ac^9)^3 = 4^3·a^3·c^(9·3) = 64a^3c^27

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Read 2 more answers

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:

Pre-Algebra

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

Algebra I

Functions
Function Notation
Coordinates (x, y)

Calculus

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:

Step 1: Define

Identify

Point (0, 18)

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

Step 2: Find General Solution

Use integration

[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: $\displaystyle y = \frac{t^2}{16} + C$