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3 years ago

Which equation accurately represents this statement? Select three options.

Mathematics

2 answers:

N76 [4]3 years ago

5 0

Answer:

X = 2

Step-by-step explanation:

4.9x - (-3) = 12.8

4.9x = 12.8 + (-3)

4.8x - 12.8 = -3

Can be three different forms

To solve:

4.9x = 12.8 - 3

4.9x = 9.8

x = 2

atroni [7]3 years ago

3 0

Complete question:

Which equation accurately represents this statement? Check all that apply.

Negative 3 less than 4.9 times a number, x, is the same as 12.8

A: -3-4.9x=12.8

B: 4.9x-(-3)=12.8

C: 3+4.9x=12.8

D: (4.9-3)x=12.8

E: 12.8=4.9x+3

Answer:

B: 4.9x-(-3)=12.8

C: 3+4.9x=12.8

E: 12.8=4.9x+3

Step-by-step explanation:

The word equation can be expressed mathematically as follows

negative 3 → -3

less than 4.9 times a number , x → 4.9 x - (-3)

is same as 12.8 → 4.9 x - (-3) = 12.8

The equation can be expressed as

4.9 x - (-3) = 12.8

when you open the bracket, you multiply -3 time - 1 . MInus times minus is equals to plus. so you get

4.9 x + 3 = 12.8

12.8 = 4.9x+3

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Which expression is equivalent to −4.6+(−0.4)+(−7.2) ?

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Answer: A. -4.6-(0.4+7.2)

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4 years ago

Find y' if y= In (x2 +6)^3/2 y'=

Schach [20]

Answer:

$\displaystyle y' = \frac{3xln(x^2 + 6)^{\frac{1}{2}}}{x^2 + 6}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Factoring

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

ln Derivative: $\displaystyle \frac{d}{dx} [lnu] = \frac{u'}{u}$

Step-by-step explanation:

Step 1: Define

$\displaystyle y = ln(x^2 + 6)^{\frac{3}{2}}$

Step 2: Differentiate

[Derivative] Chain Rule: $\displaystyle y' = \frac{d}{dx}[ln(x^2 + 6)^{\frac{3}{2}}] \cdot \frac{d}{dx}[ln(x^2 + 6)] \cdot \frac{d}{dx}[x^2 + 6]$
[Derivative] Chain Rule [Basic Power Rule]: $\displaystyle y' = \frac{3}{2}ln(x^2 + 6)^{\frac{3}{2} - 1} \cdot \frac{d}{dx}[ln(x^2 + 6)] \cdot \frac{d}{dx}[x^2 + 6]$
[Derivative] Simplify: $\displaystyle y' = \frac{3}{2}ln(x^2 + 6)^{\frac{1}{2}} \cdot \frac{d}{dx}[ln(x^2 + 6)] \cdot \frac{d}{dx}[x^2 + 6]$
[Derivative] ln Derivative: $\displaystyle y' = \frac{3}{2}ln(x^2 + 6)^{\frac{1}{2}} \cdot \frac{1}{x^2 + 6} \cdot \frac{d}{dx}[x^2 + 6]$
[Derivative] Basic Power Rule: $\displaystyle y' = \frac{3}{2}ln(x^2 + 6)^{\frac{1}{2}} \cdot \frac{1}{x^2 + 6} \cdot (2 \cdot x^{2 - 1} + 0)$
[Derivative] Simplify: $\displaystyle y' = \frac{3}{2}ln(x^2 + 6)^{\frac{1}{2}} \cdot \frac{1}{x^2 + 6} \cdot (2x)$
[Derivative] Multiply: $\displaystyle y' = \frac{3ln(x^2 + 6)^{\frac{1}{2}}}{2} \cdot \frac{1}{x^2 + 6} \cdot (2x)$
[Derivative] Multiply: $\displaystyle y' = \frac{3ln(x^2 + 6)^{\frac{1}{2}}}{2(x^2 + 6)} \cdot (2x)$
[Derivative] Multiply: $\displaystyle y' = \frac{3(2x)ln(x^2 + 6)^{\frac{1}{2}}}{2(x^2 + 6)}$
[Derivative] Multiply: $\displaystyle y' = \frac{6xln(x^2 + 6)^{\frac{1}{2}}}{2(x^2 + 6)}$
[Derivative] Factor: $\displaystyle y' = \frac{2(3x)ln(x^2 + 6)^{\frac{1}{2}}}{2(x^2 + 6)}$
[Derivative] Simplify: $\displaystyle y' = \frac{3xln(x^2 + 6)^{\frac{1}{2}}}{x^2 + 6}$

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

Unit: Derivatives

Book: College Calculus 10e

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4w+8-8=48-8
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l=10+8, l=18
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