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3 years ago

Solve for X. Please help ASAP. I have no idea how to do this and it’s due really soon!!

Mathematics

1 answer:

Margarita [4]3 years ago

8 0

Answer:

x=4

Step-by-step explanation:

14=3x+2

Move all terms that don't contain x to the right side and solve.

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Hello! I need some help. Have a blessed day!

kondor19780726 [428]

Answer:

false is the correct answer of it if the answer going wrong sorry

7 0

3 years ago

Read 2 more answers

What is the derivative of 1/square root 4x.

Bumek [7]

Answer:

$\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{-1}{4x^\bigg{\frac{3}{2}}}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Exponential Properties

Exponential Property [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Property [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Differentiation

Derivatives
Derivative Notation

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

Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

Step 1: Define

Identify.

$\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg]$

Step 2: Differentiate

Simplify: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \bigg( \frac{1}{2\sqrt{x}} \bigg)'$
Rewrite [Derivative Property - Multiplied Constant]: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( \frac{1}{\sqrt{x}} \bigg)'$
Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( \frac{1}{x^\Big{\frac{1}{2}}} \bigg)'$
Rewrite [Exponential Rule - Rewrite]: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( x^\bigg{\frac{-1}{2}} \bigg)'$
Derivative Rule [Basic Power Rule]: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( \frac{-1}{2} x^\bigg{\frac{-3}{2}} \bigg)$
Simplify: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{-1}{4} x^\bigg{\frac{-3}{2}}$
Rewrite [Exponential Rule - Rewrite]: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{-1}{4x^\bigg{\frac{3}{2}}}$

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

Unit: Differentiation

5 0

3 years ago

At a pressure 41 kPa, the gas in a cylinder has a volume of 16 liters. Assuming temperature remains the same, if the volume of t

DiKsa [7]

If temperature (T) and amount of gas (n) remain constant, but pressure (P) and volume (V) change, then the ideal gas law: PV = nRT becomes
P1V1 = P2V2 --> (41)(16) = P2 (4)
--> P2 (4) = 656
P2 = 656/4 = 164 kPa

6 0

4 years ago

Multiply or divide as indicated. Leave your answer with no factors in the denominator. p⁻⁴q⁵r⁶/p⁻³qr⁻²

KIM [24]

Answer: $\frac{q^4r^8}{p}$