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3 years ago

P-4=-9+p i don't know how to determine whether its an infinite solution, or n solution.

2 answers:

lara [203]3 years ago

6 0

Subtract P from both sides and be left with -4 = -9, which means no solutions.

adoni [48]3 years ago

4 0

♥ Solve:
Subtract "P" from both sides.
Answer:
-4 = -9
Which means
No solutions.
Final answer: No solutions.

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gayaneshka [121]

Your answer would be $3,395.00. Hope this helps! :)

5 0

3 years ago

NeX [460]

Answer:

$\displaystyle \frac{dy}{dx} \bigg| \limit_{(1, 4)} = 2$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Equality Properties

Algebra I

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Implicit Differentiation

Basic Power Rule:

Step-by-step explanation:

Step 1: Define

$\displaystyle \sqrt{x} - \sqrt{y} = -1$

Point (1, 4)

Step 2: Differentiate

[Function] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle x^{\frac{1}{2}} - y^{\frac{1}{2}} = -1$
[Implicit Differentiation] Basic Power Rule: $\displaystyle \frac{1}{2}x^{\frac{1}{2} - 1} - \frac{1}{2}y^{\frac{1}{2} - 1}\frac{dy}{dx} = 0$
[Implicit Differentiation] Simplify Exponents: $\displaystyle \frac{1}{2}x^{\frac{-1}{2}} - \frac{1}{2}y^{\frac{-1}{2}}\frac{dy}{dx} = 0$
[Implicit Differentiation] Rewrite [Exponential Rule - Rewrite]: $\displaystyle \frac{1}{2x^{\frac{1}{2}}} - \frac{1}{2y^{\frac{1}{2}}}\frac{dy}{dx} = 0$
[Implicit Differentiation] Isolate y terms: $\displaystyle -\frac{1}{2y^{\frac{1}{2}}}\frac{dy}{dx} = -\frac{1}{2x^{\frac{1}{2}}}$
[Implicit Differentiation] Isolate $\displaystyle \frac{dy}{dx}$ : $\displaystyle \frac{dy}{dx} = \frac{2y^{\frac{1}{2}}}{2x^{\frac{1}{2}}}$
[Implicit Differentiation] Simplify: $\displaystyle \frac{dy}{dx} = \frac{y^{\frac{1}{2}}}{x^{\frac{1}{2}}}$

Step 3: Evaluate

Substitute in point [Derivative]: $\displaystyle \frac{dy}{dx} = \frac{(4)^{\frac{1}{2}}}{(1)^{\frac{1}{2}}}$
Exponents: $\displaystyle \frac{dy}{dx} = \frac{2}{1}$
Division: $\displaystyle \frac{dy}{dx} = 2$