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2 years ago

Which of the following has no solution?

1 answer:

8 0

The first one.
This statement is saying that x is less than zero and greater than zero at the same time. This is not possible. A number cannot be both negative and positive.
It's not the second one because this one includes equal to zero. It can be lass than or equal to zero and greater that or equal to zero because it can be ZERO.
It's not the third because this one is an OR statement. It can be less than or equal to zero OR greater than or equal to zero.
I hope you understand

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Answer:

Step-by-step explanation:

47+34=9k

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Solve for

faust18 [17]

Given equation : n(17+x)=34x−r.

We need to solve it for x.

Distributing n over (17+x) on left side, we get

17n + nx = 34x - r.

Adding r on both sides, we get

17n+r + nx = 34x - r+r.

17n + r + nx = 34x.

Subtracting nx from both sides, we get

17n + r + nx-nx = 34x-nx

17n + r = 34x -nx.

Factoring out gcf x on right side, we get

17x + r = x(34-n).

Dividing both sides by (34-n), we get

$\frac{(17x + r)}{(34-n)} = \frac{x(34-n)}{(34-n)}$

$\frac{(17x + r)}{(34-n)} = x$

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Implicit differentiation Please help

Anvisha [2.4K]

Answer:

$y''(-1) =8$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Algebra I

Factoring

Calculus

Implicit Differentiation

The derivative of a constant is equal to 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Product Rule: $\frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

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

Quotient Rule: $\frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Step-by-step explanation:

Step 1: Define

-xy - 2y = -4

Rate of change of the tangent line at point (-1, 4)

Step 2: Differentiate Pt. 1

Find 1st Derivative

Implicit Differentiation [Product Rule/Basic Power Rule]: $-y - xy' - 2y' = 0$
[Algebra] Isolate y' terms: $-xy' - 2y' = y$
[Algebra] Factor y': $y'(-x - 2) = y$
[Algebra] Isolate y': $y' = \frac{y}{-x-2}$
[Algebra] Rewrite: $y' = \frac{-y}{x+2}$

Step 3: Find y

Define equation: $-xy - 2y = -4$
Factor y: $y(-x - 2) = -4$
Isolate y: $y = \frac{-4}{-x-2}$
Simplify: $y = \frac{4}{x+2}$

Step 4: Rewrite 1st Derivative

[Algebra] Substitute in y: $y' = \frac{-\frac{4}{x+2} }{x+2}$
[Algebra] Simplify: $y' = \frac{-4}{(x+2)^2}$

Step 5: Differentiate Pt. 2

Find 2nd Derivative

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[Derivative] Simplify: $y'' = \frac{8}{(x+2)^3}$

Step 6: Find Slope at Given Point

[Algebra] Substitute in x: $y''(-1) = \frac{8}{(-1+2)^3}$
[Algebra] Evaluate: $y''(-1) =8$

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