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3 years ago

I do not understand theses... Please help!

Mathematics

2 answers:

oee [108]3 years ago

5 0

Y = x + 1
y = 2x - 1

basically we r going to sub in one y for another
x + 1 = 2x - 1
1 + 1 = 2x - x
2 = x

now we sub 2 in for x in either of the original equations to find y
y = x + 1
y = 2 + 1
y = 3

so ur solution is (2,3)

Klio2033 [76]3 years ago

5 0

Y =x+1
y=2x-1
-------------subtract
0 = -x +2

-x + 2 = 0
x =2

y =x+1
y =2+1
y= 3

answer
solution (2,3)

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2 years ago

Are the point (-1, 1) and (3, 1/5) solutions to the equation 3x-5y=8

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3 years ago

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

Differentiate [Quotient Rule/Basic Power Rule]: $y'' = \frac{0(x+2)^2 - 8 \cdot 2(x + 2) \cdot 1}{[(x + 2)^2]^2}$
[Derivative] Simplify: $y'' = \frac{8}{(x+2)^3}$