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2 years ago

Plz help out!! And no this is not a test :)

Mathematics

2 answers:

jolli1 [7]2 years ago

5 0

9514 1404 393

Answer:

B = (-4, 5)

Step-by-step explanation:

Reflection over the origin changes the signs of the coordinates.

(x, y) ⇒ (-x, -y)

(4, -5) ⇒ (-4, 5)

Point B is (-4, 5).

vodomira [7]2 years ago

4 0

Answer:

(-4,5)

Step-by-step explanation:

i do not know

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3 years ago

The original price of a new pair of shoes is $40. If there is a 20% discount on all shoes, how much is deducted from the origina

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4 0

3 years ago

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The graph of which function(s) will contain the point (4, 12)? y = x + 8 y = 3x y = 2x y = x + 6

Alex777 [14]

Answer:

y=x+8 and y=3x

Step-by-step explanation:

The Given point is (4,12)

now we can substitute the point in each of the function and check wether it lies on it or not.

A. y=x+8

substitute y=12,x=4

12=4+8=12 which is true

B. y=3x

substitute y=12,x=4

12=3*4=12 which is true

C. y=2x

substitute y=12,x=4

12=2*4=8 which is not true

D. y=x+6

substitute y=12,x=4

12=4+6=10 . which is not true

Therefore option A and B are correct.

4 0

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

Step 6: Find Slope at Given Point

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

6 0

3 years ago

Read 2 more answers