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OverLord2011 [107]
2 years ago
11

Which graph represents the inverse sine function?

Mathematics
1 answer:
hichkok12 [17]2 years ago
6 0
Answer: A.
The inverse sine function is written as sin−1(x) or arcsin(x). Inverse functions swap x- and y-values, so the range of inverse sine is − π/2 to π/2 and the domain is −1 to 1. When evaluating problems, use identities or start from the inside function.
(REFER TO CHART BELOW)

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Help anyone can help me do the question 4,I will mark brainlest.​
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Hello,

Step-by-step explanation:

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Need to solve for Y by filling in the blanks<br>(y +44)°, 86, (3y)°​
lapo4ka [179]
Answer:

1.89

Solution:

We have -44x - 86 = -3

First, we move the non x terms to the RHS of the equation. To do this, we add 86 to both sides.

Note: if a = b, then a + c = b + c
Therefore, -44x - 86 + 86 = -3 + 86

or, -44x = 83

Next, divide both sides by -44, which is the coefficent of x

Note: if a = b, then a ÷ c = b ÷ c
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Explaination:

This is right I guess
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Show that d^2y/dx^2=-2x/y^5, if x^3 + y^3=1
aalyn [17]

Answer:

y³ + x³ = 1

First, differentiate the first time, term by term:

{3y^{2}.\frac{dy}{dx} + 3x^{2}} = 0 \\\\{3y^{2}.\frac{dy}{dx} = -3x^{2}} \\\\\frac{dy}{dx} = \frac{-3x^{2}}{3y^{2}} \\\\\frac{dy}{dx} = \frac{-x^{2}}{y^{2}}

↑ we'll substitute this later (4th step onwards)

Differentiate the second time:

3y^{2}.\frac{dy}{dx} + 3x^{2} = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} + 6x = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{-x^{2} }{y^{2} })^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{x^{4} }{y^{4} }) = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + \frac{6x^{4} }{y^{3} } = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} = - 6x - \frac{6x^{4} }{y^{3} } \\\\

3y^{2}.\frac{d^{2} y}{dx^{2}} =  - \frac{- 6xy^{3} - 6x^{4} }{y^{3}} \\\\\frac{d^{2} y}{dx^{2}} =  - \frac{- 6xy^{3} - 6x^{4} }{3y^{2}. y^{3}} \\\\\frac{d^{2} y}{dx^{2}} =  - \frac{- 2xy^{3} - 2x^{4} }{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} =  - \frac{-2x (y^{3} + x^{3})}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} =  - \frac{-2x (1)}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} =  - \frac{-2x}{y^{5}}

3 0
2 years ago
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