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DedPeter [7]
3 years ago
8

Point A(-3,-2) is reflected in the line y = x. What are the coordinates of A?

Mathematics
1 answer:
Gelneren [198K]3 years ago
8 0

Answer:

A' = (2,-3)

Step-by-step explanation:

Given

A = (-3,2)

Reflection: y = x

Required

Determine the new coordinates of A

For y = x reflection, the new coordinate will be:

A' = (y,x)

So, for

A = (-3,2)

A' will be:

A' = (2,-3)

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-5x-3y=22 help plz ​
Lerok [7]

Answer:

<h2>(-2, -4)</h2>

Step-by-step explanation:

Put the coordinates of the points to the equation, and check it:

(-4, -1) → x= -4, y = -1

L = -5(-4) - 3(-1) =20 + 3 = 23

R = 22

L ≠ R

(-1, -4) → x= -1, y = -4

L = -5(-1) - 3(-4) = 5 + 12 = 17

R = 22

L ≠ R

(-2, -4) → x= -2, y = -4

L = -5(-2) - 3(-4) = 10 + 12 = 22

R = 22

L = R    CORRECT :)

(-4, -2) → x = -4, y = -2

L = -5(-4) - 3(-2) = 20 + 6 = 26

R = 22

L ≠ R

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3 years ago
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Step-by-step explanation:

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3 years ago
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Find the area of a circle with a diameter of 8.
Kaylis [27]

Answer:

50.27

Step-by-step explanation:

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What is the derivative of this function: F(x)=(5e^4x)+(e^-x^6)
Fed [463]

Answer:

\dfrac{dF(x)}{dx} =20e^{4x}-6x^5e^{x^{-6}}

Step-by-step explanation:

The derivative of F(x) is calculated as follows:

\dfrac{dF(x)}{dx}=\dfrac{d}{dx} [(5e^{4x})+(e^{-x^6})]

\dfrac{dF(x)}{dx}=\dfrac{d}{dx} [(5e^{4x})]+\dfrac{d}{dx} [(e^{-x^6})]

\dfrac{dF(x)}{dx}=5\dfrac{d}{dx} [(e^{4x})]+\dfrac{d}{dx} [(e^{-x^6})]

using the chain rule we find that

\dfrac{d}{dx} [(e^{4x})]= \dfrac{d}{d(4x)} [(e^{4x})]+ \dfrac{d}{dx} [4x] = 4e^{4x},

\dfrac{d}{dx} [(e^{-x^6})] = \dfrac{d}{d(-x^6)} [(e^{-x^6})]+\dfrac{d}{dx} [(-x^6})]= -6x^5e^{-x^6};

therefore,

\dfrac{dF(x)}{dx}=5\dfrac{d}{dx} [(e^{4x})]+\dfrac{d}{dx} [(e^{-x^6})] =5(4e^{4x})-6x^5e^{x^{-6}}

\boxed{\dfrac{dF(x)}{dx} =20e^{4x}-6x^5e^{x^{-6}}}

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