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snow_lady [41]
2 years ago
5

Name the coordinates for the reflection. Reflect point (4, -1) over the x-axis.

Mathematics
1 answer:
inysia [295]2 years ago
6 0

Answer:

i just took the test its (4,1)

Step-by-step explanation:

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3:18-1:33=188 so I think that’s ur answer
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if trapezoid ABCD was reflected over the y-axis, reflected over the x-axis, and rotated 180°, where would point A’lie
maria [59]

Answer:

<em>Answer: (second option) (-4,1)</em>

Step-by-step explanation:

<u>Transformations of Points and Shapes</u>

The image shows a trapezoid ABCD where point A has coordinates (-4,1).

The following transformations are performed:

Reflection over the y-axis: If a point (x,y) is reflected over the y-axis, it becomes (-x,y). Thus point A becomes A'=(4,1)

Reflection over the x-axis: If a point (x,y) is reflected over the x-axis, it becomes (x,-y). Thus point A' becomes A''=(4,-1)

Rotation 180°: If a point (x,y) is rotated 180°, it becomes (-x,-y). Thus point A'' becomes A'''=(-4,1)

Answer: (second option) (-4,1)

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2 years ago
Find the exact surface area of the figure.
puteri [66]
The surface area is A = 297.33
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Find the equation of a line passing through the given point and perpendicular to the given equation (-5,-2) and y=-2x+2
liq [111]

Answer:

y=1/2x+ 1/2


Step-by-step explanation:


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2 years ago
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Find y'' by implicit differentiation. 4x^3 + 5y^3 = 4
iVinArrow [24]
\bf 4x^3+5y^3=4\implies 12x^2+15y^2\cfrac{dy}{dx}=0\implies 4x^2+5y^2\cfrac{dy}{dx}=0&#10;\\\\\\&#10;5y^2\cfrac{dy}{dx}=-4x^2\implies \boxed{\cfrac{dy}{dx}=\cfrac{-4x^2}{5y^2}}\\\\&#10;-------------------------------\\\\&#10;\cfrac{d^2y}{dx^2}=\stackrel{quotient~rule}{\cfrac{-8x(5y^2)~~-~~(-4x^2)(10y)\left( \frac{dy}{dx} \right)}{(5y^2)^2}}&#10;\\\\\\&#10;\cfrac{d^2y}{dx^2}=\cfrac{-40xy^2+(40x^2y)\left( \frac{-4x^2}{5y^2} \right)}{25y^4}

\bf \cfrac{d^2y}{dx^2}=\cfrac{-40xy^2+(40x^2y)\left( \frac{-4x^2}{5y^2} \right)}{25y^4}&#10;\\\\\\&#10;\cfrac{d^2y}{dx^2}=\cfrac{-40xy^2-\frac{160x^4y}{5y^2}}{25y^4}\implies &#10;\cfrac{d^2y}{dx^2}=\cfrac{\frac{-200xy^4-160x^4y}{5y^2}}{25y^4}&#10;\\\\\\&#10;\cfrac{d^2y}{dx^2}=\cfrac{-200xy^4-160x^4y}{(25y^4)(5y^2)}\implies &#10;\cfrac{d^2y}{dx^2}=\cfrac{-200xy^4-160x^4y}{125y^6}&#10;\\\\\\&#10;\cfrac{d^2y}{dx^2}=\cfrac{-40xy^4-32x^4y}{25y^6}
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3 years ago
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