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Stolb23 [73]
3 years ago
11

Can someone help me with this question, please. The problem is attached.

Mathematics
1 answer:
fomenos3 years ago
3 0

It is a reflection across the y axis, then a translation to the right 2 units.  


I hope this helps!  If you have more you want help with I can help.  

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Find the area of the shape
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27 pi

Step-by-step explanation:

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What is the zero of y=-4x^2+8x-28
tangare [24]

Solution, y-4x^2+8x-28=0:\quad \mathrm{Parabola\:with\:vertex\:at}\:\left(h,\:k\right)=\left(1,\:24\right),\:\mathrm{and\:focal\:length}\:|p|=\frac{1}{16}

Steps:

\mathrm{Rewrite}\:y-4x^2+8x-28=0\:\mathrm{in\:the\:standard\:form}

\mathrm{Rewrite\:as}, y=4x^2-8x+28

\mathrm{Complete\:the\:square}\:4x^2-8x+28:\quad 4\left(x-1\right)^2+24, y=4\left(x-1\right)^2+24

\mathrm{Subtract\:}24\mathrm{\:from\:both\:sides}, y-24=4\left(x-1\right)^2

\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}4, \frac{1}{4}\left(y-24\right)=\left(x-1\right)^2

\mathrm{Rewrite\:in\:standard\:form}, 4\cdot \frac{1}{16}\left(y-24\right)=\left(x-1\right)^2

\mathrm{Therefore\:parabola\:properties\:are:} \left(h,\:k\right)=\left(1,\:24\right),\:p=\frac{1}{16}

\mathrm{Hope\:This\:Helps!!!}

\mathrm{-\:austint1414}

8 0
3 years ago
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