3 years ago

A field is shaped like the figure shown. What is the area of the field? Use 3.14

1 answer:

uysha [10]3 years ago

6 0

There is no figure, i cant help you sorry

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Harman [31]

Answer:

h=-6 k=-30

Step-by-step explanation:

$y=x^2+12x+6$

$\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=ax^2+bx+c\:\mathrm{is}\:x_v=-\frac{b}{2a}$

$a=1,\:b=12,\:c=6$

$x_v=-\frac{b}{2a}$

$x_v=-\frac{12}{2\cdot \:1}\\$

$x_v=-6\\$

$\mathrm{Plug\:in}\:\:x_v=-6\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}$

$y_v=\left(-6\right)^2+12\left(-6\right)+6\\y_v=-30\\\left(-6,\:-30\right)$

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2 years ago

olasank [31]

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3 years ago

Neporo4naja [7]

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