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3 years ago

Y = –3x2 – 12 Axis of symmetry:

Mathematics

1 answer:

kicyunya [14]3 years ago

5 0

<h3>Answer: x = 0</h3>

Work Shown:

y = -3x^2 - 12 is the same as y = -3x^2 + 0x + (-12)

Compare this to y = ax^2 + bx + c, and we'll find that

a = -3

b = 0

c = -12

Plug the 'a' and 'b' values into the formula below and simplify

h = -b/(2a)

h = -0/(2*(-3))

h = 0

The x coordinate of the vertex is x = 0, so the axis of symmetry is also x = 0.

The axis of symmetry lays perfectly on top of the y axis.

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The figures below are made out of circles, semicircles, quarter circles, and a square. Find the area and the perimeter of each f

WITCHER [35]

Answer:

Part 1) $A=36(\pi-2)\ cm^2$

Part 2) $P=6(\pi+2\sqrt{2})\ cm$

Step-by-step explanation:

<u><em>The picture of the question in the attached figure</em></u>

Part 1) Find the area

we know that

The area of the shape is equal to the area of a quarter of circle minus the area of an isosceles right triangle

$A=\frac{1}{4}\pi r^{2}-\frac{1}{2}(b)(h)$

we have that the base and the height of triangle is equal to the radius of the circle

$r=12\ cm\\b=12\ cm\\h=12\ cm$

substitute

$A=\frac{1}{4}\pi (12)^{2}-\frac{1}{2}(12)(12)\\A=(36\pi-72)\ cm^2$

simplify

Factor 36

$A=36(\pi-2)\ cm^2$

Part 2) Find the perimeter

The perimeter of the figure is equal to the circumference of a quarter of circle plus the hypotenuse of the right triangle

The circumference of a quarter of circle is equal to

$C=\frac{1}{4}(2\pi r)=\frac{1}{2}\pi r$

substitute the given values

$C=\frac{1}{2}\pi (12)\\C=6\pi\ cm$

Applying the Pythagorean Theorem

The hypotenuse of right triangle is equal to

$AC=\sqrt{12^2+12^2}\\AC=\sqrt{288}\ cm$

simplify

$AC=12\sqrt{2}\ cm$

Find the perimeter

$P=(6\pi+12\sqrt{2})\ cm$

simplify

Factor 6

$P=6(\pi+2\sqrt{2})\ cm$

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