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3 years ago

ANSWER THIS ALEGBRA QUESTION IF YOU ARE GREAT AT MATH ‼

Mathematics

1 answer:

Eduardwww [97]3 years ago

3 0

Answer:

C) (5, ∞)

General Formulas and Concepts:

Alg I

Range is the set of y-values that are outputted by function f(x).

Step-by-step explanation:

According to the graph, we have a horizontal asymptote at y = 5. Therefore, the graph never actually touches/reaches the y-value 5. Therefore, 5 is not included in our range:

(5, ∞) or y ≥ 5

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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:

The picture of the question in the attached figure

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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