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3 years ago

Help please as fast as you can

Mathematics

2 answers:

Tatiana [17]3 years ago

4 0

The Answer is D no problem

Ksju [112]3 years ago

3 0

Answer: D, the fourth option.

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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3 years ago

Find the number of gallons in 6 liters of solution(1gal=3.79l)

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3 years ago

Billy and Sam started with the same number of baseball cards in their collections. Billy collected 3 cards per week and now has

Elina [12.6K]

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Step-by-step explanation:

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2 years ago

Please help ASAP! Thanks! For the figures below, assume they are made of semicircles, quarter circles and squares. For each shap

bazaltina [42]

Answer:

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

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

Step-by-step explanation:

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

<em>The circumference of a quarter of circle is equal to</em>

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

substitute the given values

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

$C=6\pi\ cm$

The hypotenuse of right triangle is equal to (applying the Pythagorean Theorem)

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

$AC=\sqrt{288}\ cm$

simplify

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

Find the perimeter

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

simplify

Factor 6

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

6 0

3 years ago