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3 years ago

There below! Can somebody help me please?

Mathematics

2 answers:

Montano1993 [528]3 years ago

4 0

Answer: C. 24 units

Step-by-step explanation:

The circumference of a circle is 2πr. Let's say x = 1/3r. Thus, the area of circle A relative to circle B is 1:3. Thus, circle B is 3 times larger than circle A. So, simply divide 64π by 4 to get 16π. Then multiply 16π*3=48π. For the circumference of a circle to be 48, the radius would have to be 24.

Hope it helps <3

babymother [125]3 years ago

3 0

Answer:

The answer is C, 24 units.

Step-by-step explanation:

Let $r_A$ be the radius is circle A and $r_B$ be the radius of circle B.

Radius of circle A is 1/3 of the radius of circle B, or:

$r_A=\frac{1}{3}r_B$

$3r_A=r_B$

The sum of their circumferences is $64\pi$ . In other words:

$2\pi r_A +2\pi r_B = 64\pi$

Substitute $r_B$ :

$2\pi(r_A +r_B)=64\pi$

$2\pi (r_A +3r_A)=64\pi$

$4r_A=32, r_A=8$

$r_B=3r_A, r_B=3(8)=24$

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Virty [35]

Answer:

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Step-by-step explanation:

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3 years ago

Roberto and Maria Santanos spend approximately $11.50 per week to wash and dry their family's clothes at a local coin laundry. A

Elena L [17]

Answer:

68 weeks

Step-by-step explanation:

1 week costs $11.50.

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3 years ago

Bradley and Kelly are out flying kites at a park one afternoon. A model of Bradley and Kelly's kites are shown below on the coor

Alex787 [66]

The correct answer is:

C. They are similar because the corresponding sides of kites KELY and BRAD all have the relationship 2:1.

Using the distance formula,

$d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

the lengths of the sides of BRAD are:

$\text{BR}=\sqrt{(7-6)^2+(3-4)^2}=\sqrt{1^2+(-1)^2}=\sqrt{1+1}=\sqrt{2}\\\\\text{RA}=\sqrt{(6-3)^2+(4-3)^2}=\sqrt{3^2+1^2}=\sqrt{9+1}=\sqrt{10}\\\\\text{AD}=\sqrt{(3-6)^2+(3-2)^2}=\sqrt{(-3)^2+1^2}=\sqrt{9+1}=\sqrt{10}\\\\\text{DB}=\sqrt{(6-7)^2+(2-3)^2}=\sqrt{(-1)^2+(-1)^2}=\sqrt{1+1}=\sqrt{2}$

The lengths of the sides of KELY are:

$\text{KE}=\sqrt{(2-0)^2+(11-9)^2}=\sqrt{2^2+2^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}\\\\\text{EL}=\sqrt{(0-2)^2+(9-3)^2}=\sqrt{(-2)^2+6^2}=\sqrt{40}=2\sqrt{10}\\\\\text{LY}=\sqrt{(2-4)^2+(3-9)^2}=\sqrt{(-2)^2+(-6)^2}=\sqrt{40}=2\sqrt{10}\\\\\text{YK}=\sqrt{(4-2)^2+(9-11)^2}=\sqrt{2^2+(-2)^2}=\sqrt{8}=2\sqrt{2}$