Answer this Quickly Please 50 points !!!!!!!!!!!!!!!

The radius of Circle A is 6 mm. The radius of Circle B is 4 mm greater than the radius of Circle A. The radius of Circle C is 2

labwork [276]

Answer:

Area of circle A =113.14 mm²

Area of circle b = 314.29 mm²

Area of circle C = 452.57 mm²

Area of circle A = 254.57 mm²

2.25 times

Step-by-step explanation:

Area of a circle = nr²

where n = 22/7

r = radius

Circle A's radius = 6mm

Circle B's radius = 6mm + 4mm = 10mm

Circle C's radius = 10mm + 2mm = 12mm

Circle D's radius = 12mm - 3mm = 9mm

Area of circle A = (22/7) x 6² = 113.14 mm²

Area of circle b = (22/7) x 10² = 314.29 mm²

Area of circle C = (22/7) x 12² = 452.57 mm²

Area of circle A = (22/7) x 9² = 254.57 mm²

Number of times the area of circle D is greater than that circle A = Area of circle D / Area of circle A

254.57 mm² / 113.14 mm² = 2.25 times

4 0

3 years ago

radium-226, which has a half-life of 1620 years, is used in medicine for treatment of disease. find the value of k for radium-22

sammy [17]

There will be a 25 grams left after 3240 years.

The half-life of Radium-226 is 1620 years. If we divide 3240 by 1620 the answer is 2. So that means the amount of Radium-226 will divide in half twice.

100 / 2 = 50 / 2 = 25

6 0

3 years ago

I need help with number 17

zimovet [89]

Answer:

3 planes parallel to FL is EK, CI , BH, AJ

Because, this is prism in which all length segments should be equal .

7 0

2 years ago

Suppose an=3an-1 -5an-2 -4an-3. and a4= -7, a5=-36, a6=-85 , find a1 , a2, a3

maria [59]

Answer:

$a_1=4\\\\a_2=0\\\\a_3=3$

Explanation:

The equation is:

$a_n=3a_{n-1}-5a_{n-2}-4a_{n-3}$

and

$a_4=-7, a_5=-36, a_6=-85$

1. Subsittute n = 6 into the equation:

$a_6=3a_{5}-5a_{4}-4a_{3}$

Now subsititute the known values:

$-85=3(-36)-5(-7)-4a_{3}$

You can solve for $a_3$

$-85=-108+35-4a_{3}\\\\4a_3=12\\\\a_3=12/4\\\\a_3=3$

2. Substitute n = 5 into the equation:

$a_5=3a_{4}-5a_{3}-4a_{2}$

Substitute the known values and solve for $a_2$

$-36=3(-7)-5(3)-4a_{2}\\\\4a_2=-21-15+36=0\\\\a_2=0$

3. Substitute n = 4 into the equation, subsitute the known values and solve for $a_1$

$a_4=3a_{3}-5a_{2}-4a_{1}$

$-7=3(3)-5(0)-4a_{1}\\\\4a_1=9+7=16\\\\a_1=4$

7 0

3 years ago

If you answer God will bless you:

VashaNatasha [74]

Answer:

600 m³

Step-by-step explanation:

Given:

length of swimming pool = 30 m
width of swimming pool = 10 m
depth at deep end = 3 m
depth at shallow end = 1 m

To find the volume of the water in the pool, find the area of a cross section (see attached image) and multiply it by the width of the pool.

Area of cross section

= area of rectangle - area of triangle

= (3 × 30) - [1/2 × 30 × (3 - 1)]

= 90 - 30

= 60 m²

Volume of pool

= area of cross section × width

= 60 × 10

= 600 m³

6 0

2 years ago