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bogdanovich [222]
3 years ago
6

A toxic radioactive substance with a density of 2 milligrams per square centimeter is detected in the ventilating ducts of a nuc

lear processing building that was used 55 years ago. If the​ half- life of the substance is 20 ​years, what was the density of the substance when it was deposited 55 years​ ago?
Mathematics
1 answer:
Zolol [24]3 years ago
3 0

Answer:

60.5 milligrams per square centimeter First, determine how many half lives have expired by dividing the time by the half-life. So: 55/20 = 2.75 That means that only 2^(-2.75) = 0.148650889 = 14.8650889% of the original substance remains. So just divide the amount remaining by 0.148650889 to get the original amount. 9 / 0.148650889 = 60.5445419 So originally, there was 60.5 milligrams per square centimeter 55 years ago.

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find the perimeter (p) of a rectangle with a length (1) of 12m and a width (w) of 8.4m using the formula p=21+2w
NikAS [45]

Answer:

The answer is 40.8 m.

Step-by-step explanation:

To find the perimeter of the rectangle, use the formula P= 2(l)+ 2(w). Plug in the information given in the question, and the equation will look like

P= 2(12m) + 2(8.4m)

Solve the equation, and the final answer will be 40.8 m.

4 0
3 years ago
Please Help! Will mark brainiliest!!<br><br> 1. y varies directly as x and inversely as z.
Paladinen [302]

Answer:

x = 5

z =2

y =4.4

Step-by-step explanation:

y varies directly as x

y = kx

and inversely as z

y = kx/z

Putting in the numbers in the first line of the table, we can solve for k

13.75 = k * 25/4

Multiply by 4

13.75*4 = 25k

55 =25k

Divide by 25

55/25 = 25k/25

2.2 = k

Now we can fill in the table

y = 2.2 x/z

Row 2 we need to find x

1 = 2.2 x/11

Multiply by 11

11 = 2.2 x

Divide by 2.2

11/2.2 = 2.2x

5 =x

Row 3 we need to find z

18.7 = 2.2 *17/z

18.7 = 37.4 /z

Multiply by z

18.7z = 37.4

Divide by 18.7

18.7/18.7z = 37.4/18.8

z =2

Row 4 we need to find y

y = 2.2 *10/5

y = 22/5

y =4.4

8 0
3 years ago
If street lights are placed at most 145
Paraphin [41]

Answer:

Step-by-step explanation:

d=4 miles=4×1760 yards=7040 yards=7040×3 ft

number of lights=21120 ft

number of lights=21120/145 +1≈145.7+1=146+1=147 lights

7 0
2 years ago
The volume of the cone shown is 5 cubic inches. What is the height of a cone with the same base diameter but a volume of 10 cubi
Alisiya [41]

Answer:

The height of the big cone is double the one in the small cone

h2 = 2h1

Step-by-step explanation:

Given that:

  • The volume of the small cone: 5 cubic inches
  • The volume of the big cone: 10 cubic inches

As we know, the volume of a cone is as following:

V = (1/3)*area of the base*height

If the base diameter are unchanged => area of the base of the two cones are  unchanged and from the given information, the volume of the big cone is double the volume of the small cone. So the height of the big cone is double the one in the small cone

<=> h2 = 2h1

3 0
3 years ago
Please help....this is difficult for me to figure out..
Nitella [24]

Answer:

z would be 0.4 !

Step-by-step explanation:

Hope this helps :)))

6 0
3 years ago
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