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ludmilkaskok [199]
3 years ago
8

Plz answer don't send link ​

Mathematics
1 answer:
avanturin [10]3 years ago
8 0
Here you go the answer is in the picture

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The half-life of the radioactive element unobtanium-43 is 10 seconds. If 48 grams of unobtanium-43 are initially present, how ma
natka813 [3]

Answer:

At time, 10 seconds, the mass remaining is 24 grams

At time, 20 seconds, the mass remaining is 12 grams

At time, 30 seconds, the mass remaining is 6 grams

At time, 40 seconds, the mass remaining is 3 grams

At time, 50 seconds, the mass remaining is 1.5 grams

Step-by-step explanation:

Given;

initial mass of the radioactive element  = 48 grams

half life, t = 10 seconds

time of decay (s)                          remaining mass of the  radioactive element

0 ------------------------------------------------- 48 grams

10------------------------------------------------- 24 grams

20 ------------------------------------------------- 12 grams

30 -------------------------------------------------- 6 grams

40 --------------------------------------------------- 3 grams

50 ----------------------------------------------------- 1.5 grams

60------------------------------------------------------0.75 grams

Thus;

At time, 10 seconds, the mass remaining is 24 grams

At time, 20 seconds, the mass remaining is 12 grams

At time, 30 seconds, the mass remaining is 6 grams

At time, 40 seconds, the mass remaining is 3 grams

At time, 50 seconds, the mass remaining is 1.5 grams

7 0
3 years ago
13 POINTS 13 Points 13 POINTS 13 Points 13 POINTS 13 Points 13 POINTS 13 Points
ira [324]

Answer:

8

Step-by-step explanation:

  • because 16-8≈ 8
5 0
3 years ago
Read 2 more answers
WILL MARK BRAINLIEST
Nutka1998 [239]

Answer:

a² + b² = 68

a3 + b3 = 520

Step-by-step explanation:

Given :

a + b = 10 (1)

ab = 16 (2)

A. Find a² + b²

(a + b)² = a² + 2ab + b² (3)

Substitutite the values of (1) and (2) into (3)

(10)² = a² + 2(16) + b²

100 = a² + 32 + b²

Subtract 32 from both sides

100 - 32 = a² + b²

a² + b² = 68

B. a^3 + b^3

(a + b)^3 = a^3 + b^3 + 3ab(a + b)

(10)^3 = a^3 + b^3 + 3*16(10)

1000 = a^3 + b^3 + 480

a^3 + b^3 = 1000 - 480

a3 + b3 = 520

5 0
3 years ago
if you take my age and divide it by any off number greater than 1 and less than 9 you will get a remainder of 1
postnew [5]

Answer:

no???

Step-by-step explanation:

6 0
3 years ago
Please answer this correctly
Tanya [424]

Answer:

All make numbers make that statement true

Step-by-step explanation:

6406358 + 8550780= 14957138

14957138-5598256= 9358882

therefore is you add any number to 5598256 that is greater than 9358882 then the statement will be true. All those numbers given are larger than 9358882

6 0
3 years ago
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