Answer:
0.75 mg
Step-by-step explanation:
From the question given above the following data were obtained:
Original amount (N₀) = 1.5 mg
Half-life (t₁/₂) = 6 years
Time (t) = 6 years
Amount remaining (N) =.?
Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:
Half-life (t₁/₂) = 6 years
Time (t) = 6 years
Number of half-lives (n) =?
n = t / t₁/₂
n = 6/6
n = 1
Finally, we shall determine the amount of the sample remaining after 6 years (i.e 1 half-life) as follow:
Original amount (N₀) = 1.5 mg
Half-life (t₁/₂) = 6 years
Number of half-lives (n) = 1
Amount remaining (N) =.?
N = 1/2ⁿ × N₀
N = 1/2¹ × 1.5
N = 1/2 × 1.5
N = 0.5 × 1.5
N = 0.75 mg
Thus, 0.75 mg of the sample is remaining.
Answer:
Adult=58
Step-by-step explanation:
c=child, a=adult
6.4c+9.7a=1145 equation 1
c+a=149 equation 2
a=149-c modified equation 2 to isolate a
6.4c+9.7(149-c)=1145 substitute value of a from equation 1 into equation 2
6.4c+1445.3-9.7c=1145
-3.3c=-300.3
c=91
solve for a
c+a=149
91+a=149
a=58
Check answer:
6.4c+9.7a=1145
6.4(91)+9.7(58)=1145
582.40+562.60=1145
1145=1145
Answer:
B
r=9.7
C=60.95
Step-by-step explanation:
Given diameter of 19.4
r = d/2
19.4/2
r=9.7
C= 2(pi)r
2(pi)(9.7)
60.95 = C
Answer:
The smaller number is 0.6666... repeating, and the larger number is 1.3333... repeating.
Step-by-step explanation:
Answer:
D) x² -x - 10 =0
Step-by-step explanation:
Given that
L.C.M
⇒ 
⇒ 9 x + 54 = 9 x² - 36
⇒ 9 x² - 36 - 9x - 54 =0
⇒ 9 x² - 9x - 90 =0
⇒ 9(x² -x - 10 ) =0
⇒ x² -x - 10 =0
