Answer:
A = 40.0 (3 sig. fig)
C = 40
c = 15
Step-by-step explanation:
Using sin law,
sinA/15 = sin100/23
A = 40.0 (3 sig. fig)
Using angle sum of triange,
C = 180 - 40.0 - 100 = 40
Since angleA = angleC, by side equal opp angles, c = 15
I'm assuming that this is the complete question.
If f(x) = 3 – 2x and g(x)=1/(x+5), what is the value of (f/g)(8)? a) –169 b) –1 c) 13 d) 104
x = 8
f(x) = 3 -2xf(8) = 3 - 2(8) = 3 - 16 = -13
g(x) = 1/(x+5)g(8) = 1/(8+5) = 1/13
(f/g)(8)f(8)/g(8) = -13/ (1/13) = -13 * 13 = -169 Choice A :)
Answer: The range is 55.
Step-by-step explanation:
3/9 will fit on one bus so it will take 3 buses to fit 9/9 of the students, which 9/9 is the whole group.
Answer:
1.) 8.09g ; 2) 206.7 years
Step-by-step explanation:
Given the following :
Half-life(t1/2) of Uranium-232 = 68.9 years
a) If you have a 100 gram sample, how much would be left after 250 years?
Initial quantity (No) = 100g
Time elapsed (t) = 250 years
Find the quantity of substance remaining (N(t))
Recall :
N(t) = No(0.5)^(t/t1/2)
N(250) = 100(0.5)^(250/68.9)
N(250) = 100(0.5)^3.6284470
N(250) = 100 × 0.0808590
= 8.0859045
= 8.09g
2) If you have a 100 gram sample, how long would it take for there to be 12.5 grams remaining?
Using the relation :
N / No = (1/2)^n
Where N = Amount of remaining or left
No = Original quantity
n = number of half-lifes
N = 12.5g ; No = 100g
12.5 / 100 = (1/2)^n
0.125 = (1/2)^n
Converting 0.125 to fraction
(1/8) = 1/2^n
8 = 2^n
2^3 = 2^n
n = 3
Recall ;
Number of half life's (n) = t / t1/2
t = time elapsed ; t1/2 = half life
3 = t / 68.9
t = 3 × 68.9
t = 206.7 years
