Answer:
25°
Step-by-step explanation:
These types of figures have 3 angles, which add up to 180°. If you found the 2 angles already, you can find the last one by adding both angles together and using 180° to deduct it with the total degree of thd 2 angles.
127 + 28 = 155
180 - 155 = 25
Answer:
I think the answer is D.) 7/8 mile
Step-by-step explanation:
If you make the mixed numbers improper fractions, 2 and 1/8 is 17/8, 1 and 1/4 (2/8) is 10/8, then you subtract the numerators and keep the denomitators the same to get 7/8 mile (sorry if I spelt something wrong)
Answer:
c. 10
Step-by-step explanation:
Eliminate parentheses using the distributive property.
3(b+4) -2(2b+3) = -4 . . . . given
3b +12 -4b -6 = -4 . . . . . parentheses eliminated
-b +6 = -4 . . . . . . . . . . . . collect terms
-b = -10 . . . . . . . . . . . . . . subtract 6
b = 10 . . . . . . . . . . . . . . . . multiply by -1
_____
The rules of equality require that any operation you perform on one side of the equal sign must also be done on the other side. So when we say "subtract 6", we assume you know that means "subtract 6 from both sides of the equation", for example.
Step-by-step explanation:
see the answer in the pic
Answer:
Q(t) = Q_o*e^(-0.000120968*t)
Step-by-step explanation:
Given:
- The ODE of the life of Carbon-14:
Q' = -r*Q
- The initial conditions Q(0) = Q_o
- Carbon isotope reaches its half life in t = 5730 yrs
Find:
The expression for Q(t).
Solution:
- Assuming Q(t) satisfies:
Q' = -r*Q
- Separate variables:
dQ / Q = -r .dt
- Integrate both sides:
Ln(Q) = -r*t + C
- Make the relation for Q:
Q = C*e^(-r*t)
- Using initial conditions given:
Q(0) = Q_o
Q_o = C*e^(-r*0)
C = Q_o
- The relation is:
Q(t) = Q_o*e^(-r*t)
- We are also given that the half life of carbon is t = 5730 years:
Q_o / 2 = Q_o*e^(-5730*r)
-Ln(0.5) = 5730*r
r = -Ln(0.5)/5730
r = 0.000120968
- Hence, our expression for Q(t) would be:
Q(t) = Q_o*e^(-0.000120968*t)
