What was the instructions given
3 nickels and 1 quarter add to $0.40.
The person has 11 times that amount, so has 11 quarters and 33 nickels.
The equation of the graph of 
is (d) on a coordinate plane, an absolute value graph starts at (0, 0) and goes up through (4, 2).
The equation of the graph is represented as:
See attachment for the graph of
From the attached graph, the curve passes through points (0,0) and (4,2).
Hence, the equation of the graph of is (d)
Read more about graphs and functions at:
brainly.com/question/3939432
Answer:
Step-by-step explanation:
Hello,
Pls post the full question next time so that we can make sure what is expected.
I assume that you want to solve for (x,y)
(1) 2x + 3y = 7
(2) 3x + 2y = 8
2*(2)-3*(1) gives
2*3x + 2*2y - 3*2x - 3*3y = 2*8 - 3*7 = 16 - 21 = -5
6x - 6x + 4y -9y = -5
-5y=-5
y = 1
and we can replace in (1)
2x + 3*1 = 7
2x = 7 - 3 = 4
x = 2
hope this helps
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)
