Answer:
its 70 and 58
Step-by-step explanation:
The pattern: -2, -4, -6, -8, -10, -12...
Answer:
i don't
know math
Step-by-step explanation:
<em>-3x-y=10</em>
<em>4x-4y=8</em>
<em>or</em>
<em>y = -3x - 10 </em>
<em>P(0,-10) and P(-2, -4) on this line: Plot and connect with the Line.</em>
<em></em>
<em> </em>
<em>y = x - 2</em>
<em>P(0, -2) and (2,0) on this line: Plot and connect with the Line.</em>
<em>P(-2,-4) is the ordered pair that is the solution for this system of EQs</em>
<em>On may CHECK by substituting x= -2 and y = -4 into the EQs of these Lines</em>
<em></em>
<em>Hope this helps!!!</em>
Answer:
The slope would be 1/4.
Step-by-step explanation:
To find the perpendicular slope first you have to set this equation to y
to do this you use the subtraction prop of equality and you will get 12x-9= -3y
then you divide by -3 on both sides to get y by itself
you will get -4x+3=y
now that you know the slope is -4, you must find the negative reciprocal slope of this which means you have to flip the numerator and denominator and make the sign its opposite.
The perpendicular slope would then be 1/4.
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)
