Answer:
Step-by-step explanation:
Answer:
a) y-8 = (y₀-8) , b) 2y -5 = (2y₀-5)
Explanation:
To solve these equations the method of direct integration is the easiest.
a) the given equation is
dy / dt = and -8
dy / y-8 = dt
We change variables
y-8 = u
dy = du
We replace and integrate
∫ du / u = ∫ dt
Ln (y-8) = t
We evaluate at the lower limits t = 0 for y = y₀
ln (y-8) - ln (y₀-8) = t-0
Let's simplify the equation
ln (y-8 / y₀-8) = t
y-8 / y₀-8 =
y-8 = (y₀-8)
b) the equation is
dy / dt = 2y -5
u = 2y -5
du = 2 dy
du / 2u = dt
We integrate
½ Ln (2y-5) = t
We evaluate at the limits
½ [ln (2y-5) - ln (2y₀-5)] = t
Ln (2y-5 / 2y₀-5) = 2t
2y -5 = (2y₀-5)
c) the equation is very similar to the previous one
u = 2y -10
du = 2 dy
∫ du / 2u = dt
ln (2y-10) = 2t
We evaluate
ln (2y-10) –ln (2y₀-10) = 2t
2y-10 = (2y₀-10)
Step-by-step explanation:
Exponential Functions:
y=ab^x
y=ab
x
a=\text{starting value = }1600
a=starting value = 1600
r=\text{rate = }5.25\% = 0.0525
r=rate = 5.25%=0.0525
\text{Exponential Growth:}
Exponential Growth:
b=1+r=1+0.0525=1.0525
b=1+r=1+0.0525=1.0525
\text{Write Exponential Function:}
Write Exponential Function:
y=1600(1.0525)^x
y=1600(1.0525)
x
Put it all together
\text{Plug in time for x:}
Plug in time for x:
y=1600(1.0525)^{25}
y=1600(1.0525)
25
y= 5750.0628984
y=5750.0628984
Evaluate
y\approx 5750.06
y≈5750.06
18= 0.5 (b)(h)
36=(6c)(c-1)
6c^2-6c-36=0
6(c^2-c-6)=0
6(c-3)(c+2)=0
c= 3 or c=-2 but you cant use the negative because your measurement can not be negative, so c=3.
now plug into the original equation.
base is c-1 3-1=2
height is 6c= 6(3)=18
Okay, so the lines are the same, so the answer would be the last question.
m is another word for slope, and the slopes are both -1.
the y-intercepts are also both 5.
I can't see whatever graph is on the problem, but I am pretty sure this would be the answer.
I hope this helps.
Your best decision would be to find a subject that the entire grade has so it is not biased such as a course class like science or language arts depending on what grade they are referring to if this is not the answer you where looking for I am sorry hope I helped :)
