For the first two, put it them over 100. So number 1 would be 34/100 or 17/50 reduced. Number 2 is 77.5/100 and since it is a decimal, I would keep it at over 100. Number 3 put 8/10 or 4/5 reduced and number 4 is different. Think of it as money and how a quarter is in relation to a dollar. You have 4 quarters in a dollar, so since you have .25%, it is 1/4.
By definition of absolute value, you have
or more simply,
On their own, each piece is differentiable over their respective domains, except at the point where they split off.
For <em>x</em> > -1, we have
(<em>x</em> + 1)<em>'</em> = 1
while for <em>x</em> < -1,
(-<em>x</em> - 1)<em>'</em> = -1
More concisely,
Note the strict inequalities in the definition of <em>f '(x)</em>.
In order for <em>f(x)</em> to be differentiable at <em>x</em> = -1, the derivative <em>f '(x)</em> must be continuous at <em>x</em> = -1. But this is not the case, because the limits from either side of <em>x</em> = -1 for the derivative do not match:
All this to say that <em>f(x)</em> is differentiable everywhere on its domain, <em>except</em> at the point <em>x</em> = -1.
The area of the circle shown would not be 5pi. If you look closer, the area is 25pi, therefore the answer is 3. I hope this helps!
Answer:
Step-by-step explanation:
Given that,
y' = 17y ( 1-y^7)
Let y=1
Then, y' = 0 for all t
Then show that it is the only stable equilibrium point so that as y→1, t→∞ with any initial value.
So, the graph solution will be
y(0) = 1 and this will be an horizontal line
If, y(0) > 1 then, y' < 0 by inspecting the first equation, so the graph is has decreasing solution.
Likewise, if y(0) < 1 then, y' > 0 and the graph is increasing.
So no matter the initial condition, graph of the solution will be asymptotic to the horizontal line above.
This make the limit be 1.
This shows that x = 1 is a stable equilibrium.
Decimal: Divide 16 by 25: 0.64
Percent: move the decimal 2 places to the left: 64, then add a percent sign: 64%
