Points H, G and F are collinear. Point G is between Hand G. Solve for x. HG = x, GF = 1, HF = 10 *
1 answer:
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Answer:
sink at 1.28 g/cm^3
v = 4/3
v = 4/3
v =11.49 /9 =1.28
Step-by-step explanation:
That one correct so yes um cause the app said answer must be above 20 characters
Applying our power rule gets us our first derivative,
simplifying a little bit,
looking for critical points,
We can apply more factoring.
I hope this next step isn't too confusing.
We want to factor out the smallest power of x from both terms,
and also the 2 from each.
When you divide x^(-2/3) out of x^(1/3),
it leaves you with x^(3/3) or simply x.
Then apply your Zero-Factor Property,
and solve for x in each case to find your critical points.
Apply your First Derivative Test to further classify these points. You should end up finding that x=-1/2 is an relative extreme value, while x=0 is not.
Let's come back to this,
and take our second derivative.
Looking for inflection points,
Again, pulling out the smaller power of x, and fractional part,
And again, apply your Zero-Factor Property, setting each factor to zero and solving for x in each case. You should find that x=0 and x=1 are possible inflection points.
Applying your Second Derivative Test should verify that both points are in fact inflection points, locations where the function changes concavity.
simplifying a little bit,
looking for critical points,
We can apply more factoring.
I hope this next step isn't too confusing.
We want to factor out the smallest power of x from both terms,
and also the 2 from each.
When you divide x^(-2/3) out of x^(1/3),
it leaves you with x^(3/3) or simply x.
Then apply your Zero-Factor Property,
and solve for x in each case to find your critical points.
Apply your First Derivative Test to further classify these points. You should end up finding that x=-1/2 is an relative extreme value, while x=0 is not.
Let's come back to this,
and take our second derivative.
Looking for inflection points,
Again, pulling out the smaller power of x, and fractional part,
And again, apply your Zero-Factor Property, setting each factor to zero and solving for x in each case. You should find that x=0 and x=1 are possible inflection points.
Applying your Second Derivative Test should verify that both points are in fact inflection points, locations where the function changes concavity.