I'm not sure where to apply the trig identity.
I got sin(x)/ (cos(x)/sin(x))^2 - sin(x)/ cos(x)*cos(x)
Where does the identity come into play?
Thanks for the quick response
Trigonometry Basic (URGENT) please help !
Homework Statement
SIMPLIFY:
sin(x)/cot^2(x) - sin(x)/cos^2(x)
Homework Equations
Trigonometric identities I think
The Attempt at a Solution
I got sin(x)cot^-2(x) - sin(x) sec^2(x) ...
but the book says the answer is...
thanks i got all the answers except for 1)
so here's what i did:
dh/dt = 2.5 cm
h=11.5
dA/dt= 4cm
A= 99
and i need to find the rate of change of db/dt when h= 11.5
first i found out what the base would be so I plugged h=11.5 into A= 1/2bh and got b=17.2173913
Next I took...
yeah. i did do that already ...
orig equation =25600 +300x+x^2
avg cost= 25600/x +300 +x
to find the minimal avg cost , take derivative of avg cost equation and set it to 0.
... could you explain what i did wrong instead of just saying average cost is c(x)/x ? i already knew that
For one of my semester finals, my teacher wanted us to write a paper on the New Deal. He did not specify any sort of topic except to be very analytical. We were required to read several texts and incorporate it into our writing.
the first text is FDR's speech in October 31, 1936 in Madison...
Thanks Mark . I understand that the minimum is -5 and the maximum is 4 because of the endpoints of the interval. and I also know that the inflection point is at x=0 y=0. But for the concavity, are there specific numbers as to where f(x) is concave down/up on the region from ___ to ___ ...
8.
for number 8, did you mean the minimum is at (0,9) or (1,9)? Then does that mean the maximum will be at (6,129)?
9.
I really need to learn how to do that so it won't be so confusing next time.
Following your advice, I canceled out the e^8x and got the quadratic equation f"(x)=...
5.
Okay I tried again this time using the quadratic formula instead, to factor it.
This time I got x= sqrt(-32)/4 and x= sqrt(32)/4 .
Next, I plugged x= sqrt(-32)/4 into the original and got the (x,y) coords: [ (-sqrt(32)/4), -3.4635785 ] , which will be my minimum
Then I plugged...
Ah sorry mark, to be bothering again... I emailed the professor and he told me my number 1 and number 3 is wrong. but he said i did a great job on number 3 and 4, so it's all thanks to you :) !
question number one answer isn't -29.0952... i keep getting this same answer but i might be doing a...
8.
f(x)=5x^2-10x+9
f'(x)= 10x-10
I set it to =0 and got x=1.
Then I made a table
which means that (0,9) is minimum and (6,129) is maximum?
9. f"(x)= [2x*8e^8x+2e^8x]+[8e^8x*2x + x^2*64e^8x]
I keep getting this as a second derivative and it's so long so I don't think it...
Happy Thanksgiving :) !
4. Thank you so much for explaining so clearly. I know what my teacher is talking about and I took a look at this before going to school today and understood everything in class. If it wasn't for your explanation, i would still be completely clueless !
5. may i ask what...
thank you for responding.
8. f(x)= 5x^2-10x+9
f'(x)= 10x-10
f'(x)=0 --> 10x-10=0 --> which will give me x=1 only. how will i know the max and min given that x=1 ? I plug it back in and am i missing another root?
9. the inflection point is when it changes from concave up...