Answer:
9ms
Step-by-step explanation:
Answer:
I think it would be D
Step-by-step explanation:
20 x 3 = 60 60+35= 95 20/4 = 5
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.
-13+21=14
The temperature is 14 degrees during your afternoon.
Answer: 195.5 or 12 7/32
Step-by-step explanation:
There is no letter tetha in the table so I use α instead. However it is not sence to final result.
The expression is:
(sinα+cosα)/(cosα*(1-cosα))
Lets divide the nominator and denominator by cosα
(sinα/cosα+cosα/cosα)/(cosα*(1-cosα)/cosα)= (tanα+1)/(1-cosα)=
=(8/15+1)/(1-cosα)= 23/(15*(1-cosα)) (1)
As known cos²α=1-sin²α (divide by cos²α both sides of equation)
cos²a/cos²α=1/cos²α-sin²α/cos²α
1=1/cos²α-tg²α
1/cos²α=1+tg²α
cos²α=1/(1+tg²α)
cosα=sqrt(1/(1+tg²α))= +-sqrt(1/(1+64/225))=+-sqrt(225/(225+64))=
=+-sqrt(225/289)=+-15/17 (2)
Substitute in (1) cosα by (2):
1st use cosα=15/17
1) 23/(15*(1-cosα)) =23/(15*(1-15/17))= 23*17/2=195.5
2-nd use cosα=-15/17
2)23/(15*(1-cosα)) =23/(15*(1+15/17))= 23*17/32=12 7/32
