Answer:
3x + 4y - 32= 0
Step-by-step explanation:
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Answer:
£13,819 to the nearest pound is still £13,819.
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
Answer:
Using either method, we obtain: 
Step-by-step explanation:
a) By evaluating the integral:
![\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdt%7D%20%5Cint%5Climits%5Et_0%20%7B%5Csqrt%5B8%5D%7Bu%5E3%7D%20%7D%20%5C%2C%20du)
The integral itself can be evaluated by writing the root and exponent of the variable u as: ![\sqrt[8]{u^3} =u^{\frac{3}{8}](https://tex.z-dn.net/?f=%5Csqrt%5B8%5D%7Bu%5E3%7D%20%3Du%5E%7B%5Cfrac%7B3%7D%7B8%7D)
Then, an antiderivative of this is: 
which evaluated between the limits of integration gives:
and now the derivative of this expression with respect to "t" is:
b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:
"If f is continuous on [a,b] then
is continuous on [a,b], differentiable on (a,b) and 
Since this this function 
is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:
