Question that needs to be solved:You use 4 gallons of water on 30 plants in your garden. At that rate, how much water will it take to water 45 plants?=> 4 gallons is to 30 plants.=> 4 gallons / 30 plants = 0.13 gallons per plants.Now, it needs to water around 45 plants, let us solve how much will it need => 0.13 gallons * 45 plants = 6 gallons<span>Thus, to water 45 plants you need to have 6 gallons of water</span>
Answer:
w = 10
Step-by-step explanation:
![\frac{4}{5}w=8~(Given)\\\\\frac{5}{4}(\frac{4}{5}w)=\frac{5}{4}(8)~(Multiply~5/4~on~both~sides)\\\\w=10~(Simplify)](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B5%7Dw%3D8~%28Given%29%5C%5C%5C%5C%5Cfrac%7B5%7D%7B4%7D%28%5Cfrac%7B4%7D%7B5%7Dw%29%3D%5Cfrac%7B5%7D%7B4%7D%288%29~%28Multiply~5%2F4~on~both~sides%29%5C%5C%5C%5Cw%3D10~%28Simplify%29)
Answer:A. -1.11x^2+ 3.74+ 10.64
Step-by-step explanation:
a. p. e. x. (just took the test)
![(2x+1)^{\cot x}=\exp\left(\ln(2x+1)^{\cot x}\right)=\exp\left(\cot x\ln(2x+1)\right)=\exp\left(\dfrac{\ln(2x+1)}{\tan x}\right)](https://tex.z-dn.net/?f=%282x%2B1%29%5E%7B%5Ccot%20x%7D%3D%5Cexp%5Cleft%28%5Cln%282x%2B1%29%5E%7B%5Ccot%20x%7D%5Cright%29%3D%5Cexp%5Cleft%28%5Ccot%20x%5Cln%282x%2B1%29%5Cright%29%3D%5Cexp%5Cleft%28%5Cdfrac%7B%5Cln%282x%2B1%29%7D%7B%5Ctan%20x%7D%5Cright%29)
where
![\exp(x)\equiv e^x](https://tex.z-dn.net/?f=%5Cexp%28x%29%5Cequiv%20e%5Ex)
.
By continuity of
![e^x](https://tex.z-dn.net/?f=e%5Ex)
, you have
![\displaystyle\lim_{x\to0^+}\exp\left(\dfrac{\ln(2x+1)}{\tan x}\right)=\exp\left(\lim_{x\to0^+}\dfrac{\ln(2x+1)}{\tan x}\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bx%5Cto0%5E%2B%7D%5Cexp%5Cleft%28%5Cdfrac%7B%5Cln%282x%2B1%29%7D%7B%5Ctan%20x%7D%5Cright%29%3D%5Cexp%5Cleft%28%5Clim_%7Bx%5Cto0%5E%2B%7D%5Cdfrac%7B%5Cln%282x%2B1%29%7D%7B%5Ctan%20x%7D%5Cright%29)
As
![x\to0^+](https://tex.z-dn.net/?f=x%5Cto0%5E%2B)
in the numerator, you approach
![\ln1=0](https://tex.z-dn.net/?f=%5Cln1%3D0)
; in the denominator, you approach
![\tan0=0](https://tex.z-dn.net/?f=%5Ctan0%3D0)
. So you have an indeterminate form
![\dfrac00](https://tex.z-dn.net/?f=%5Cdfrac00)
. Provided the limit indeed exists, L'Hopital's rule can be used.
![\displaystyle\exp\left(\lim_{x\to0^+}\dfrac{\ln(2x+1)}{\tan x}\right)=\exp\left(\lim_{x\to0^+}\dfrac{\frac2{2x+1}}{\sec^2x}\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cexp%5Cleft%28%5Clim_%7Bx%5Cto0%5E%2B%7D%5Cdfrac%7B%5Cln%282x%2B1%29%7D%7B%5Ctan%20x%7D%5Cright%29%3D%5Cexp%5Cleft%28%5Clim_%7Bx%5Cto0%5E%2B%7D%5Cdfrac%7B%5Cfrac2%7B2x%2B1%7D%7D%7B%5Csec%5E2x%7D%5Cright%29)
Now the numerator approaches
![\dfrac21=2](https://tex.z-dn.net/?f=%5Cdfrac21%3D2)
, while the denominator approaches
![\sec^20=1](https://tex.z-dn.net/?f=%5Csec%5E20%3D1)
, suggesting the limit above is 2. This means