Answer:
the answer will be x^3 + 2x^2 + 9x
Step-by-step explanation:
as :
volume = l* w* h
= X * (X + 2) * (X + 3)
= X ^ 2 + 2 X (X + 3)
= x ^2 (X + 3) + 2x (X + 3)
= x^3 + 3x +2x ^2 + 6x
= x^3 + 2x^2 + 9x
Answer:
Two expression are:
![\sin A = \frac{40}{41}](https://tex.z-dn.net/?f=%5Csin%20A%20%3D%20%5Cfrac%7B40%7D%7B41%7D)
![\cos B = \frac{40}{41}](https://tex.z-dn.net/?f=%5Ccos%20B%20%3D%20%5Cfrac%7B40%7D%7B41%7D)
Step-by-step explanation:
Let rt triangle ABC.
Now, labeling the parts of a right triangle.
Let the right angle be labelled C and the hypotenuse c and
A and B denote the other two angles, and a and b the sides opposite them respectively.
In a right triangle ABC,
the trigonometric ratio is given by:
![\sin A = \frac{opposite}{Hypotenuse} = \frac{a}{c}=\frac{40}{41}](https://tex.z-dn.net/?f=%5Csin%20A%20%3D%20%5Cfrac%7Bopposite%7D%7BHypotenuse%7D%20%3D%20%5Cfrac%7Ba%7D%7Bc%7D%3D%5Cfrac%7B40%7D%7B41%7D)
![\cos B= \frac{Adjacent side}{Hypotenuse} = \frac{a}{c}=\frac{40}{41}](https://tex.z-dn.net/?f=%5Ccos%20B%3D%20%5Cfrac%7BAdjacent%20side%7D%7BHypotenuse%7D%20%3D%20%5Cfrac%7Ba%7D%7Bc%7D%3D%5Cfrac%7B40%7D%7B41%7D)
Therefore, the two expression which is equivalent to
is ; ![\sin A = \frac{40}{41}](https://tex.z-dn.net/?f=%5Csin%20A%20%3D%20%5Cfrac%7B40%7D%7B41%7D)
![\cos B = \frac{40}{41}](https://tex.z-dn.net/?f=%5Ccos%20B%20%3D%20%5Cfrac%7B40%7D%7B41%7D)
Answer:
i dont know if it is a true or false question bc there isnt enough info
Step-by-step explanation:
so ima say true
First, rewrite
![x^y=e^{\ln x^y}=e^{y\ln x}](https://tex.z-dn.net/?f=x%5Ey%3De%5E%7B%5Cln%20x%5Ey%7D%3De%5E%7By%5Cln%20x%7D)
![5^{x-y}=e^{\ln5^{x-y}} = e^{\ln(5)(x-y)}](https://tex.z-dn.net/?f=5%5E%7Bx-y%7D%3De%5E%7B%5Cln5%5E%7Bx-y%7D%7D%20%3D%20e%5E%7B%5Cln%285%29%28x-y%29%7D)
Now, differentiate both sides using the chain rule:
![\dfrac{\mathrm d\left(e^{y\ln x}\right)}{\mathrm dx}=\dfrac{\mathrm d\left(e^{\ln(5)(x-y)}\right)}{\mathrm dx}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%5Cleft%28e%5E%7By%5Cln%20x%7D%5Cright%29%7D%7B%5Cmathrm%20dx%7D%3D%5Cdfrac%7B%5Cmathrm%20d%5Cleft%28e%5E%7B%5Cln%285%29%28x-y%29%7D%5Cright%29%7D%7B%5Cmathrm%20dx%7D)
![e^{y\ln x}\dfrac{\mathrm d(y\ln x)}{\mathrm dx}=e^{\ln(5)(x-y)}\dfrac{\mathrm d(\ln(5)(x-y))}{\mathrm dx}](https://tex.z-dn.net/?f=e%5E%7By%5Cln%20x%7D%5Cdfrac%7B%5Cmathrm%20d%28y%5Cln%20x%29%7D%7B%5Cmathrm%20dx%7D%3De%5E%7B%5Cln%285%29%28x-y%29%7D%5Cdfrac%7B%5Cmathrm%20d%28%5Cln%285%29%28x-y%29%29%7D%7B%5Cmathrm%20dx%7D)
![x^y\left(\dfrac{\mathrm dy}{\mathrm dx}\ln x+y\dfrac{\mathrm d(\ln x)}{\mathrm dx}\right)=\ln(5)5^{x-y}\left(\dfrac{\mathrm d(x)}{\mathrm dx}-\dfrac{\mathrm dy}{\mathrm dx}\right)](https://tex.z-dn.net/?f=x%5Ey%5Cleft%28%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%5Cln%20x%2By%5Cdfrac%7B%5Cmathrm%20d%28%5Cln%20x%29%7D%7B%5Cmathrm%20dx%7D%5Cright%29%3D%5Cln%285%295%5E%7Bx-y%7D%5Cleft%28%5Cdfrac%7B%5Cmathrm%20d%28x%29%7D%7B%5Cmathrm%20dx%7D-%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%5Cright%29)
![x^y\left(\ln x\dfrac{\mathrm dy}{\mathrm dx}+\dfrac yx\right)=\ln(5)5^{x-y}\left(1-\dfrac{\mathrm dy}{\mathrm dx}\right)](https://tex.z-dn.net/?f=x%5Ey%5Cleft%28%5Cln%20x%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%2B%5Cdfrac%20yx%5Cright%29%3D%5Cln%285%295%5E%7Bx-y%7D%5Cleft%281-%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%5Cright%29)
![\left(x^y\ln x+\ln(5)5^{x-y}\right)\dfrac{\mathrm dy}{\mathrm dx}=\ln(5)5^{x-y}-yx^{y-1}](https://tex.z-dn.net/?f=%5Cleft%28x%5Ey%5Cln%20x%2B%5Cln%285%295%5E%7Bx-y%7D%5Cright%29%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%3D%5Cln%285%295%5E%7Bx-y%7D-yx%5E%7By-1%7D)
![\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\ln(5)5^{x-y}-yx^{y-1}}{x^y\ln x+\ln(5)5^{x-y}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%3D%5Cdfrac%7B%5Cln%285%295%5E%7Bx-y%7D-yx%5E%7By-1%7D%7D%7Bx%5Ey%5Cln%20x%2B%5Cln%285%295%5E%7Bx-y%7D%7D)
Answer:
11?
Step-by-step explanation:
Hope I'm right! (: