Answer:
The height of First Copy is 20 cm.
Step-by-step explanation:
Given:
Original Height = 25 cm
First copy is reduce to 80% of Original height.
Therefore to find the height of first copy can be find out by multiplying 80 with original height and then dividing by 100.
∴ Height of First Copy = ![\frac{80}{100} \times \textrm{original height}= 0.8\times 25 cm = 20cm](https://tex.z-dn.net/?f=%5Cfrac%7B80%7D%7B100%7D%20%5Ctimes%20%5Ctextrm%7Boriginal%20height%7D%3D%200.8%5Ctimes%2025%20cm%20%3D%2020cm)
Hence Height of first copy is 20 cm
Now Another copy is reduce to 85% of First Copy.
Therefore to find the height of another copy can be find out by multiplying 85 with first copy height and then dividing by 100.
∴ Height of Another Copy = ![\frac{85}{100} \times \textrm{first copy height}= 0.85\times 20 cm = 17cm](https://tex.z-dn.net/?f=%5Cfrac%7B85%7D%7B100%7D%20%5Ctimes%20%5Ctextrm%7Bfirst%20copy%20height%7D%3D%200.85%5Ctimes%2020%20cm%20%3D%2017cm)
Hence Height of Another copy is 17 cm
Answer:
answer to what? free points?
Step-by-step explanation:
Answer:
the answer is 12
i used 5 12 13 special trangle
and there is 2 angle 56 and 34
so the biggest angle meet the 12
and the little angle meet the 5
hopefully its correct
Answer:
if the symbol there between the 3 and the x is a period, it is a different way of showing multiplication or the *(×)sign.
Step-by-step explanation:
4-3*x
=1x
=x
but if it's not, I'll have to check again. this would not be th answer
In matrix form, the system is
![\dfrac{\mathrm d}{\mathrm dt}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1&1\\4&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dt%7D%5Cbegin%7Bbmatrix%7Dx%5C%5Cy%5Cend%7Bbmatrix%7D%3D%5Cbegin%7Bbmatrix%7D1%261%5C%5C4%261%5Cend%7Bbmatrix%7D%5Cbegin%7Bbmatrix%7Dx%5C%5Cy%5Cend%7Bbmatrix%7D)
First find the eigenvalues of the coefficient matrix (call it
).
![\det(\mathbf A-\lambda\mathbf I)=\begin{vmatrix}1-\lambda&1\\4&1-\lambda\end{vmatrix}=(1-\lambda)^2-4=0\implies\lambda^2-2\lambda-3=0](https://tex.z-dn.net/?f=%5Cdet%28%5Cmathbf%20A-%5Clambda%5Cmathbf%20I%29%3D%5Cbegin%7Bvmatrix%7D1-%5Clambda%261%5C%5C4%261-%5Clambda%5Cend%7Bvmatrix%7D%3D%281-%5Clambda%29%5E2-4%3D0%5Cimplies%5Clambda%5E2-2%5Clambda-3%3D0)
![\implies\lambda_1=-1,\lambda_=3](https://tex.z-dn.net/?f=%5Cimplies%5Clambda_1%3D-1%2C%5Clambda_%3D3)
Find the corresponding eigenvector for each eigenvalue:
![\lambda_1=-1\implies(\mathbf A+\mathbf I)\vec\eta_1=\vec0\implies\begin{bmatrix}2&1\\4&2\end{bmatrix}\begin{bmatrix}\eta_{1,1}\\\eta_{1,2}\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}](https://tex.z-dn.net/?f=%5Clambda_1%3D-1%5Cimplies%28%5Cmathbf%20A%2B%5Cmathbf%20I%29%5Cvec%5Ceta_1%3D%5Cvec0%5Cimplies%5Cbegin%7Bbmatrix%7D2%261%5C%5C4%262%5Cend%7Bbmatrix%7D%5Cbegin%7Bbmatrix%7D%5Ceta_%7B1%2C1%7D%5C%5C%5Ceta_%7B1%2C2%7D%5Cend%7Bbmatrix%7D%3D%5Cbegin%7Bbmatrix%7D0%5C%5C0%5Cend%7Bbmatrix%7D)
![\lambda_2=3\implies(\mathbf A-3\mathbf I)\vec\eta_2=\vec0\implies\begin{bmatrix}-2&1\\4&-2\end{bmatrix}\begin{bmatrix}\eta_{2,1}\\\eta_{2,2}\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}](https://tex.z-dn.net/?f=%5Clambda_2%3D3%5Cimplies%28%5Cmathbf%20A-3%5Cmathbf%20I%29%5Cvec%5Ceta_2%3D%5Cvec0%5Cimplies%5Cbegin%7Bbmatrix%7D-2%261%5C%5C4%26-2%5Cend%7Bbmatrix%7D%5Cbegin%7Bbmatrix%7D%5Ceta_%7B2%2C1%7D%5C%5C%5Ceta_%7B2%2C2%7D%5Cend%7Bbmatrix%7D%3D%5Cbegin%7Bbmatrix%7D0%5C%5C0%5Cend%7Bbmatrix%7D)
![\implies\vec\eta_1=\begin{bmatrix}1\\-2\end{bmatrix},\vec\eta_2=\begin{bmatrix}1\\2\end{bmatrix}](https://tex.z-dn.net/?f=%5Cimplies%5Cvec%5Ceta_1%3D%5Cbegin%7Bbmatrix%7D1%5C%5C-2%5Cend%7Bbmatrix%7D%2C%5Cvec%5Ceta_2%3D%5Cbegin%7Bbmatrix%7D1%5C%5C2%5Cend%7Bbmatrix%7D)
Then the system has general solution
![\begin{bmatrix}x\\y\end{bmatrix}=C_1\vec\eta_1e^{\lambda_1t}+C_2\vec\eta_2e^{\lambda_2t}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7Dx%5C%5Cy%5Cend%7Bbmatrix%7D%3DC_1%5Cvec%5Ceta_1e%5E%7B%5Clambda_1t%7D%2BC_2%5Cvec%5Ceta_2e%5E%7B%5Clambda_2t%7D)
or
![\begin{cases}x(t)=C_1e^{-t}+C_2e^{3t}\\y(t)=-2C_1e^{-t}+2C_2e^{3t}\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Dx%28t%29%3DC_1e%5E%7B-t%7D%2BC_2e%5E%7B3t%7D%5C%5Cy%28t%29%3D-2C_1e%5E%7B-t%7D%2B2C_2e%5E%7B3t%7D%5Cend%7Bcases%7D)
Given that
and
, we have
![\begin{cases}1=C_1+C_2\\2=-2C_1+2C_2\end{cases}\implies C_1=0,C_2=2](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D1%3DC_1%2BC_2%5C%5C2%3D-2C_1%2B2C_2%5Cend%7Bcases%7D%5Cimplies%20C_1%3D0%2CC_2%3D2)
so that the system has particular solution
![\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}e^{3t}\\2e^{3t}\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7Dx%5C%5Cy%5Cend%7Bbmatrix%7D%3D%5Cbegin%7Bbmatrix%7De%5E%7B3t%7D%5C%5C2e%5E%7B3t%7D%5Cend%7Bbmatrix%7D)