Product of 4 and a number translates to 4x because the word product means multiplication.
We then need to subtract one from that which would be 4x - 1.
The word is translates to "equals" so 4x - 1 = 11 is the translation of the whole problem.
Now solve that equation.
4x - 1 = 11
1st add +1 +1
2nd simplify 4x = 12
3rd divide. 4 4
x = 3
Answer:
jerry 5% saves more than tom
Step-by-step explanation:
tom saves 35%
and jerry saves 40%
Answer: The required solution is
![y=(-2+t)e^{-5t}.](https://tex.z-dn.net/?f=y%3D%28-2%2Bt%29e%5E%7B-5t%7D.)
Step-by-step explanation: We are given to solve the following differential equation :
![y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)](https://tex.z-dn.net/?f=y%5E%7B%5Cprime%5Cprime%7D%2B10y%5E%5Cprime%2B25y%3D0%2C~~~~~~~y%280%29%3D-2%2C~~y%5E%5Cprime%280%29%3D11~~~~~~~~~~~~~~~~~~~~~~~~%28i%29)
Let us consider that
be an auxiliary solution of equation (i).
Then, we have
![y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.](https://tex.z-dn.net/?f=y%5Eprime%3Dme%5E%7Bmt%7D%2C~~~~~y%5E%7B%5Cprime%5Cprime%7D%3Dm%5E2e%5E%7Bmt%7D.)
Substituting these values in equation (i), we get
![m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.](https://tex.z-dn.net/?f=m%5E2e%5E%7Bmt%7D%2B10me%5E%7Bmt%7D%2B25e%5E%7Bmt%7D%3D0%5C%5C%5C%5C%5CRightarrow%20%28m%5E2%2B10y%2B25%29e%5E%7Bmt%7D%3D0%5C%5C%5C%5C%5CRightarrow%20m%5E2%2B10m%2B25%3D0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~%5B%5Ctextup%7Bsince%20%7De%5E%7Bmt%7D%5Cneq0%5D%5C%5C%5C%5C%5CRightarrow%20m%5E2%2B2%5Ctimes%20m%5Ctimes5%2B5%5E2%3D0%5C%5C%5C%5C%5CRightarrow%20%28m%2B5%29%5E2%3D0%5C%5C%5C%5C%5CRightarrow%20m%3D-5%2C-5.)
So, the general solution of the given equation is
![y(t)=(A+Bt)e^{-5t}.](https://tex.z-dn.net/?f=y%28t%29%3D%28A%2BBt%29e%5E%7B-5t%7D.)
Differentiating with respect to t, we get
![y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.](https://tex.z-dn.net/?f=y%5E%5Cprime%28t%29%3D-5e%5E%7B-5t%7D%28A%2BBt%29%2BBe%5E%7B-5t%7D.)
According to the given conditions, we have
![y(0)=-2\\\\\Rightarrow A=-2](https://tex.z-dn.net/?f=y%280%29%3D-2%5C%5C%5C%5C%5CRightarrow%20A%3D-2)
and
![y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.](https://tex.z-dn.net/?f=y%5E%5Cprime%280%29%3D11%5C%5C%5C%5C%5CRightarrow%20-5%28A%2BB%5Ctimes0%29%2BB%3D11%5C%5C%5C%5C%5CRightarrow%20-5A%2BB%3D11%5C%5C%5C%5C%5CRightarrow%20%28-5%29%5Ctimes%28-2%29%2BB%3D11%5C%5C%5C%5C%5CRightarrow%2010%2BB%3D11%5C%5C%5C%5C%5CRightarrow%20B%3D11-10%5C%5C%5C%5C%5CRightarrow%20B%3D1.)
Thus, the required solution is
![y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.](https://tex.z-dn.net/?f=y%28t%29%3D%28-2%2B1%5Ctimes%20t%29e%5E%7B-5t%7D%5C%5C%5C%5C%5CRightarrow%20y%28t%29%3D%28-2%2Bt%29e%5E%7B-5t%7D.)
Answer:
For the first question the answer is 7/8 and for the second 1 the answer is 10/10 or 1/10
Step-by-step explanation:
you subtract 1/2 from 1 3/4 and get 7/8 same for the bottom question subtracted 1/4 from 2 1/2 for you answer.