62 triangle a and 59 triangle b
(<u>−1</u>
2 )(n^3)+
<u>1</u>
2 n^2+4.6n+(−
<u>1</u>
2)(n^3)+
<u>1</u>
2 n^2+4.5n
=
<u>−1</u>
2 n^3+
1
2 n^2+4.6n+
−1
2 n^3+
1
2 n^2+4.5n
Combine Like Terms:
=
<u>−1</u>
2 n^3+
<u>1</u>
2 n^2+4.6n+
<u>−1</u>
2 n^3+
<u>1</u>
2 n^2+4.5n
=(<u>−1</u>
2 n^3+
<u>−1</u>
2 n^3)+(
<u>1</u>
2 n^2+
<u>1</u>
2 n^2)+(4.6n+4.5n)
=−n^3+n^2+9.1n
Answer:
=−n^3+n^2+9.1n
Everything underlined means its a fraction/divided hope this helps <em>:D</em>
Answer: The required solution is
Step-by-step explanation: We are given to solve the following differential equation :
Let us consider that
be an auxiliary solution of equation (i).
Then, we have
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
Differentiating with respect to t, we get
According to the given conditions, we have
and
Thus, the required solution is
Answer:
X= -5/3 Y= 10/3
Let me know if you need the work to be shown
Answer:
y = 3x + 3
Step-by-step explanation:
If the entire line, or rather, the graph, is translated upwards by 5 units; the slope stays the same, but the y-intercept rises by 5 as well. Thus, the equation of the new graph would be;
y = 3x (-2+5)
y = 3x + 3
