I think it is 5 but I am not for sure
Answer:
Step-by-step explanation:
Okay so I will explain the first one.
2 to the power of 4 = 2x2x2x2
2 to the power of 6 = 2x2x2x2x2x2
2 to the power of 4 = 2x2x2x2
2^4 = 16
2^6 = 64
When you are multiplying powers you add.
+
When you subtract you takeaway.
-
So you would do
4+6+4
= 14
Meaning 2 to the power of 14
2
So what is 2 to the power of 14.
You would do
2x2x2x2x2x2x2x2x2x2x2x2x2x2
The answer is 16384
You can also use a calculator
GOOD LUCK FOR YOUR TEST!
Equation of an ellipse
→having center (0,0) , vertex (
and covertex
and focus
is given by:

As definition of an ellipse is that locus of all the points in a plane such that it's distance from two fixed points called focii remains constant.
Consider two points (a,0) and (-a,0) on Horizontal axis of an ellipse:
Distance from (a,0) to (c,0) is = a-c = 
Distance from (-a,0) to (c,0) is = a + c = 
a -c + a +c
= a + a
= 2 a →(Option A )
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
![5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.](https://tex.z-dn.net/?f=5m%5E2e%5E%7Bmt%7D%2B3me%5E%7Bmt%7D-2e%5E%7Bmt%7D%3D0%5C%5C%5C%5C%5CRightarrow%20%285m%5E2%2B3y-2%29e%5E%7Bmt%7D%3D0%5C%5C%5C%5C%5CRightarrow%205m%5E2%2B3m-2%3D0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~%5B%5Ctextup%7Bsince%20%7De%5E%7Bmt%7D%5Cneq0%5D%5C%5C%5C%5C%5CRightarrow%205m%5E2%2B5m-2m-2%3D0%5C%5C%5C%5C%5CRightarrow%205m%28m%2B1%29-2%28m%2B1%29%3D0%5C%5C%5C%5C%5CRightarrow%20%28m%2B1%29%285m-1%29%3D0%5C%5C%5C%5C%5CRightarrow%20m%2B1%3D0%2C~~~~~5m-1%3D0%5C%5C%5C%5C%5CRightarrow%20m%3D-1%2C~%5Cdfrac%7B1%7D%7B5%7D.)
So, the general solution of the given equation is

Differentiating with respect to t, we get

According to the given conditions, we have

and
![y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.](https://tex.z-dn.net/?f=y%5E%5Cprime%280%29%3D2.8%5C%5C%5C%5C%5CRightarrow%20-A%2B%5Cdfrac%7BB%7D%7B5%7D%3D2.8%5C%5C%5C%5C%5CRightarrow%20-5A%2BB%3D14%5C%5C%5C%5C%5CRightarrow%20-5A-A%3D14~~~~~~~~~~~~~~~~~~~~~~~~~~~%5B%5Ctextup%7BUisng%20equation%20%28ii%29%7D%5D%5C%5C%5C%5C%5CRightarrow%20-6A%3D14%5C%5C%5C%5C%5CRightarrow%20A%3D-%5Cdfrac%7B14%7D%7B6%7D%5C%5C%5C%5C%5CRightarrow%20A%3D-%5Cdfrac%7B7%7D%7B3%7D.)
From equation (ii), we get

Thus, the required solution is
