What property is m+0=m

Mathematics

2 answers:

RSB [31]3 years ago

8 0

Answer:additional

Step-by-step explanation:

Mama L [17]3 years ago

7 0

Identity property (for other problems look in the middle of my page)

You might be interested in

The vertices of a triangle are located at X(-2,-1), Y(2,5) and Z(4,3). What is the perimeter of this triangle?

Lana71 [14]

If you plot the points you will find you have to use the distance formula. Pick out 2 pairs label them (x1,y1) (x2,y2) then plug it into the distance formula. The repeat for the other sides. So you should get (2,5) (4,3) plug into distance formula what is get it distance between those two points, then (4,3) (-2,-1) plug into distance formula to get an answer, lastly (-2,-1) (2,5) plug in. Now you have three answers add all together and here is your perimeter of a triangle.

3 0

3 years ago

I will give you ALL of my 50 points if you did this

netineya [11]

If you send me the points I’ll do it in the comments

3 0

3 years ago

Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview

svetlana [45]

Answer: The required solution is

$y=(-2+t)e^{-5t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.$

So, the general solution of the given equation is

$y(t)=(A+Bt)e^{-5t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.$

According to the given conditions, we have

$y(0)=-2\\\\\Rightarrow A=-2$

and

$y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.$