All categories

3 years ago

Pls help helpful answer

Mathematics

1 answer:

zalisa [80]3 years ago

4 0

Answer:

as you can see in both triangles, comparing to each other, x = 90 angle

w = 52, because when 90 + 38 + w = 180 for a triangle

z=10yd

u = 62yd

v = 79 yd

please, also make sure these answers are correct from anyone, but ihave answer correct :)

Hope it helps

You might be interested in

Look at the picture help please

Fofino [41]

X = 39ft
It’s the same units as house A it’s just a reflection

3 0

3 years ago

Jeff bought 10 cookies and ate 4 of them. He ate

zmey [24]

he ate 40 percent of the cookies

8 0

3 years ago

Read 2 more answers

Need help!!!!!!!!!!!!!!!

icang [17]

Answer:

no error original system so good by

8 0

3 years ago

Help? pls? (10 points) (image down below to make it easier to understand.)

dusya [7]

This is middle school??? AinT no way

7 0

2 years ago

Read 2 more answers

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

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

Thus, the required solution is

$y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.$

6 0

3 years ago