Answer:
1/5 ig
Step-by-step explanation:
Snap filter T-T ik ik it makes mwe look irresistible :P
Answer: It will take her 125 days to save 500, so technically 126 to save more than 500
Step-by-step explanation:
500/4
Answer:
Ok, so first, at the end of the equation, It says "+ 4" , right? So you put a point on positive 4 on the y-axis (the one that goes up and don, the vertical line.). Then, 2x means that's your slope, but since your graph only goes up to five, you gotta count down ward. So, go down from your point on the y-axis (4) <em><u>twice</u></em> and over to the <u>LEFT</u> (<--) <u><em>once.</em></u> Plot your point where you end up. then use a straightedge like a ruler, and connect your two points on the graph, and you have your line!
Hope this helps!
X = number of singles matches
y = number of doubles mnathes
so we have the equation x + y = 13.......(1)
also
2x + 4y = 38 ...........(2)
(1) and (2) are the required equations
b) Find how many maches are in progress by solivng the above equations:-
x + y = 13 multiply this by -2:-
-2x - 2y = -26
2x + 4y = 38 Adding to removes the terms in x:-
2y = 12
y = 6
so x = 13-6 = 7
So the answer is 6 double matches and 7 singles are in progress.
In the ODE, solve for
:
![t^2y''-2y'+(3+t)y=0\implies y''=\dfrac{2y'+(3+t)y}{t^2}](https://tex.z-dn.net/?f=t%5E2y%27%27-2y%27%2B%283%2Bt%29y%3D0%5Cimplies%20y%27%27%3D%5Cdfrac%7B2y%27%2B%283%2Bt%29y%7D%7Bt%5E2%7D)
The Wronskian is then
![W=\begin{vmatrix}y_1&y_2\\{y_1}'&{y_2}'\end{vmatrix}=y_1{y_2}'-{y_1}'y_2](https://tex.z-dn.net/?f=W%3D%5Cbegin%7Bvmatrix%7Dy_1%26y_2%5C%5C%7By_1%7D%27%26%7By_2%7D%27%5Cend%7Bvmatrix%7D%3Dy_1%7By_2%7D%27-%7By_1%7D%27y_2)
Differentiating the Wronskian gives
![W'=({y_1}'{y_2}'+y_1{y_2}'')-({y_1}''y_2+{y_1}'{y_2}')=y_1{y_2}''-{y_1}''y_2](https://tex.z-dn.net/?f=W%27%3D%28%7By_1%7D%27%7By_2%7D%27%2By_1%7By_2%7D%27%27%29-%28%7By_1%7D%27%27y_2%2B%7By_1%7D%27%7By_2%7D%27%29%3Dy_1%7By_2%7D%27%27-%7By_1%7D%27%27y_2)
Substitute
into the equation for
, then substitute
into
:
![W'=y_1\dfrac{2{y_2}'+(3+t)y_2}{t^2}-y_2\dfrac{2{y_1}'+(3+t)y_1}{t^2}](https://tex.z-dn.net/?f=W%27%3Dy_1%5Cdfrac%7B2%7By_2%7D%27%2B%283%2Bt%29y_2%7D%7Bt%5E2%7D-y_2%5Cdfrac%7B2%7By_1%7D%27%2B%283%2Bt%29y_1%7D%7Bt%5E2%7D)
![\implies W'=\dfrac{2W}{t^2}](https://tex.z-dn.net/?f=%5Cimplies%20W%27%3D%5Cdfrac%7B2W%7D%7Bt%5E2%7D)
which is another separable ODE; we have
![\dfrac{\mathrm dW}W=\dfrac2{t^2}\,\mathrm dt\implies \ln|W|=-\dfrac2t+C\implies W=Ce^{-2/t}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20dW%7DW%3D%5Cdfrac2%7Bt%5E2%7D%5C%2C%5Cmathrm%20dt%5Cimplies%20%5Cln%7CW%7C%3D-%5Cdfrac2t%2BC%5Cimplies%20W%3DCe%5E%7B-2%2Ft%7D)
Given that
, we find
![3=Ce^{-2/2}\implies C=3e](https://tex.z-dn.net/?f=3%3DCe%5E%7B-2%2F2%7D%5Cimplies%20C%3D3e)
so that
![W(y_1,y_2)(t)=3e^{1-2/t}](https://tex.z-dn.net/?f=W%28y_1%2Cy_2%29%28t%29%3D3e%5E%7B1-2%2Ft%7D)
and so
![W(y_1,y_2)(6)=\boxed{3e^{2/3}}](https://tex.z-dn.net/?f=W%28y_1%2Cy_2%29%286%29%3D%5Cboxed%7B3e%5E%7B2%2F3%7D%7D)