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3241004551 [841]
1 year ago
13

mr.davis divided his money in the ratio 4:6 between james and brett. james got the smaller amount of $540. how much more did bre

tt receive? how much in total was the money?
Mathematics
1 answer:
STALIN [3.7K]1 year ago
4 0
4x-\ Jamese's\ money\\\\
6x-\ Brett's\ money\\\\\
4x=540\ \ \ |Divide\ by\ 4\\
x=135\\\\
6x=6*135=810\$\\\\\
Brett\ received\ 810\$.
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. Solve the following initial value problem: (t2−20t+51)dydt=y (t2−20t+51)dydt=y with y(10)=1y(10)=1. (Find yy as a function of
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Answer:

y=(\frac{t-17}{t-3})^{\frac{1}{14}}

Step-by-step explanation:

We are given that initial value problem

t^2-20t+51)\frac{dy}{dt}=y

\frac{dy}{y}=\frac{dt}{t^2-20t+51}

\frac{dy}{y}=\frac{dt}{t^2-3t-17t+51}

\frac{dy}{y}=\frac{dt}{(t(t-3)-17(t-3)}

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\frac{1}{(t-3)(t-17)}=\frac{A(t-17)+B(t-3)}{(t-3)(t-17)}

1=A(t-17)+B(t-3)...(1)

Substitute t-3=0

t=3

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t=17

Substitute t=3 in equation (1)

1=A(3-17)+0

1=-14A

A=-\frac{1}{14}

Substitute t=17

1=B(17-3)

1=14B

B=\frac{1}{14}

Substitute the values of A and B

\frac{1}{(t-3)(t-17)}=-\frac{1}{14}(\frac{1}{t-3})+\frac{1}{14}(\frac{1}{t-17})

\int\frac{dy}{y}=-\frac{1}{14}\int\frac{dt}{t-3}+\frac{1}{14}\int\frac{dt}{t-17}

ln y=-\frac{1}{14}ln\mid{t-3}\mid+\frac{1}{14}\mid{t-17}\mid+ln C

By using formula:\frac{dx}{x}=ln x+C

ln y=\frac{1}{14}(-ln\mid{t-3}\mid+ln\mid{t-17}\mid)+ln C

Using formula:ln x-ln y=ln \frac{x}{y}

ln y=\frac{1}{14}(ln\mid{\frac{t-17}{t-3}}\mid)+ln C

ln y=\frac{1}{14}ln\mid{\frac{t-17}{t-3}}\mid+ ln C

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0=0+ln C

Because ln 1=0

lnC=0

C=e^0=1

Because ln x=y\implies x=e^y

Substitute the value of C

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lny^{14}=ln\mid\frac{t-17}{t-3}\mid

By using identity blog a= loga^b

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