The sum of Sally and Beth's age is 60. Six years ago, Sally was 3 times as old as Beth. What are their present ages?

Mathematics

2 answers:

Thepotemich [5.8K]3 years ago

4 0

Sally is 45 and Beth is 15.

45 + 15 = 60

15*3 = 45

pav-90 [236]3 years ago

3 0

Answer: Present age of Beth is 24 years.

And Present age of Sally becomes 36 years.

Step-by-step explanation:

Let the present age of Beth be 'x'.

Six years ago,

Age of Beth becomes (x-6) years

Age of Sally becomes

$3(x-6)\ yrs$

Sum of Sally and Beth's age = 60

According to question, it becomes

$x-6+3(x-6)=60+6+6\\\\x-6+3x-18=72\\\\4x-24=72\\\\4x=72+24\\\\4x=96\\\\x=\frac{96}{4}=24\ yrs$

So, Present age of Beth is 24 years.

And Present age of Sally becomes 60-24=36 years.

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swat32

I have the same question! have you figured it out ??

3 0

3 years ago

D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$