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VARVARA [1.3K]
3 years ago
12

Carson is g years old Haley is 2 yrs younger then Carson. find the sum of their ages in terms of g

Mathematics
2 answers:
BaLLatris [955]3 years ago
6 0

Answer:

2g-2

.....................

N76 [4]3 years ago
3 0

Answ<em>e</em><em>r</em><em> </em>

<em>2</em><em>g</em><em> </em><em>-</em><em>2</em>

Step-by-step explanation:

<h2>Carson is =g</h2><h2>Harley = 2-g</h2>
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Consider the differential equation x^2 y''-xy'-3y=0. If y1=x3 is one solution use redution of order formula to find a second lin
Anastasy [175]

Suppose y_2(x)=y_1(x)v(x) is another solution. Then

\begin{cases}y_2=vx^3\\{y_2}'=v'x^3+3vx^2//{y_2}''=v''x^3+6v'x^2+6vx\end{cases}

Substituting these derivatives into the ODE gives

x^2(v''x^3+6v'x^2+6vx)-x(v'x^3+3vx^2)-3vx^3=0

x^5v''+5x^4v'=0

Let u(x)=v'(x), so that

\begin{cases}u=v'\\u'=v''\end{cases}

Then the ODE becomes

x^5u'+5x^4u=0

and we can condense the left hand side as a derivative of a product,

\dfrac{\mathrm d}{\mathrm dx}[x^5u]=0

Integrate both sides with respect to x:

\displaystyle\int\frac{\mathrm d}{\mathrm dx}[x^5u]\,\mathrm dx=C

x^5u=C\implies u=Cx^{-5}

Solve for v:

v'=Cx^{-5}\implies v=-\dfrac{C_1}4x^{-4}+C_2

Solve for y_2:

\dfrac{y_2}{x^3}=-\dfrac{C_1}4x^{-4}+C_2\implies y_2=C_2x^3-\dfrac{C_1}{4x}

So another linearly independent solution is y_2=\dfrac1x.

3 0
3 years ago
If Fahmida randomly picks a jelly bean out of the jar, what is the probability that the jelly bean will be green (find the simpl
Neko [114]

Answer:

1/5

Step-by-step explanation:

you divide the amount of green jelly beans by the total amount of jelly beans in the jar, which gives you 10/50 or 1/5.

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For what value of the number k is the following function differentiable at x=2
d1i1m1o1n [39]

Answer:

For f(x) to be differentiable at 2, k = 5.

Step-by-step explanation:

For f(x) to be differentiable at x = 2, f(x) has to be continuous at 2.

For f(x) to be continuous at 2, the limit of f(2 – h) = f(2) = f(2 + h) as h tends to 0.

Now,

f(2 – h) = 2(2 – h) + 1 = 4 – 2h + 1 = 5 – 2h.

As h tends to 0, lim (5 – 2h) = 5

Also

f(2 + h) = 3(2 + h) – 1 = 6 + 3h – 1 = 5 + 3h

As h tends to 0, lim (5 + 3h) = 5.

So, for f(2) to be continuous k = 5

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