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lakkis [162]
3 years ago
10

Translate the graph of f(x)=2x+3 up 2 units.

Mathematics
1 answer:
lord [1]3 years ago
4 0
If you translate the graph up 2 units, the new equation would be 2x+5
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Find the solution of the given initial value problem in explicit form. y′=(1−5x)y2, y(0)=−12 Enclose numerators and denominators
Nata [24]

Answer:

The solution is y=-\frac{12}{12x-30x^2+1}.

Step-by-step explanation:

A first order differential equation y'=f(x,y) is called a separable equation if the function f(x,y) can be factored into the product of two functions of x and y:

f(x,y)=p(x)h(y)

where p(x) and h(y) are continuous functions.

We have the following differential equation

y'=(1-5x)y^2, \quad y(0)=-12

In the given case p(x)=1-5x and h(y)=y^2.

We divide the equation by h(y) and move dx to the right side:

\frac{1}{y^2}dy\:=(1-5x)dx

Next, integrate both sides:

\int \frac{1}{y^2}dy\:=\int(1-5x)dx\\\\-\frac{1}{y}=x-\frac{5x^2}{2}+C

Now, we solve for y

-\frac{1}{y}=x-\frac{5x^2}{2}+C\\-\frac{1}{y}\cdot \:2y=x\cdot \:2y-\frac{5x^2}{2}\cdot \:2y+C\cdot \:2y\\-2=2yx-5yx^2+2Cy\\y\left(2x-5x^2+2C\right)=-2\\\\y=-\frac{2}{2x-5x^2+2C}

We use the initial condition y(0)=-12 to find the value of C.

-12=-\frac{2}{2\left(0\right)-5\left(0\right)^2+2C}\\-12=-\frac{1}{c}\\c=\frac{1}{12}

Therefore,

y=-\frac{2}{2x-5x^2+2(\frac{1}{12})}\\y=-\frac{12}{12x-30x^2+1}

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Answer:

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The answer to this question would be: 2*52! -1

In this case, the card is either picked up by Naoko or they are not picked up. That means there are two possible outcomes for each card. For 52 card, the possibility should be:
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