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maks197457 [2]
2 years ago
14

Please help! Hurry!!!!

Mathematics
2 answers:
SVEN [57.7K]2 years ago
8 0

Expression:  =====>   2x^2    +   3x    +   5

Terms ======>   2x^2,    3x  ====> A number or product of a number and variables raised to a power.

Coefficient =====>   2x^2,    3x  ====> Numerical factor of the term.

Constant   =====>   5   =====> Term without a Variable





Number One is Done For You!!!    




Hope that helps!!!                            : )   Sorry I couldn't help with the rest!!!

statuscvo [17]2 years ago
4 0

for number one there are three terms

the coefficient is 2

and the constant is 5

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Answer:

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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)

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We have the following differential equation

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

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

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

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

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-\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}

4 0
2 years ago
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