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Furkat [3]
2 years ago
13

2m^2+2m-12=0 in vertex form

Mathematics
1 answer:
oksian1 [2.3K]2 years ago
5 0

Answer:

y = 2(m + 1/2)^ -25/2

Step-by-step explanation:

vertex form is:

y = 2m^2 + 2m - 12


y + 12 + ( )___ =( ) 2m^2 + 2m + ___


Because a is not one, you must divide the entire right side by the a value, which is 2. Then place the 2 in the parenthesis.


y + 12 + (2)__ = (2) m^2 + m + __


Take half of b, b being one.

Half of be is 1/2. Now square it to get 1/4. Place that value in the blanks.


y + 12 + (2)1/4 = (2) m^2 + m + 1/4


Now simplify the left side and factor the right side.


y+ 12 + 1/2 = 2(m + 1/2)^2


y + 25/2 = 2(m + 1/2)^2


Subtract 25/2 from both sides.


y = 2(m + 1/2)^ -25/2

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\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}

Combining the fractions on the right gives

\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}

\implies 1 = (2a+b)x + 3a

\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23

so that

\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}

2. a. The given ODE is separable as

x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}

Using the result of part (1), integrating both sides gives

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C

Given that y = 1 when x = 1, we find

\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)

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\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|

\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|

\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}

2. b. When x = 9, we get

y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}

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