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3 years ago

X =A.1B.3C.7Picture Provided Below

Mathematics

1 answer:

Yakvenalex [24]3 years ago

7 0

Answer:

B 3

Step-by-step explanation:

We can use the secant-secant formula:

(whole secant) x (external part) = (whole secant) x (external part)

9 * 4 = 12 *x

36 = 12x

Divide each side by 12

36/12 =12x/12

3 =x

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1. Express <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%282x%2B3%29%20%7D" id="TexFormula1" title="\frac{1}{x(2x+3) }" a

katovenus [111]

1. Let a and b be coefficients such that

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

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

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

so the particular solution to the ODE is

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)$

We can solve this explicitly for y :

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)$

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

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

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

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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Naddika [18.5K]

Apply to Row 2 : Row 2 + Row 1

2x + 2y + 3z = 0

y + 4z = -3

2x + 3y + 3z = 5

Apply to Row 3: Row 3 - Row 1

2x + 2y + 3z = 0

y + 4z = -3

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Apply to Row 3: Row 3 - Row 2

2x + 2y + 3z = 0

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Simplify rows

2x + 2y + 3z = 0

y + 4z = -3

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Note that the matrix is in echelon form now. The next steps are for back substitution.

Apply to Row 2: Row 2 - 4 Row 3

2x + 2y + 3z = 0

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z = -2

Apply to Row 1: Row 1 - 3 Row 3

2x + 2y = 6

y = 5

z = -2

Apply to Row 1: Row 1 - 2 Row 2

2x = -4

y = 5

z = 2

Simplify the rows

x = -2

y = 5

z = -2

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