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3 years ago

6

A ball is launched from ground level directly upward at 39.2 m/s. The height can be modeled by the function f(t)=-4.9t^2+39.2t.

For how long Is the ball at or above a height of 34.3 meters

1 answer:

Elden [556K]3 years ago

6 0

Answer:yellog

Step-by-step explanation:

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Answer:

$\frac{2x + 3}{(x- 1)(x - 4)} = \frac{-5}{3(x- 1)} + \frac{11}{3(x - 4)}$

Step-by-step explanation:

Given the rational expression: $\frac{2x + 3}{x^2 - 5x + 4}$ , to express this in simplified form, we would need to apply the concept of partial fraction.

Step 1: factorise the denominator

$x^2 - 5x + 4$

$x^2 - 4x - x + 4$

$(x^2 - 4x) - (x + 4)$

$x(x - 4) - 1(x - 4)$

$(x- 1)(x - 4)$

Thus, we now have: $\frac{2x + 3}{(x- 1)(x - 4)}$

Step 2: Apply the concept of Partial Fraction

Let,

$\frac{2x + 3}{(x- 1)(x - 4)}$ = $\frac{A}{x- 1} + \frac{B}{x - 4}$

Multiply both sides by (x - 1)(x - 4)

$\frac{2x + 3}{(x- 1)(x - 4)} * (x - 1)(x - 4)$ = $(\frac{A}{x- 1} + \frac{B}{x - 4}) * (x - 1)(x - 4)$

$2x + 3 = A(x - 4) + B(x - 1)$

Step 3:

Substituting x = 4 in $2x + 3 = A(x - 4) + B(x - 1)$

$2(4) + 3 = A(4 - 4) + B(4 - 1)$

$8 + 3 = A(0) + B(3)$

$11 = 3B$

$\frac{11}{3} = B$

$B = \frac{11}{3}$

Substituting x = 1 in $2x + 3 = A(x - 4) + B(x - 1)$

$2(1) + 3 = A(1 - 4) + B(1 - 1)$

$2 + 3 = A(-3) + B(0)$

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Step 4: Plug in the values of A and B into the original equation in step 2

$\frac{2x + 3}{(x- 1)(x - 4)} = \frac{A}{x- 1} + \frac{B}{x - 4}$

$\frac{2x + 3}{(x- 1)(x - 4)} = \frac{-5}{3(x- 1)} + \frac{11}{3(x - 4)}$

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Answer:

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