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3 years ago

Given s(x) = 2x - 3 and t(x) = 5x + 4. find the formula and domain for s(x) over t(x) and w(x) = t(x) over s(x)

1 answer:

6 0

V(x) = (2x - 3)/(5x + 4) The domain is all Real numbers except x = -4/5, because if x = -4/5 the denominator would be zero and you cannot divide by zero.{x | x ∈ R, x ≠ -4/5} w(x) = (5x + 4)/(2x - 3)similarly, x ≠ 3/2so, {x| x ∈ R, x ≠ 3/2}

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elena-14-01-66 [18.8K]

Answer:

The equation of line Le is 5 y + x + 89 = 0

Step-by-step explanation:

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The line Li passes through point (1,2)

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or, 2 = 5 × 1 + c

Or, c = 2 - 5 = -3

∴ Eq of line Li is

y = 5 x - 3

Another line Le is perpendicular to line Li

Let the slope of line Le = M

The line Le meet at points where x = 4

So, for perpendicular property

Products of slopes = - 1

So, m × M = - 1

or, M = - $\frac{1}{m}$

Or. M = - $\frac{1}{5}$

Now equation of line Le is y = M x + c

∵ Line Le meet the line Li at x = 4

So, y = 5 × 4 - 3

I.e y = 20 - 3

Or, y = 17

Now, equation of line Le with slope - $\frac{1}{5}$ and passes through points ( 4 , 17 ) is

y = M x + c

or, 17 = - $\frac{1}{5}$ × 4 + c

or, 17 + $\frac{4}{5}$ = c

So, c = $\frac{85+4}{5}$

Or, c = - $\frac{89}{5}$

∴ Equation of line Le is y = - $\frac{1}{5}$ x - $\frac{89}{5}$

ie. 5 y = - x - 89

or, 5 y + x + 89 = 0

Hence The equation of line Le is 5 y + x + 89 = 0 answer

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