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3 years ago

Give the excluded values for the rational equation. (3x)/(x-4)-(x+3)/(x+4)=(2x+7)/(x^(2)-16)

Mathematics

2 answers:

scoray [572]3 years ago

8 0

$\frac{3x}{x-4} - \frac{x+3}{x-4} = \frac{2x+7}{x^2-16}$

We factor the denominators

Factor x^2 - 16

x^2 - 4^2

We use a^2 - b^2 = (a+b)(a-b)

so x^2 - 4^2 = (x+4)(x-4)

Replace it in the given equation

$\frac{3x}{x-4} - \frac{x+3}{x-4} = \frac{2x+7}{(x+4)(x-4)}$

Excluded values are the values that makes the denominator 0

we have (x-4) and (x+4) in the denominator

We set the denominator =0 and solve for x

x-4 =0

Add 4 on both sides

x= 4

x+4=0

subtract 4 onboth side

so x= -4

Excluded values are x=-4 and x=4

meriva3 years ago

8 0

Answer is x =-4 and x =4

Explanation:

$\frac{3x}{x-4} -\frac{x+3}{x+4}=\frac{2x+7}{x^2-16}\\Let us take LCD\\\frac{3x(x+3)-(x+3)(x-4)}{x^2-16}= \frac{2x+7}{x^2-16}$

We find that there is a common denominator.

This should not be equal to 0.

i.e. x ≠4 or x ≠-4

Thus x=4 and x=-4 are the excluded values for the rational equation.

Once we exclude these values, we cancel and equate numerator

3x(x+4)-(x+3)(x-4) = 2x+7

Simplify to get 3x^2+12x-x^2+x-12 = 2x+7

2x^2+11x-19 =0

Thus when we exclude those values, we et a quadratic equation with 2 solutions.

simplify to get

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Answer:

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number of Lilies = 75

Step-by-step explanation:

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Let

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y = number of carnations

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Answer:

The standard equation of the sphere is $(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = 53$

Step-by-step explanation:

From the question, the end point are (3,-2,4) and (7,12,4)

Since we know the end points of the diameter, we can determine the center (midpoint of the two end points) of the sphere.

The midpoint can be calculated thus

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Hence,

Midpoint = $(\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2}, \frac{z_{1} + z_{2} }{2})$

Midpoint = $(\frac{3 + 7 }{2}, \frac{-2+12 }{2}, \frac{4 + 4 }{2})$

Midpoint = $(\frac{10 }{2}, \frac{10}{2}, \frac{8 }{2})\\$

Midpoint = $(5, 5, 4)$

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Now, we will determine the distance (diameter) of the sphere

The distance is given by

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$d = \sqrt{16 +196 + 0$

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$d = 2\sqrt{53}$

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To find the radius, r

From $Radius = \frac{Diameter}{2}$

$Radius = \frac{2\sqrt{53} }{2}$

∴ $Radius = \sqrt{53}$

r = $\sqrt{53}$

Now, we can write the standard equation of the sphere since we know the center and the radius

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Radius of the sphere is $\sqrt{53}$

The equation of a sphere of radius r and center $(h,k,l)$ is given by

$(x-h)^{2} + (y-k)^{2} + (z-l)^{2} = r^{2}$

Hence, the equation of the sphere of radius $\sqrt{53}$ and center $(5, 5, 4)$ is

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This is the standard equation of the sphere

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