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MakcuM [25]
3 years ago
14

What are the discontinuities of the function f(x)= (x^2 - 36)/(4x - 24)

Mathematics
1 answer:
cricket20 [7]3 years ago
6 0
When x = -6  the denominator  = 0
There is a hole in the graph at (-6,0)
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Which is greater 6.0 or 0.6
Tamiku [17]

Answer:

6.0

Step-by-step explanation: It is 6.0 because 0.6 is a decimal number and 6.0 is a whole number and can I be brainliest please



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cluponka [151]
B is the correct anwser
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What is the answer of 8(3v+3w-2)
stepan [7]

Answer:24v+24w-16

Step-by-step explanation:

8(3v+3w-2)

8 times 3 =24

So 24v+24w

Then 8 times 2=16

So the answer would be 24v+24w-16

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horrorfan [7]

Answer:

p=\frac{181}{301}

Step-by-step explanation:

30100 have dogs, 18100 have cats.

The question simply asks for a union scenario thus the law of AND & OR is applicable.

Therefore:-Those who have both cats and dogs partially have cats.

P(X=both \ cat \ and \ dog)=\frac{18100}{30100}\\p=\frac{181}{301}

8 0
3 years ago
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Someone help me out please
uysha [10]

1 step (B): raise both sides of the equation to the power of 2.

(\sqrt{x+3}-\sqrt{2x-1})^2=(-2)^2,\\   (x+3)-2\sqrt{x+3}\cdot \sqrt{2x-1}+(2x-1)=4,\\ 3x+2-2\sqrt{x+3}\cdot \sqrt{2x-1}=4.

2 step (A): simplify to obtain the final radical term on one side of the equation.

-2\sqrt{x+3}\cdot \sqrt{2x-1}=4-3x-2,\\ -2\sqrt{x+3}\cdot \sqrt{2x-1}=2-3x,\\ 2\sqrt{x+3}\cdot \sqrt{2x-1}=3x-2.

3 step (F): raise both sides of the equation to the power of 2 again.

(2\sqrt{x+3}\cdot \sqrt{2x-1})^2=(3x-2)^2,\\ 4(x+3)(2x-1)=(3x-2)^2.

4 step (E): simplify to get a quadratic equation.

4(2x^2-x+6x-3)=(3x)^2-2\cdot 3x\cdot 2+2^2,\\ 8x^2+20x-12=9x^2-12x+4,\\ x^2-32x+16=0.

5 step (D): use the quadratic formula to find the values of x.

D=(-32)^2-4\cdot 16=1024-64=960, \\ \sqrt{D} =8\sqrt{5} ,\\ x_{1,2}=\dfrac{32\pm 8\sqrt{5}}{2} =16\pm 4\sqrt{5}.

6 step (C): apply the zero product rule.

x^2-32x+16=(x-16-4\sqrt{5}) (x-16+4\sqrt{5}) ,\\ (x-16-4\sqrt{5}) (x-16+4\sqrt{5}) =0,\\ x_1=16+4\sqrt{5} ,x_2=16-4\sqrt{5}.

Additional 7 step: check these solutions, substituting into the initial equation.

3 0
3 years ago
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