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Katen [24]
3 years ago
10

I already named the shape I just need help solving for x​

Mathematics
1 answer:
Art [367]3 years ago
6 0

Answer:

3 =x

Step-by-step explanation:

The sides are all equal length

x+5 = 3x-1

Subtract x from each side

x+5 -x = 3x-1 -x

5 = 2x-1

Add 1 to each side

5+1 = 2x-1+1

6 = 2x

Divide by 2

6/2 = 2x/2

3 =x

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Hope this helps!

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Let f be the function given by f(x)= (x-1)(x^2-4)/x^2-a. For what positive values of a is f continuous for all real numbers x?
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Answer:

a =1 and a=4.

Step-by-step explanation:

The function is

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Now, for a a positive real we have that x=\sqrt{a} will annulate the denominator, i.e

(\sqrt{a})^2-a = a-a = 0. But, if a = 1 we have:

f(x)=\frac{(x-1)(x^2-4)}{x^2-1} = \frac{(x-1)(x^2-4)}{(x-1)(x+1)}=\frac{(x^2-4)}{x+1}

so, the value x=\sqrt{a} = \sqrt{1} = 1 won't annulate the denominator.

Now, for a = 4 we have:

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4 0
3 years ago
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Answer:

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3 years ago
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