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HACTEHA [7]
3 years ago
6

Find three solutions of the equation y = 7x – 5.

Mathematics
1 answer:
soldi70 [24.7K]3 years ago
5 0
 y=7x-5 This is the same as saying y-7x = -5 Well easy. You just make up numbers that make this work. Let x be 1 y-7=-5 y=-5+7 y=2 So one ordered pair is (1,2) Let x=2 y-14=-5 y=9 So another ordered pair is (2,9) Let x=3 y-21=-5 y=-5+21 y=16 So another ordered pair is (3,16) 
hope it helps
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7 0
3 years ago
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

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

If we want f(x) to be continuous the denominator needs to be different to 0, otherwise f(x) will be indeterminate.

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:

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

so, the value x=\sqrt{a} = \sqrt{4} = 2 won't annulate the denominator.

In conclusion, for a=1 or a=1, the function will be continuos for all real numbers, since the denominator will never be 0.

4 0
3 years ago
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