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3 years ago

State how many imaginary and real zeros the function has.

Mathematics

1 answer:

Klio2033 [76]3 years ago

8 0

Answer:

2 imaginary; 2 real

Step-by-step explanation:

You can factor it as ...

f(x) = (x^2 -16)(x^2 +1) . . . . . . . . . x^2-16 is the difference of squares

= (x -4)(x +4)(x^2 +1)

The two linear factors have real zeros; the quadratic factor has two imaginary zeros. There are 2 imaginary and 2 real zeros.

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F(x) = 4/x-9<br> g(x)=4/x + 9<br> find f(g(x))

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Hello,

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3 years ago

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3 years ago

The GCD(a, b) = 9, the LCM(a, b)=378. Find the least possible value of a+b.

denis-greek [22]

$\mathrm{gcd}(a,b)=9\implies9\mid a\text{ and }9\mid b\implies9\mid a+b$

which means there is some integer $k$ for which $a+b=9k$ .

Because $9\mid a$ and $9\mid b$ , there are integers $n_1,n_2$ such that $a=9n_1$ and $b=9n_2$ , and

$\mathrm{lcm}(a,b)=\mathrm{lcm}(9n_1,9n_2)=9\mathrm{lcm}(n_1,n_2)=378\implies\mathrm{lcm}(n_1,n_2)=42$

We have $42=2\cdot3\cdot7$ , which means there are four possible choices of $n_1,n_2$ :

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which is to say there are also four corresponding choices for $a,b$ :

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18, 189
27, 126
54, 63

whose sums are:

387
207
153
117

So the least possible value of $a+b$ is 117.

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Answer:

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Step-by-step explanation:

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2 years ago

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