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4 years ago

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Write the system of equations associated with the augmented matrix 1 0 0 10 0 1 0 2 0 0 1 4

1 answer:

alukav5142 [94]4 years ago

3 0

Try this solution:if given (1;0;0;10); (0;1;0;2) and (0;0;1;4), then the requirement system of equations is: $\left \{ \begin{array}{ccc}1*x_1+0*x_2+0*x_3=10\\0*x_1+1*x_2+0*x_3=2\\0*x_1+0*x_2+1*x_3=4 \end{array} \ =\ \textgreater \$ $\left \{ \begin{array}{ccc}x_1=10\\x_2=2\\x_3=4\end{array}$

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Consider the function f(x)=9-x^2/x^2-4 For which intervals is f(x) positive? Check ALL that apply.

solong [7]

Answer:

a. f (x) < 0 for x ∈ (-∞ ,-3)

b. f (x) > 0 for x ∈ (-3,-2)

c. f (x) < 0 for x ∈ (-2,2)

d. f (x) > 0 for x ∈ (2,3)

e.f (x) < 0 for x ∈ (3,∞)

Step-by-step explanation:

Here, the given function is: $f(x)= \frac{9-x^2}{x^2-4}$

Now, to check for the sign of f(x) at x = k, put the value of x from the given interval.

We get:

<u>a. (-infinity, -3) </u>

put k = -4 from the given interval

We get $f(-4)= \frac{9-(-4)^2}{(-4)^2-4} = \frac{9-16}{16-4} = \frac{-7}{12}$

⇒ f (x) < 0 for x ∈ (-∞ ,-3)

b. (-3, -2)

put k = -2.5 from the given interval

We get $f(-2.5)= \frac{9-(-2.5)^2}{(-2.5)^2-4} = \frac{9-6.25}{6.25-4} = \frac{2.75}{2.25} > 0$

⇒ f (x) > 0 for x ∈ (-3,-2)

c. (-2, 2)

put k = 0 from the given interval

We get $f(0)= \frac{9-(0)^2}{(0)^2-4} = \frac{9}{-4} = -\frac{9}{4} < 0$

⇒ f (x) < 0 for x ∈ (-2,2)

d. (2, 3)

put k =2.5 from the given interval

We get $f(2.5)= \frac{9-(2.5)^2}{(2.5)^2-4} = \frac{9-6.25}{6.25-4} = \frac{2.75}{2.25} > 0$

⇒ f (x) > 0 for x ∈ (2,3)

e. (infinity, 3)

put k = 4 from the given interval

We get $f(4)= \frac{9-(4)^2}{(4)^2-4} = \frac{9-16}{16-4} = \frac{-7}{12}$

⇒ f (x) < 0 for x ∈ (3,∞)

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4 years ago