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mihalych1998 [28]
3 years ago
11

Unit Test Unit Test Active 1 2 3 4 5 6 7 CO Given fix) = 17- x2, what is the average rate of change in f(x) over the interval [1

, 5]? -6, -1/2, 1/4, 1​
Mathematics
1 answer:
sveticcg [70]3 years ago
6 0

Answer:

Step-by-step explanation:

Average rate of change is the same thing as the slope. This is a quadratic so the slope is not something that is constant like it is in a line. Here you have the x coordinates of 1 and 5; each one of these x coordinates has a y coordinate that goes with it. If we draw a line from one of these coordinates to another,  that line will have a slope. That is the slope we are trying to find. Thus, we need the y coordinates that go with each of these x coordinates. To do that, plug x into the equation and do the math to find y:

Let's start with x = 1.

f(1) = 17-(1)^2 so

f(1) = 16 and the coordinate is (1, 16).

f(5) = 17-(5)^2 so

f(5) = -8 and the coordinate is (5, -8). Now we apply the slope formula:

m=\frac{-8-16}{5-1}=\frac{-24}{4}=-6 So the answer is -6.

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Answer:

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1 year ago
What are the solutions of x^2-2x+5=0
sweet [91]

The solution of x^{2}-2 x+5=0 are 1 + 2i and 1 – 2i

<u>Solution:</u>

Given, equation is x^{2}-2 x+5=0

We have to find the roots of the given quadratic equation

Now, let us use the quadratic formula

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}  --- (1)

<em><u>Let us determine the nature of roots:</u></em>

Here in x^{2}-2 x+5=0 a = 1 ; b = -2 ; c = 5

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Now plug in values in eqn 1, we get,

x=\frac{-(-2) \pm \sqrt{(-2)^{2}-4 \times 1 \times 5}}{2 \times 1}

On solving we get,

x=\frac{2 \pm \sqrt{4-20}}{2}

x=\frac{2 \pm \sqrt{-16}}{2}

x=\frac{2 \pm \sqrt{16} \times \sqrt{-1}}{2}

we know that square root of -1 is "i" which is a complex number

\begin{array}{l}{\mathrm{x}=\frac{2 \pm 4 i}{2}} \\\\ {\mathrm{x}=1 \pm 2 i}\end{array}

Hence, the roots of the given quadratic equation are 1 + 2i and 1 – 2i

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