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jonny [76]
3 years ago
11

Rounding Numbers (Level 1) Jan 29, 4:29:32 PM Round 5.729 to the nearest tenth.

Mathematics
1 answer:
andrey2020 [161]3 years ago
4 0

Answer:i believe it is 5.7

Step-by-step explanation:

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4 years ago
Which statement describes the inverse of m(x) = x2 – 17x?
stealth61 [152]

Answer:

The correct option is;

The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}

Step-by-step explanation:

The given information is that m(x) = x² - 17·x

The above equation can be written in the form;

y = x² - 17·x

Therefore;

0 = x² - 17·x - y

From the general solution of a quadratic equation, 0 = a·x² + b·x + c we have;

x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}

By comparison to the equation,0 = x² - 17·x - y, we have;

a = 1, b = -17, and c = -y

Substituting the values of a, b and c into the formula for the general solution of a quadratic equation, we have;

x = \dfrac{-(-17)\pm \sqrt{(-17)^{2}-4\times (1) \times (-y)}}{2\times (1)} = \dfrac{17\pm \sqrt{289+4\cdot y}}{2}

Which can be simplified as follows;

x =  \dfrac{17\pm \sqrt{289+4\cdot y}}{2}= \dfrac{17}{2} \pm \dfrac{1}{2}  \times \sqrt{289+4\cdot y}} = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +\dfrac{4\cdot y}{4} }}

And further simplified as follows;

x = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +y }} = \dfrac{17}{2} \pm \sqrt{y + \dfrac{289}{4} }}

Interchanging x and y in the function of the inverse, m⁻¹(x), we have;

m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}

We note that the maximum or minimum point of the function, m(x) = x² - 17·x found by differentiating the function and equating the result to zero, gives;

m'(x) = 2·x - 17 = 0

x = 17/2

Similarly, the second derivative is taken to determine if the given point is a maximum or minimum point as follows;

m''(x) = 2 > 0, therefore, the point is a minimum point on the graph

Therefore, as x increases past the minimum point of 17/2, m⁻¹(x) increases to give;

The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }} to increase m⁻¹(x) above the minimum.

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In a 180-kilogram sample of ore, there was 3.2% metal. How many kilograms of metal were in the sample?
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Step-by-step explanation:


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