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Alenkinab [10]
4 years ago
8

Help please I need awenser

Mathematics
1 answer:
Bumek [7]4 years ago
7 0
It affect the sum by being the higher number
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If possible, give an example or if not possible, explain why<br>a whole number less than 0​
aliya0001 [1]

Answer:

Not possible

Step-by-step explanation:

By definition, all whole numbers are non-negative.

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2 years ago
How do you evaluate 3ab
dimaraw [331]
Lets say a=2 and b=5  you now plug in you numbers to get 3x2x5 we know 3x2=6 so now take 6 and multiply it by 5 so 6x5=30  you can also set this equal to 0 3ab=0 then divide by either a or b ill show you both if you divide by a you are left with 3b=a if you divide by b than you get 3a=b
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3 years ago
How do you convert improper fractions to proper fractions
Jobisdone [24]

Answer:

Step- for example, 27/6. The fraction bar means you need to divide 27 by 6.

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5 0
3 years ago
Read 2 more answers
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.

8 0
3 years ago
Someone please help me with this! Yes
Snowcat [4.5K]
The answer is letter d
5 0
3 years ago
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