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Liula [17]
2 years ago
10

What is the minimum value for g(x)=x2−10x 16? enter your answer in the box.

Mathematics
1 answer:
motikmotik2 years ago
3 0

The minimum value for g(x)=x² - 10x + 16 is -9

<h3>How to determine the minimum value?</h3>

The function is given as:

g(x)=x² - 10x + 16

Differentiate the function

g'(x) = 2x - 10

Set the function 0

2x - 10 = 0

Add 10 to both sides

2x = 10

Divide by 2

x = 5

Substitute 5 for x in g(x)

g(5)=5² - 10*5 + 16

Evaluate

g(5) = -9

Hence, the minimum value for g(x)=x² - 10x + 16 is -9

Read more about quadratic functions at:

brainly.com/question/7784687

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Starting more simply, if we wanted to know how many students like pink in general, that's 68/100. We could do that for each single category and the fractions would add together to equal 1. Now say we wanted to know something about that 68/100 people. That 68 is our new 100%, or another way of looking at it is if we take however many people like pink and don't like black and those that do like black, they will equal 68/68.

The number of people that like pink but don't like black is 41/68 and those that like pink and black are 27/68. 27+41=68 For the question of your problem it is asking about those that do not like pink which you can tell from the table or use from my saying 68/100 like pink is 32. Now you can split that into those that do or don't like black, and the two results will equal 32/32.

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How can i differentiate this equation?
Dmitry_Shevchenko [17]

\bf y=\cfrac{2x^2-10x}{\sqrt{x}}\implies y=\cfrac{2x^2-10x}{x^{\frac{1}{2}}} \\\\\\ \cfrac{dy}{dx}=\stackrel{\textit{quotient rule}}{\cfrac{(4x-10)(\sqrt{x})~~-~~(2x^2-10x)\left( \frac{1}{2}x^{-\frac{1}{2}} \right)}{\left( x^{\frac{1}{2}} \right)^2}} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~(2x^2-10x)\left( \frac{1}{2\sqrt{x}} \right)}{\left( x^{\frac{1}{2}} \right)^2} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~\left( \frac{2x^2-10x}{2\sqrt{x}} \right)}{x}


\bf\cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~\left( \frac{2x^2-10x}{2\sqrt{x}} \right)}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{ \frac{(4x-10)(\sqrt{x})(2\sqrt{x})~~-~~(2x^2-10x)}{2\sqrt{x}}}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})(2\sqrt{x})~~-~~(2x^2-10x)}{2x\sqrt{x}}


\bf \cfrac{dy}{dx}=\cfrac{(4x-10)2x~~-~~(2x^2-10x)}{2x\sqrt{x}}\implies \cfrac{dy}{dx}=\cfrac{8x^2-20x~~-~~(2x^2-10x)}{2x\sqrt{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{8x^2-20x~~-~~2x^2+10x}{2x\sqrt{x}} \implies \cfrac{dy}{dx}=\cfrac{6x^2-10x}{2x\sqrt{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{2x(3x-5)}{2x\sqrt{x}}\implies \cfrac{dy}{dx}=\cfrac{3x-5}{\sqrt{x}}

8 0
3 years ago
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