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3 years ago

Please help ! I’ll mark you as brainliest if correct

Mathematics

1 answer:

Nata [24]3 years ago

3 0

Answer:

Step-by-step explanation:

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I don't know how to do the questions on the page can you help me

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3 years ago

If 12x=4(x+3) then x equals

Delvig [45]

3x=x+3 2x=3 x=3/2 Move x terms to one side, then solve

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3 years ago

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I need help <br><br> 50/50+40-60

natulia [17]

50 divided by 50= 1
1 plus 40= 41
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So your final answer is -19.
Hope it helps.

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3 years ago

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Drag the point around to graph more horizontal lines. What is the general equation of a horizontalline?

svetlana [45]

A horizontal line is one for which the value of y is the same for the entire length of the line. Therefore this type of line can be expressed as below:

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2 years ago

Use f’( x ) = lim With h ---> 0 [f( x + h ) - f ( x )]/h to find the derivative at x for the given function. 5-x²

beks73 [17]

<h2>Answer:</h2>

The derivative of the function f(x) is:

$f'(x)=-2x$

<h2>Step-by-step explanation:</h2>

We are given a function f(x) as:

$f(x)=5-x^2$

We have:

$f(x+h)=5-(x+h)^2\\\\i.e.\\\\f(x+h)=5-(x^2+h^2+2xh)$

( Since,

$(a+b)^2=a^2+b^2+2ab$ )

Hence, we get:

$f(x+h)=5-x^2-h^2-2xh$

Also, by using the definition of f'(x) i.e.

$f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$

Hence, on putting the value in the formula:

$f'(x)= \lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-(5-x^2)}{h}\\\\\\f'(x)=\lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-5+x^2}{h}\\\\i.e.\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2-2xh}{h}\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2}{h}+\dfrac{-2xh}{h}\\\\f'(x)=\lim_{h \to 0} -h-2x\\\\i.e.\ on\ putting\ the\ limit\ we\ obtain:\\\\f'(x)=-2x$

Hence, the derivative of the function f(x) is:

$f'(x)=-2x$

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3 years ago

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