Which equations represent the line that is perpendicular to the line 5x - 2y = -6 and passes through the point

Mathematics

1 answer:

Vinvika [58]3 years ago

6 0

<u>Answer</u>

See below

Step-by-step explanation:

The given line has equation

$5x - 2y = - 6$

The slope intercept form is

$y = \frac{5}{2} x + 3$

The slope of this line is

$m = \frac{5}{2}$

The line perpendicular to this line has slope

$- \frac{2}{5}$

If this line passes through (5,-4), the point-slope form is

$y + 4 = - \frac{2}{5} (x - 5)$

The slope-intercept form is

$y = - \frac{2}{5} x- 2$

The standard form is

2x+5y=-10

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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:

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