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4 years ago

The given line segment has a midpoint at (-1, -2).

Mathematics

1 answer:

madreJ [45]4 years ago

8 0

Answer:

y = -4x -6

Step-by-step explanation:

The given segment has a rise if 1 for a run of 4, so a slope of ...

m = rise/run = 1/4

The desired perpendicular has a slope that is the negative reciprocal of this:

m = -1/(1/4) = -4

A point that has a rise of -4 for a run of 1 from the given midpoint is ...

(-1, -2) +(1, -4) = (0, -6) . . . . . . . the y-intercept of the bisector

So, our perpendicular bisector has a slope of m=-4 and a y-intercept of b=-6. Putting these in the slope-intercept form equation, we find the line to be ...

y = mx +b

y = -4x -6

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3 years ago

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sladkih [1.3K]

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3 years ago

Suppose the population of a certain city is 4164 thousand. It is expected to decrease to 3460 thousand in 50 years. Find the per

lilavasa [31]

Answer:

16.9%

Step-by-step explanation:

Given data

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3 years ago

Can someone please help me with this exercise? I'm having problems with point B.

Margaret [11]

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4 years ago

Find the derivative of <img src="https://tex.z-dn.net/?f=tan%5E%7B-1%7D%20x" id="TexFormula1" title="tan^{-1} x" alt="tan^{-1} x

sladkih [1.3K]

$\huge{\color{magenta}{\fbox{\textsf{\textbf{Answer}}}}}$

$\frak {\huge{ \frac{1}{1 + {x}^{2} } }}$

Step-by-step explanation:

$\sf let \: f(x) = { \tan }^{ - 1} x \\ \\ \sf f(x + h) = { \tan}^{ - 1} (x + h)$

$\sf f'(x) = \frac{f(x+h) - f(x) }{h}$

$\sf \implies \lim_{ h \to 0 } \frac{ { \tan }^{ - 1}(x + h) - { \tan}^{ - 1}x }{h} \\ \\ \\ \sf \implies \lim_ {h \to 0} \frac{ { \tan}^{ - 1} \frac{x + h - x}{1 + (x + h)x} }{h}$

By using

$\sf { \tan}^{ - 1} x - { \tan}^{ - 1} y = \\ \sf { \tan}^{ - 1} \frac{x - y}{1 + xy} formula$

$\sf \implies \large \lim_{h \to0 } \frac{ { \tan}^{ - 1} \frac{h}{1 + hx + {x}^{2} } }{h} \\ \\ \\ \sf \implies \large{\lim_{h \to0} } \frac{ { \tan}^{ - 1} \frac{h}{1 + hx + {x}^{2} } }{ \frac{h}{1 + hx + {x}^{2} } \times (1 + hx + {x}^{2} )} \\ \\ \\ \sf \implies \large \lim_{h \to0} \frac{ { \tan}^{ - 1} \frac{h}{1 + hx + {x}^{2} } }{ \frac{h}{1 + hx + {x}^{2} } } + \lim_{h \to0} \frac{1}{1 + hx + {x}^{2} }$

Now putting the value of h = 0

$\sf \large \implies 0 + \frac{1}{1 + 0 + {x}^{2} } \\ \\ \\ \purple{ \boxed { \implies \frac{1}{1 + {x}^{2} } }}$

6 0

2 years ago