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Rufina [12.5K]
3 years ago
9

What's the slope of the line containing the pair of points (9,-7) and (9,-10)

Mathematics
1 answer:
Contact [7]3 years ago
4 0
\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~{{ 9}} &,&{{ -7}}~) 
%  (c,d)
&&(~{{ 9}} &,&{{ -10}}~)
\end{array}
\\\\\\
% slope  = m
slope \implies 
\cfrac{\stackrel{rise}{{{ y_2}}-{{ y_1}}}}{\stackrel{run}{{{ x_2}}-{{ x_1}}}}\implies \cfrac{-10-(-7)}{9-9}\implies \cfrac{-10+7}{9-9}\implies \stackrel{und efined}{\cfrac{-3}{0}}

with an undefined slope, that means is a vertical line, check the picture below.

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Answer:

A≅562.45π u²

Step-by-step explanation:

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evaluated (60,77), then

A=\frac{8\pi }{3}(77^{3/2}-60^{3/2})\\A=\frac{8\pi }{3}(675.67-161.75})\\A=562.45\pi

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Is 4x+5<2 an equation give reason
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No

Step-by-step explanation:

This is not an equation since it does not contain an equal (=) sign and instead contains a less than (<) sign making it an inequality.

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3 years ago
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Which of the following steps is needed to inscribe a circle in a triangle?
xxTIMURxx [149]

Construct two angle bisectors of the triangle

Answer:

Option (D) is correct

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

We have to contruct two angle bisectors of the triangle.

various steps to inscribe the circle in a triangle are

  1. First draw a triangle
  2. Draw angle bisector of one angle
  3. Draw angle bisector of another angle
  4. Extent these angle bisectors, where they meet that point will be called the incenter
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6 0
3 years ago
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which of the following is the equation of a line in slope-intercept from for a line with slope =2 and y -intercept at (0 3?
serg [7]

Answer:

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Step-by-step explanation:

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6 0
3 years ago
A particle moves along a line with acceleration a (t) = -1/(t+2)2 ft/sec2. Find the distance traveled by the particle during the
amm1812

Answer:

s(t)=(ln|7|+ln|2|)\,ft

Step-by-step explanation:

Acceleration is second derivative of distance and are related as:

a(t)=\frac{d^2s}{dt^2}\\\\\frac{d^2s}{dt^2}=\frac{-1}{(t+2)^2}\\

Integrating both sides w.r.to t

v(t)=\frac{ds}{dt}=\frac{1}{t+2} +C\\

Using initial value

v(0)=\frac{1}{2}\\\\\frac{1}{2}=\frac{1}{0+2} +C\\\\C=0\\\\\frac{ds}{dt}=\frac{1}{t+2}

We have to calculate the distance covered in time interval [0,5], so:

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3 years ago
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