Use iteration to determine the real zeros, the the nearest tenth.

Mathematics

1 answer:

FinnZ [79.3K]3 years ago

5 0

I plugged the equation in to my graphing calculator and found the zeros at (-2.7,0) and (2.7,0)

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9 + 5m = 5m - 1 Please help! Thanks :))

vfiekz [6]

Answer:

No solution

Step-by-step explanation:

In pic

(Credits: Symbolab)

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Vinil7 [7]

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larisa86 [58]

We know that
1) Luis bought stock at ---------------> $66.80
2) The next day, the price increased $13.85--------> $66.80+$13.85=$80.65
3) This new price changed by −2 3 /4 % the following day
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the answer is
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3 years ago

Identify the center and the radius of a circle that has a diameter with endpoints at (−5, 9) and (3, 5)

marusya05 [52]

Check the picture below, so the circle looks more or less like that one.

well, the center of it is simply the Midpoint of those two points, and its radius is simply half-the-distance between them.

$~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-5}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{5}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 3 -5}{2}~~~ ,~~~ \cfrac{ 5 + 9}{2} \right)\implies \left( \cfrac{-2}{2}~~,~~\cfrac{14}{2} \right)\implies \stackrel{center}{(-1~~,~~7)} \\\\[-0.35em] ~\dotfill$

$~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-5}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[3 - (-5)]^2 + [5 - 9]^2}\implies d=\sqrt{(3+5)^2+(-4)^2} \\\\\\ d=\sqrt{8^2+16}\implies d=\sqrt{80}\implies d=4\sqrt{5}~\hfill \stackrel{\textit{half the diameter}}{\cfrac{4\sqrt{5}}{2}\implies \underset{radius}{2\sqrt{5}}}$

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2 years ago