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saul85 [17]
2 years ago
15

Which calculation results in the best estimate of 148% of 203? (200) (140 percent) = (200) (100 percent + 40 percent) = (200) (1

00 percent) + (200) (40 percent) = (200) (1.0) + (200) (0.4) = 200 + 80 = 280 (210) (140 percent) = (210) (100 percent + 40 percent) = (210) (100 percent) + (210) (40 percent) = (210) (1.0) + (210) (0.4) = 210 + 84 = 294 (200) (150 percent) = (200) (100 percent + 50 percent) = (200) (100 percent) + (200) (50 percent) = (200) (1) + (200) (one-half) = 200 + 100 + 300 (210) (150 percent) = (210) (100 percent + 50 percent) = (210) (100 percent) + (210) (50 percent) = (210) (1) + (210) (one-half) = 210 + 105 = 315
Mathematics
2 answers:
nikitadnepr [17]2 years ago
6 0

✅The answer is C✅

IamSugarBee

frez [133]2 years ago
5 0

Answer:

the best calculation for an estimate would be to do 150% of 203 which is 300

Step-by-step explanation:

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so we have three points, A, B and C, if indeed AC is the diameter of the circle, then half the distance of AC is its radius, and the midpoint of AC is the center of the circle, morever, since B is also on the circle, the distance from B to the center must be the same radius distance.

in short, half the distance of AC must be equals to the distance of B to the midpoint of AC, if indeed AC is the diameter.

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad C(\stackrel{x_2}{12}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{12+7}{2}~~,~~\cfrac{3+4}{2} \right)\implies \left( \cfrac{19}{2}~~,~~\cfrac{7}{2} \right)=M\impliedby \textit{center of the circle}

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\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AM=\sqrt{\left( \frac{19}{2}-7 \right)^2+\left( \frac{7}{2}-4 \right)^2} \\\\\\ AM=\sqrt{\left( \frac{5}{2}\right)^2+\left( -\frac{1}{2} \right)^2}\implies \boxed{AM\approx 2.549509756796392} \\\\[-0.35em] ~\dotfill

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{10}~,~\stackrel{y_1}{6})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}}) \\\\\\ BM=\sqrt{\left( \frac{19}{2}-10 \right)^2+\left( \frac{7}{2}-6 \right)^2} \\\\\\ BM=\sqrt{\left( -\frac{1}{2}\right)^2+\left( -\frac{5}{2} \right)^2}\implies \boxed{BM\approx 2.549509756796392}

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Answer:

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Step-by-step explanation:

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