See picture^ Can someone pls help with this question, easy points.

Mathematics

2 answers:

gavmur [86]3 years ago

4 0

Answer:

Given below

Step-by-step explanation:

Jose's method

Distance on bus: 2.85 x 1000= 2850m

Total distance= 320m + 2850m

= 3170 m

Pam's method

Distance to bus stop = 320 ÷ 100 = 0.32 km

Total distance= 0.32 km + 2.85 km

= 3.17 km

ioda3 years ago

3 0

Answer with step by step explanation

Joe's method

$distance \: \: \: on \: \: bus \\ = 2.85 \times 1000 = 2850m \\ total \: \: distance \\ = 320m + 2850m \\ = 3170m$

Pam's method

$320 \div 1000 = 0.32km \\ total \: \: \: distance \\ = 0.32km + 2.85km \\ = 3.17km$

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A circle passes through points A(7,4), B(10,6), C(12,3). Show that AC must be the diameter of the circle.

Artist 52 [7]

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.

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

now, let's check the distance from say A to the center, and check the distance of B to the center, if it's indeed the center, they'll be the same and thus AC its diameter.

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

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