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KatRina [158]
2 years ago
5

Jace wants to save $260 to buy a bird bath. He has already saved $178, how much more does he need to save? Write an equation and

solve.
Do I subtract multiply, or add or divide?
Mathematics
2 answers:
Nat2105 [25]2 years ago
6 0
You subtract “260-178” and then you’ll get your answer i’m pretty sure
IRINA_888 [86]2 years ago
5 0
You have to subtract 178 from 260 to get your answer that equals 82.
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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.
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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}

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}

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