1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Likurg_2 [28]
2 years ago
7

Could use some help with this math xl problem ​

Mathematics
1 answer:
noname [10]2 years ago
8 0
Make the picture bigger, can barely see it
You might be interested in
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}

6 0
3 years ago
I need help on 4 and 5 thanks for your help and time
DanielleElmas [232]
The answer for 4. | She rode her bike 5 and a half miles and 6 miles yesterday so the longer ride is yesterday
6 > 5'1/2
  (I do not know 5. sorry.)
 I hope this helps
Have a great day!
4 0
3 years ago
A cricketer's average score for 6 innings is 12 runs. what must he score in his next innings to make his average 13?
Inessa05 [86]
The answer is 19 since 19-6
8 0
3 years ago
The graph shows the number of Calories burned running. How many Calories do you burn per​ minute?
zlopas [31]
11.2 you could do 112/10 or 22.4/2.
3 0
3 years ago
Find the area of the shaded region. Round your answer to the nearest hundredth.
valina [46]

Answer:

  • FIRSTLY GET THE AREA OF THE WHOLE CIRCLE BY USING FORMULA
  • pi \times r ^2
  • THEN AFTER THAT DIVIDE THE AREA BY 2
  • AFTER THAT UR ANSWER WILL COME

<em>PLEASE</em><em> </em><em>MARK ME AS BRAINLIEST</em><em> </em>

3 0
3 years ago
Read 2 more answers
Other questions:
  • How do you solve <br> -7x+(-21)=41
    11·1 answer
  • Help me please please with these problem
    14·2 answers
  • What possible restrictions exist on domains and ranges of arithmetic sequences?
    10·2 answers
  • Will mark as brainliest to whoever first answers! Please help!
    13·2 answers
  • Need help<br><br> 66 hundreds ÷ 10=?
    13·1 answer
  • Simplify 7x 42 / x^2 13x 42
    5·2 answers
  • Whats an equivalent ratio for 24:56
    5·1 answer
  • PLEAS HELP ME I BEG ONYL 4 QUESTIONS PLEASE PLEASE
    12·1 answer
  • The extremes in the proportion 3/4=15/20 are
    8·2 answers
  • 2.
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!