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vfiekz [6]
3 years ago
11

WHAT IS HELP PLEASE 6x2 -2x3 +x4 =? DONT ANSWER TO GET QUICK POINTS

Mathematics
2 answers:
Nuetrik [128]3 years ago
8 0
The correct answer is / =10x
blondinia [14]3 years ago
3 0

Answer:

i told you in the comment it equals 7

Step-by-step explanation:

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aalyn [17]
Wym by A B C‍♀️I'm not seeing a question
3 0
2 years ago
Need help on this as soon as possibke
miss Akunina [59]
LA = perimeter of base x height

Use Pythagorean theorem to find the base

4^2 + 6^2 = 52
Sq root of 52 is 7.2111...

4+6+7.2 = 17.2

17.2 x 8

137.6 is your answer






3 0
3 years ago
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
2 years ago
Find the value of x in the figure given that SU = 133.<br><br> A) 7<br> B) 9<br> C) 11<br> D) 8
Alekssandra [29.7K]

Answer:

The answer is 9 (B)

Step-by-step explanation:

I got 9 because you have to plug in answer (7, 9, 11, 8) in each equation than add them both together. I first plugged in 7 for x in the equation 9x-5. I got 58. I plugged in 7 for x in the equation 5x+12 and got 47. When I added both the answers together I got 105. The answer was to low because SU has to be 133. I plugged in 9 for x in both equations then added them both and got 133 for SU. I hope this helps.

4 0
2 years ago
Read 2 more answers
Answer asap
balandron [24]

Answer:

A

Step-by-step explanation:

went up a positive 4 points:)

7 0
3 years ago
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