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kogti [31]
3 years ago
10

Angels Inscribed and circumscribed.

Mathematics
1 answer:
zhenek [66]3 years ago
7 0

Remark

The diameter of the circle is AC. The hypotenuse of the right triangle is also AC. This assumes that A,B and C all lie on the circumference of the circle.

Answers

One

The center of the circle is on the diameter of the circle. In fact, the center bisects the diameter. So the center lies on AC and if the center is labeled O then AO = BO.

<em><u>Answer 1: </u></em> AC

Two

The answer is the single word diameter.

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Step-by-step explanation:

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3 years ago
How many different triangles can be constructed with angle measurements 90°, 45°, 50°?
melisa1 [442]
The angles of a triangle always add up to 180°

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4 years ago
Pls answer asap!!ty!!<br> Please label the missing angles below.
marusya05 [52]

Answer:

Triangle A: 38 degrees

Triangle B: Unknown (not enough information)

Triangle C: Unknown (not enough information)

Triangle D: 70 degrees

Triangle E: 40 degrees

Step-by-step explanation:

Work for Triangle A: 90 + 52 = 142. 180 - 142 = 38.

Work for Triangle B: Unidentifiable because there is no indicator to tell you if any of the angles/lines are equal. Generally there will be a "double lined" indicator in the corners of which a triangles angles are equal.

Work for Triangle C: Same as B.

Work for Triangle D: 90 + 20 = 110. 180 - 110 = 70.

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7 0
3 years ago
Find a set of vectors {u⃗ ,v⃗ }{u→,v→} in r4r4 that spans the solution set of the equations {w−x+2y+4z5w+2x−y+3z==0,0. {w−x+2y+4
rusak2 [61]
Given

w-x+2y+4z=0 \\ 5w+2x-y+3z=0

We can rewrite it in matrix form as:

\left[\begin{array}{cccc}1&-1&2&4\\5&2&-1&3\\0&0&0&0\\0&0&0&0\end{array}\right]   \left[\begin{array}{c}w\\x\\y\\z\end{array}\right] =\left[\begin{array}{c}0\\0\\0\\0\end{array}\right] \\  \\ \left[\begin{array}{cccc}1&-1&2&4\\5&2&-1&3\\0&0&0&0\\0&0&0&0\end{array}\right|\left.\begin{array}{c}0\\0\\0\\0\end{array}\right]\ \ \ \ \ -5R_1+R_2\rightarrow R_2
\left[\begin{array}{cccc}1&-1&2&4\\0&7&-11&-17\\0&0&0&0\\0&0&0&0\end{array}\right|\left.\begin{array}{c}0\\0\\0\\0\end{array}\right]\ \ \ \ \  \frac{1}{7} R_2\rightarrow R_2 \\  \\ \left[\begin{array}{cccc}1&-1&2&4\\0&1&-\frac{11}{7}&-\frac{17}{7}\\0&0&0&0\\0&0&0&0\end{array}\right|\left.\begin{array}{c}0\\0\\0\\0\end{array}\right]\ \ \ \ \ R_1+R_2\rightarrow R_1
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Thus, the solution set is a span of \{\left\ \textless \ -\frac{3}{7},\frac{11}{7}1,0\right\ \textgreater \ ,\left\ \textless \ - \frac{11}{7},\frac{17}{7},0,1\right\ \textgreater \ \}
7 0
3 years ago
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Answer:

The requires solution to the given differential equation is

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Step-by-step explanation:

The workings are shown in the attachment.

8 0
4 years ago
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