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Alika [10]
3 years ago
7

Which theorem explains why the circumcenter is equidistant from the vertices of a triangle

Mathematics
2 answers:
vagabundo [1.1K]3 years ago
5 0
The circumcenter is the center of the circle which goes through the three vertices of the triangle. Recall that all radii of a circle are congruent, i.e. equal to one another. So this is why the circumcenter is equidistant from the vertices of the triangle. 
Brut [27]3 years ago
3 0

Answer:

The proof is explained in step-by-step explaination.

Step-by-step explanation:

Circumcenter is the point at which the perpendicular bisector of sides of triangle intersects inside the circle.

This points lies inside the triangle as well as circle and the vertices of triangles lies on the circle. As a result the distance from circumcenter and vertices is called to be radius of the circle which is always equidistant from the center.

Hence, circumcenter is equidistant from the vertices of a triangle.  

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Which expression represents the composition [g o f o h](x) for the functions below?
asambeis [7]

Answer: 16875x³-13500x²+3600x-320

Step-by-step explanation:

[gοfοh](x) means g(f(h(x))). So you plug in h(x) into f(x) and that into g(x).

f(3x)=5(3x)-4=15x-4

g(f(3x))=5(15x-4)³

g(f(3x))=5(3375x³-2700x²+720x-64)

g(f(3x))=16875x³-13500x²+3600x-320

5 0
4 years ago
What's the answer to this problem?
WITCHER [35]
The answer to this question is < (0.36)
8 0
4 years ago
Select all that apply.
Burka [1]

For this case we propose a rule of three to obtain the number that represents 25% of 37.

So, we have:

37 --------> 100%

x -----------> 25%

Where "x" is the number that represents 25% of 37.

x = \frac {25 * 37} {100}\\x = 9.25

Finally, we have that 25% of 37 is 9.25.

Answer:

25% of 37 is 9.25.

3 0
3 years ago
Suppose the integral from 2 to 8 of g of x, dx equals 12, and the integral from 6 to 8 of g of x, dx equals negative 3, find the
Artyom0805 [142]

\displaystyle\int_2^8g(x)\,\mathrm dx=12

\displaystyle\int_6^8g(x)\,\mathrm dx=-3

Use the fact that integrals are additive on their intervals. Mathematically, if c\in[a,b], then

\displaystyle\int_a^bg(x)\,\mathrm dx=\int_a^cg(x)\,\mathrm dx+\int_c^bg(x)\,\mathrm dx

So we have

\displaystyle2\int_2^6g(x)\,\mathrm dx=2\left(\int_2^8g(x)\,\mathrm dx-\int_6^8g(x)\,\mathrm dx\right)=2(12-(-3))=30

8 0
4 years ago
A website offers free shipping to customers who order at least $75 dollars worth of merchandise. Della is ordering 2 flashlights
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It’s A. 11 or more
YW
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3 years ago
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