Mathematics College

Find x:




X=________________

Find X:X=________________

Answers

Answer 1

Assuming [tex]\overline{BC}[/tex] is a diameter, minor arc [tex]AB[/tex] has a measure of [tex]180^{\circ}-108^{\circ}=72^{\circ}[/tex].

Using the inscribed angle theorem, [tex]36=x+3 \implies x=\boxed{33}[/tex].

Related Questions

The graph shows g(x), which is a translation of f(x)=|x|. Write the function rule for g(x).
Write your answer in the form a|x-h| + k, where a, h, and k are integers or simplified fractions.
g(x)=____

Answers

From the graph it can be calculated that Function rule for g(x) g(x) = |x - 7|

What is graph of a function?

The collection of ordered pairs with the (x,y)(x,y) sign is known as the graph of the a function in mathematics, where f(x)=y. x and f seem to be Cartesian coordinates of pts in two-dimensional space, and in the typical case where they are real numbers, they form a subset of the this plane.

Instead of the pairs "((x,y),z)" and "((x,y),z)" as in the definition above, the graph for functions of two variables, i.e. functions for whom the domain consists of pairs "(x,y)," "(x,y)," is typically the set of ordered triples "(x,y,z)" and "(x,y,z)" where "A surface for a constant real-valued function of the two real variables, this set is indeed a sub - set of three-dimensional space.

Graph of a function f(x) is the collection of points of the form (x, f(x))The given graph is a translation of |x| only,
So value of a is 1
Now since the tip of the graph lies on x- axis, k = 0So the value of g(x) is
g(x) = |x - h|g(7) = 0|7 - h| = 07 - h = 0h = 7
Function rule for g(x) g(x) = = |x - 7|

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How do you find the equidistant point of a triangle?

Answers

The equidistant point of  the triangle is point O , which is shown below.

In the question ,

we have to find the equidistant point in the triangle ,

We know that the equidistant point from the vertices of the triangle is the circumcenter ,

let the circumcenter be point O .

The perpendicular bisectors of the given triangle ABC are AC and BC that intersect at point O .

We have to prove that point O lies on the perpendicular bisector of AB and it is equidistant from A , B and C .

So ,we draw OA ,OB and OC .

we know that any point on perpendicular bisector of a segment is equidistant from endpoints of the segment.

that means , OA = OC and OC = OB .

which means OA = OB .

we also know that any point that is equidistant from end points of a segment lies on its perpendicular bisector.

So, we can say that point O is on the perpendicular bisector of AB .

and

Since OA = OB = OC , the point O is equidistant from A , B and C .

Therefore , the equidistant point of the triangle is the circumcenter (point O) .

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