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1 year ago

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Circle T is shown. Line segment Q S is a diameters. Point R is opposite of point T. Lines are drawn from points Q and S to point

R to form a right triangle. Angle Q R S is a right angle.
The measure of arc QS is (4x – 18)°.

What is the value of x?
40.5
49.5
94.5
180

1 answer:

Juliette [100K]1 year ago

8 0

The value of angle x of the given cyclic segment is; 49.5°

<h3>How to find the angle of an arc?</h3>

We are given the measure of the angle of arc QS as (4x – 18)°

Now, to find the measure of arc QS, this angle is to be equal to 180° and as such;

Thus;

(4x – 18)° = 180°

4x - 18 = 180

4x = 180 + 18

4x = 198

x = 198/4

x = 49.5°

The angle subtended by the arc at the center of a circle with center C is the angle of the arc. It is denoted by. m AB, where A and B are the endpoints of the arc. With the help of the arc length formula, we can find the measure of arc angle.

The formula to measure the length of the arc is;

Arc Length Formula (if angle θ is in degrees); s = 2πr (θ/360°)

Arc Length Formula (if θ is in radians) s = ϴ × r.

Thus, the value of x of the given cyclic segment is; 49.5°

Read more about arc angle at; brainly.com/question/2005046

#SPJ1

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alisha [4.7K]

Answer:

y+9=1/5(x+3)

Step-by-step explanation:

Point slope formula: y-y1=m(x-x1)

slope=m

m=1/5

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y-(-9) =1/5(x-(-3))

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3 years ago

Check whether the relation R on the set S = {1, 2, 3} is an equivalent

kozerog [31]

Answer:

$R$ isn't an equivalence relation. It is reflexive but neither symmetric nor transitive.

Step-by-step explanation:

Let $S$ denote a set of elements. $S \times S$ would denote the set of all ordered pairs of elements of $S\!$ .

For example, with $S = \lbrace 1,\, 2,\, 3 \rbrace$ , $(3,\, 2)$ and $(2,\, 3)$ are both members of $S \times S$ . However, $(3,\, 2) \ne (2,\, 3)$ because the pairs are ordered.

A relation $R$ on $S\!$ is a subset of $S \times S$ . For any two elements $a,\, b \in S$ , $a \sim b$ if and only if the ordered pair $(a,\, b)$ is in $R\!$ .

A relation $R$ on set $S$ is an equivalence relation if it satisfies the following:

Reflexivity: for any $a \in S$ , the relation $R$ needs to ensure that $a \sim a$ (that is: $(a,\, a) \in R$ .)

Symmetry: for any $a,\, b \in S$ , $a \sim b$ if and only if $b \sim a$ . In other words, either both $(a,\, b)$ and $(b,\, a)$ are in $R$ , or neither is in $R\!$ .

Transitivity: for any $a,\, b,\, c \in S$ , if $a \sim b$ and $b \sim c$ , then $a \sim c$ . In other words, if $(a,\, b)$ and $(b,\, c)$ are both in $R$ , then $(a,\, c)$ also needs to be in $R\!$ .

The relation $R$ (on $S = \lbrace 1,\, 2,\, 3 \rbrace$ ) in this question is indeed reflexive. $(1,\, 1)$ , $(2,\, 2)$ , and $(3,\, 3)$ (one pair for each element of $S$ ) are all elements of $R\!$ .

$R$ isn't symmetric. $(2,\, 3) \in R$ but $(3,\, 2) \not \in R$ (the pairs in $\! R$ are all ordered.) In other words, $3$ isn't equivalent to $2$ under $R\!$ even though $2 \sim 3$ .

Neither is $R$ transitive. $(3,\, 1) \in R$ and $(1,\, 2) \in R$ . However, $(3,\, 2) \not \in R$ . In other words, under relation $R\!$ , $3 \sim 1$ and $1 \sim 2$ does not imply $3 \sim 2$ .

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