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Rufina [12.5K]
3 years ago
8

Triangle SQT is isosceles. The measure of angle STQ is 48°. Triangle S Q T is cut by perpendicular bisector T R. The lengths of

line segments S R and R Q are congruent. Side lengths S T and T Q are congruent. Angles S T R and R T Q are congruent. What is the measure of Angle S T R? 24° 38° 48° 76°
Mathematics
2 answers:
Aleonysh [2.5K]3 years ago
7 0

Answer:

A.24

Step-by-step explanation:

igomit [66]3 years ago
3 0

Answer:

m\angle STR=24^o

Step-by-step explanation:

we know that

m\angle STQ=m\angle STR+m\angle RTQ ---> by addition angle postulate

we have

m\angle STQ=48^o ----> given problem

m\angle STR=m\angle RTQ ----> given problem

substitute in the expression above

48^o=2m\angle STR

Divide by 2 both sides

m\angle STR=24^o

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3 years ago
which a and b family both have 8 people in there family. The ages of each term is listed below which which state meant is correc
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4 0
3 years ago
Find the midpoint and distance between each pair of points, with easy explanation and work !! will give brainliest.
rewona [7]
If we have 2 coordinates say: (x1,y1) and (x2,y2)

Then the formula for the midpoint is:

((x1+x2)/2,(y1+y2)/2)

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The midpoint is:

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6 0
2 years ago
2/3x - 11 = x/3 + 3
Grace [21]
2
3

−
1
1
=

3
+
3
2
3
x
−
11
=
x
3
+
3
32​x−11=3x​+3
2

3
−
1
1
=

3
+
3
2
x
3
−
11
=
x
3
+
3
32x​−11=3x​+3
2
Find common denominator
2

3
−
1
1
=

3
+
3
2
x
3
−
11
=
x
3
+
3
32x​−11=3x​+3
2

3
+
3
(
−
1
1
)
3
=

3
+
3
2
x
3
+
3
(
−
11
)
3
=
x
3
+
3
32x​+33(−11)​=3x​+3
3
Combine fractions with common denominator
2

3
+
3
(
−
1
1
)
3
=

3
+
3
2
x
3
+
3
(
−
11
)
3
=
x
3
+
3
32x​+33(−11)​=3x​+3
2

+
3
(
−
1
1
)
3
=

3
+
3
2
x
+
3
(
−
11
)
3
=
x
3
+
3
32x+3(−11)​=3x​+3
4
Multiply the numbers
2

+
3
(
−
1
1
)
3
=

3
+
3
2
x
+
3
(
−
11
)
3
=
x
3
+
3
32x+3(−11)​=3x​+3
2

−
3
3
3
=

3
+
3
2
x
−
33
3
=
x
3
+
3
32x−33​=3x​+3
5
Find common denominator
2

−
3
3
3
=

3
+
3
2
x
−
33
3
=
x
3
+
3
32x−33​=3x​+3
2

−
3
3
3
=

3
+
3
⋅
3
3
2
x
−
33
3
=
x
3
+
3
⋅
3
3
32x−33​=3x​+33⋅3​
6
Combine fractions with common denominator
2

−
3
3
3
=

3
+
3
⋅
3
3
2
x
−
33
3
=
x
3
+
3
⋅
3
3
32x−33​=3x​+33⋅3​
2

−
3
3
3
=

+
3
⋅
3
3
2
x
−
33
3
=
x
+
3
⋅
3
3
32x−33​=3x+3⋅3​
7
Multiply the numbers
2

−
3
3
3
=

+
3
⋅
3
3
2
x
−
33
3
=
x
+
3
⋅
3
3
32x−33​=3x+3⋅3​
2

−
3
3
3
=

+
9
3
2
x
−
33
3
=
x
+
9
3
32x−33​=3x+9​
8
Multiply all terms by the same value to eliminate fraction denominators
2

−
3
3
3
=

+
9
3
2
x
−
33
3
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x
+
9
3
32x−33​=3x+9​
3
⋅
2

−
3
3
3
=
3
(

+
9
3
)
3
⋅
2
x
−
33
3
=
3
(
x
+
9
3
)
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9
Cancel multiplied terms that are in the denominator
3
⋅
2

−
3
3
3
=
3
(

+
9
3
)
3
⋅
2
x
−
33
3
=
3
(
x
+
9
3
)
3⋅32x−33​=3(3x+9​)
2

−
3
3
=

+
9
2
x
−
33
=
x
+
9
2x−33=x+9
10
Add
3
3
33
33
to both sides of the equation
2

−
3
3
=

+
9
2
x
−
33
=
x
+
9
2x−33=x+9
2

−
3
3
+
3
3
=

+
9
+
3
3
2
x
−
33
+
33
=
x
+
9
+
33
2x−33+33=x+9+33
11
Simplify
Add the numbers
Add the numbers
2

=

+
4
2
2
x
=
x
+
42
2x=x+42
12
Subtract

x
x
from both sides of the equation
2

=

+
4
2
2
x
=
x
+
42
2x=x+42
2

−

=

+
4
2
−

2
x
−
x
=
x
+
42
−
x
2x−x=x+42−x
13
Simplify
Combine like terms
Multiply by 1
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=
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2
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