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jasenka [17]
3 years ago
6

What is the translation for B’ (6,-5) B(-5,-2)

Mathematics
1 answer:
Elena L [17]3 years ago
5 0

Answer:

-7

Step-by-step explanation:

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Given that SQ¯¯¯¯¯ bisects ∠PSR and ∠SPQ≅∠SRQ, which of the following proves that PS¯¯¯¯¯≅SR¯¯¯¯¯?
yanalaym [24]

Step-by-step explanation:

It is given that ∠SPQ≅∠SRQ

.

The figure shows the same triangles P Q S and R Q S as in the beginning of the task. Angles S P Q and S R Q are highlighted in red.

It is also given that SQ⎯⎯⎯⎯⎯

bisects ∠PSR

.

By the definition of angle bisector, ∠PSQ≅∠QSR

.

The figure shows the same triangles P Q S and R Q S as in the beginning of the task. Angles P S Q and R S Q are congruent and highlighted in red.

△PQS

and △RQS share a common side SQ⎯⎯⎯⎯⎯, and SQ⎯⎯⎯⎯⎯≅SQ⎯⎯⎯⎯⎯

by the Reflexive Property of Congruence.

The figure shows the same triangles P Q S and R Q S as in the beginning of the task. Segment S Q is highlighted in red.

Two angles, ∠SPQ

and ∠PSQ, and a nonincluded side, SQ⎯⎯⎯⎯⎯, of △PQS are congruent to two angles, ∠SRQ and ∠QSR, and a nonincluded side, SQ⎯⎯⎯⎯⎯, of △RQS

.

The figure shows the same triangles P Q S and R Q S as in the beginning of the task. Angles P S Q and R S Q are highlighted in red. Angles S P Q and S R Q are highlighted in red. Side S Q is highlighted in blue.

So, △PQS≅△RQS

by the Angle-Angle-Side (AAS) Congruence Theorem.

PS⎯⎯⎯⎯⎯

and SR⎯⎯⎯⎯⎯ are corresponding sides of congruent triangles, △PQS and △RQS. So, PS⎯⎯⎯⎯⎯≅SR⎯⎯⎯⎯⎯

by CPCTC.

The figure shows the same triangles P Q S and R Q S as in the beginning of the task. Sides S R and S P are congruent and highlighted in red.

Translate these six statements and reasons into a 2

-column proof,

1. ∠SPQ≅∠SRQ

(Given)

2. SQ⎯⎯⎯⎯⎯

bisects ∠PSR

. (Given)

3. ∠PSQ≅∠QSR

(Def. of ∠

bisect)

4. SQ⎯⎯⎯⎯⎯≅SQ⎯⎯⎯⎯⎯

(Reflex. Prop. of ≅

)

5. △PQS≅△RQS

(AAS Steps 1, 3, 4)

6. PS⎯⎯⎯⎯⎯≅SR⎯⎯⎯⎯⎯

(CPCTC)

There you go

4 0
3 years ago
Which shows another way to write 11/2 A. 11 × 2 B. 11 + 2 C. 11 × 11
poizon [28]
Another way to write 11/2 is 11 / 2, which is NOT a choice.

11/2=5.5 OR 5 1/2.

A. 11*2=22
B. 11+2=13
C. 11*11=121

Hope this helped☺☺
6 0
4 years ago
Read 2 more answers
Given: AB is diameter, BC is tangent to circle O.<br><br> Prove: AXB~ABC
Dmitry_Shevchenko [17]
Please see picture attached.

3 0
3 years ago
Read 2 more answers
Consider the quadratic equation x2 = 4x - 5. How many solutions does the equation have?
JulijaS [17]

Answer:

  • no real solutions
  • 2 complex solutions

Step-by-step explanation:

The equation can be rearranged to vertex form:

  x^2 -4x = -5 . . . . . . . . . subtract 4x

  x^2 -4x +4 = -5 +4 . . . . add 4

  (x -2)^2 = -1 . . . . . . . . . show the left side as a square

  x -2 = ±√-1 = ±i . . . . . . take the square root; the right side is imaginary

  x = 2 ± i . . . . . . . . . . . . . add 2. These are the complex solutions.

_____

<em>Comment on the question</em>

Every 2nd degree polynomial equation has two solutions. They may be real, complex, or (real and) identical. That is, there may be 0, 1, or 2 real solutions. This equation has 0 real solutions, because they are both complex.

3 0
3 years ago
What is the area of a circle with a circumference of 37.68
Pepsi [2]
A = c^2/4 x pi
A = 37.68^2/12.56
A = 113.04 is what I get
8 0
4 years ago
Read 2 more answers
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