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insens350 [35]
3 years ago
10

Finding Slopes of Perpendicular Lines

Mathematics
1 answer:
MA_775_DIABLO [31]3 years ago
7 0

Answer:

1. d - b

2. -d (-b)

3. -d + b

Step-by-step explanation:

202 on edg yakdvb

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15
alexandr1967 [171]

By using parallel lines and transversal lines concept we can prove m∠1=m∠5.

Given that, a║b and both the lines are intersected by transversal t.

We need to prove that m∠1=m∠5.

<h3>What is a transversal?</h3>

In geometry, a transversal is a line that passes through two lines in the same plane at two distinct points.

m∠1+m∠3= 180° (Linear Pair Theorem)

m∠5+m∠6=180° (Linear Pair Theorem)

m∠1+m∠3=m∠5+m∠6

m∠3=m∠6

m∠1=m∠5 (Subtraction Property of Equality)

Hence, proved. By using parallel lines and transversal lines concept we can prove m∠1=m∠5.

To learn more about parallel lines visit:

brainly.com/question/16701300

#SPJ1

7 0
2 years ago
(4x+5)(x^2+2x+3) help?
Leya [2.2K]
This is the right anwser

5 0
3 years ago
Please Help me with these questions ASAP!
Kryger [21]

Answer:

What grqde is this

Step-by-step explanation:

please I need to know the grade

4 0
3 years ago
rotation of 90 degrees counterclockwise about the origin, point O, then a reflection across the x-axis reflection across the y-a
SpyIntel [72]

Answer:

The point O can be (x,y)  or (-x,y) or (-x,-y) or ( x,-y) reason being that you haven't given point O lies in which quadrant.

Rotation through 90° counter clockwise

(x,y) = (-y,x)

(-x,y)=(-y,-x)

(-x,-y)=(y,-x)

(x,-y) =(y,x)

Then Reflection across X axis has taken place.

(-y,x) = (-y,-x)

(-y,-x)=(-y,x)

(y,-x)=(y,x)

(y,x)=(y,-x)

Then a reflection across the y-axis has taken place.

(-y,-x)=(y,-x)

(-y,x) = (y,x)

(y,x) =(-y,x)

(y,-x)=(-y,-x)

Then a translation a units to the right and b units up has taken place.

(y,-x)=(y+a,-x+b)

(y,x) =(y+a, x+b)

(-y,x) =(-y+a,x+b)

(-y,-x)=(-y+a,-x+b)

Then a rotation of 180 degrees counterclockwise about the origin has taken place.

(y+a,-x+b)=[-(y+a),-(-x+b)]

(y+a, x+b)=[- (y+a), -(x+b)]

(-y+a,x+b)=[-(-y+a),-(x+b)]

(-y+a,-x+b) =[-(-y+a),-(-x+b)]

Now Again a reflection across the Y axis has taken place.

[-(y+a),-(-x+b)]=[(y+a),-(-x+b)]

[-(y+a),-(x+b)]=[(y+a),-(-x+b)]

[-(-y+a),-(x+b)]=[(-y+a),-(x+b)]

[-(-y+a),-(-x+b)]=[(-y+a),-(-x+b)]

Totally depends on value of a and b on which quadrant these point lies.



7 0
3 years ago
The figure shows a bridge support, a cable, and the roadway of a bridge.
Murljashka [212]

Answer:

soln,

feet × feet tall

170×80

13,600

3 0
2 years ago
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