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vladimir2022 [97]
3 years ago
7

...............................................

Mathematics
1 answer:
olga2289 [7]3 years ago
3 0

Answer:

  1. OT⊥BA;T is the midpoint of BA-------Given
  2. ∠BTO and ∠ATO are right angles ---Definition of perpendicular lines
  3. ∠BTO≅∠ATO---------------------------------All right angles are congruent
  4. T is the midpoint of BA------------------ Given
  5. TA≅TB-------------------------------------------Definition of midpoint
  6. TO≅TO------------------------------------------Reflexive
  7. ΔBOT≅ΔAOT---------------------------------SAS

Step-by-step explanation:

Given that OT⊥BA and T is the midpoint of BA

As OT⊥BA,

∠BTO and ∠ATO are right angles (Definition of perpendicular lines)

⇒∠BTO≅ ∠ATO (All right angles are congruent)

T is the midpoint ofBA (given)

⇒TA≅TB (Definition of midpoint)

TO≅TO (Reflexive)

Therefore, ΔBOT≅ΔAOT (by SAS criteria ):

conditions:

  • BT=AT(side)
  • ∠BTO=∠ATO(angle)
  • TO=TO(side)
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Use slopes and y-intercepts to determine if the lines 5x-5y=-2 and -x+2y=4
HACTEHA [7]

The given pair of lines are not perpendicular.

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

The line is a curve showing the shortest distance between 2 points.

5x - 5y = -2 - - - - - (1)
Transform the equation into standard form,
5x + 2 = 5y
y = 5x /5 + 2/5
y = x + 2/5


The slope of equation 1 is m_1 = 1  and intercept c = 2 / 5


Similarly
x + 2y = 4    - - - - - - - -(2)
Transform it into standard form
y = -x/2 + 4 /2
y = -x / 2 + 2


Slope of the equation 2  m_2= -1 / 2 and intercept c = 2
Slope of line 1 * slope of line 2 = 1 * -1/2 = -1/2


Since the lines are not perpendicular because the pair of lines does not satisfy the property of perpendicular lines i.e
m_1*m_2 = -1

Thus, the given pair of lines are not perpendicular.

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Find The difference 25(d−10)−23(d+6) is
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Answer:

2d-388 or 2(d-194)

Step-by-step explanation:

25(d-10)-23(d+6)

25d-250-23d-138

25d-23d-250-138

2d-250-138

2d-388

factor out or simplify,

you get 2(d-194)

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3 years ago
Which linear equation has no solution?
olga_2 [115]

Answer:

Option A

Step-by-step explanation:

Given:

  • a. 3x-5= 3x + 5
  • b. 3x-5= 3x - 5
  • c. 3x - 5 = 2x+5
  • d. 3x-5 = 2x + 10

To find:

  • Which one of the linear equations have no solution.

Solution:

a)  3x-5= 3x + 5

Add 5 to both sides

3x-5= 3x + 5

3x - 5 + 5 = 3x + 5 + 5

Simplify

(Add the numbers)

3x - 5 + 5 = 3x + 5 + 5

3x = 3x + 5 + 5

(Add the numbers)

3x  = 3x + 5 + 5

3x = 3x + 10

Subtract 3x from both sides

3x = 3x + 10

3x - 3x = 3x + 10 - 3

Simplify

(Combine like terms)

3x -3x = 3x + 10 - 3

0 = 3x + 10 - 3

(Combine like terms)

0 = 3x + 10 - 3

0 = 10

The input is a contradiction: it has no solutions

b)  3x-5= 3x - 5

Since both sides equal, there are infinitely many solutions.

c)  3x - 5 = 2x+5

Add 5 to both sides

3x = 2x + 5 + 5

Simplify  2x + 5 + 5 to 2x + 10

3x = 2x + 10

Subtract 2x from both sides

3x - 2x = 10

Simplify 3x - 2x to x.

x = 10

d) 3x-5 = 2x + 10

Add 5 to both sides

3x = 2x + 10 + 5

Simplify 2x + 10 + 5 to 2x + 15

3x = 2x + 15

Subtract 2x from both sides

3x - 2x = 15

Simplify 3x -2x to x.

x = 15

--------------------------------------------

Answer:

As you can see all c and d both have solutions, eliminating them as options. Option B has infinite solutions leaving Option A which has no solutions.

Therefore, <u><em>Option A</em></u> is the linear equation that has no solution.

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