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love history [14]
3 years ago
11

What are the coordinates of vertex A of square ABCD?

Mathematics
2 answers:
tia_tia [17]3 years ago
6 0

Answer:

D (-2,1)

Step-by-step explanation:

zlopas [31]3 years ago
5 0

The answer is D. (-2,1)

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State whether the lines are parallel, perpendicular,or neither.
cluponka [151]

Answer:

Please check the explanation.

Step-by-step explanation:

  • Two lines are parallel if their slopes are equal.
  • Two lines are perpendicular if the product of their slope is -1

We also know that the slope-intercept form of the line equation is

y=mx+b

where m is the slope and b is the y-intercept

Given the lines

1)

  • y = 6х - 3

Comparing with y=mx+b, the slope of y = 6х - 3:

m₁=6  

  • y = - 1/6x + 7

Comparing with y=mx+b, the slope of y = - 1/6x + 7:

m₂=-1/6

As

m₁ × m₂ = -1

6 ×  - 1/6 = -1

-1 = -1

Thus, the lines y = 6х - 3 and y = - 1/6x + 7 are perpendicular.

2)

  • y = 3x + 2

Comparing with y=mx+b, the slope of y = 3x + 2:

m₁=3  

  • 2y = 6x - 6

simplifying to write in slope-intercept form

y=3x-3

Comparing with y=mx+b, the slope of y=3x-3:

m₂=3

As the slopes of y = 3x + 2 and 2y = 6x - 6 are equal.

i.e. m₁ = m₂ → 3 = 3

Thus, the lines y = 3x + 2 and 2y = 6x - 6 are paralle.

3)

  • 8x - 2y = 3

simplifying to write in slope-intercept form

y = 4x - 3/2

Comparing with y=mx+b, the slope of y = 4x - 3/2:

m₁=4  

  • x + 4y = - 1

simplifying to write in slope-intercept form

y=-1/4x-1/4

Comparing with y=mx+b, the slope of y=-1/4x-1/4:

m₂=-1/4

As

m₁ × m₂ = -1

4 ×  - 1/4 = -1

-1 = -1

Thus, the lines 8x - 2y = 3 and x + 4y = - 1 are perpendicular.

4)

  • 3x+2y = 5

simplifying to write in slope-intercept form

y = -3/2x + 5/2

Comparing with y=mx+b, the slope of y = -3/2x + 5/2:

m₁=-3/2  

  • 3y + 2x = - 3

simplifying to write in slope-intercept form

y = -2/3x - 1

Comparing with y=mx+b, the slope of y = -2/3x - 1:

m₂=-2/3

As m₁ and m₂ are neither equal nor their product is -1, hence the lines neither perpendicular nor parallel.

5)

  • y - 5 = 6x

simplifying to write in slope-intercept form

y=6x+5

Comparing with y=mx+b, the slope of y=6x+5:

m₁=6  

  • y - 6x = - 1

simplifying to write in slope-intercept form

y=6x-1

Comparing with y=mx+b, the slope of y=6x-1:

m₂=6

As the slopes of y - 5 = 6x and y - 6x = -1 are equal.

i.e. m₁ = m₂ → 6 = 6

Thus, the lines y - 5 = 6x and y - 6x = -1 are paralle.

6)

  • y = 3х + 9

Comparing with y=mx+b, the slope of y = 3х + 9:

m₁=3  

  • y = -1/3x - 4

Comparing with y=mx+b, the slope of y =  1/3x - 4:

m₂=1/3

As m₁ and m₂ are neither equal nor their product is -1, hence the lines neither perpendicular nor parallel.

8 0
2 years ago
Question in image again
Reil [10]
16in I believe irdk
Hope it helped..
7 0
3 years ago
Read 2 more answers
Eric has 7 red shirts and 11 black shirts what is the ratio of red and black shirts.
olya-2409 [2.1K]

Answer:

7:11

Step-by-step explanation:

Hope this helps! God bless.

7 0
3 years ago
Read 2 more answers
B) T is due north of C, calculate the bearing of B from C
choli [55]

Answer:

(a) 52°

(b) 322°

Step-by-step explanation:

(a) The details of the circle are;

The diameter of the circle = AOC

The center of the circle = Point O

The point the line AT cuts the circle = Point B

The point the tangent PT touches the circle = Point C

Angle ∠COB = 76°

We have that angle AOB and angle COB are supplementary angles, therefore;

∠AOB + ∠COB = 180°

∠AOB = 180° - ∠COB

∴ ∠AOB = 180° - 76° = 104°

∠AOB = 104°

OA = OB = The radius of the circle

Therefore, ΔAOB  =  An isosceles triangle

∠OAB = ∠OBA by base angles of an isosceles triangle are equal

∠AOB + ∠OAB + ∠OBA = 180° by angle summation property

∴ ∠AOB + ∠OAB + ∠OBA = ∠AOB + ∠OAB + ∠OAB = ∠AOB + 2×∠OAB = 180°

∠OAB = (180° - ∠AOB)/2

∴ ∠OAB = (180° - 104°)/2 = 38°

∠TAC = ∠OAB = 38° by reflexive property

AOC is perpendicular to tangent PT at point C, by tangent to a circle property, therefore;

∠TCA = 90° and ΔTCA = A right triangle

∠TAC + ∠ATC + ∠TCA = 180° by angle sum property

∠ATC = 180° - (∠TAC + ∠TCA)

∴ ∠ATC = 180° - (38° + 90°) = 52°

Angle ATC = 52°

(b) In ΔABC, ∠ABC = Angle subtended by the diameter = 90°

∴ ΔABC = A right triangle

∠ABC and ∠TBC are supplementary angles, therefore;

∠ABC + ∠TBC = 180°

∠TBC = 180° - ∠ABC

∴ ∠TBC = 180° - 90° = 90°

∠TCB = 180° - (∠TBC + ∠ATC)

∴ ∠TCB = 180° - (90° + 52°) = 38°

The bearing of B from C = (360° - 38°) = 322°.

7 0
2 years ago
When you have coordinates of two points A(1, 2) and B(2,6) slope is:
bearhunter [10]

Answer: 4

Step-by-step explanation:

4 0
2 years ago
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