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Lelu [443]
3 years ago
11

3. Write the equation of a line that is perpendicular to the given line and that passes through the given point.

Mathematics
1 answer:
prohojiy [21]3 years ago
3 0
To find the perpendicular line a y- 2 = 7/3 (x + 5) you must first observe that the slope is m = 7/3, therefore, the slope of the perpendicular line put of the reciprocal, that is, m = - 3/7. So the possible options are A or C, but since the point (-4, 9) must go through the line, we find that the equation that satisfies that is option C.Therefore the perpendicular line that passes through the point (-4, 9) is C. y - 9 = -3/7 (x + 4)
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Answer:

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Step-by-step explanation:

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Find the missing coefficient.<br><br> -<br> y2 − [-5y − y(-7y − 9)] − [-y (15y + 4)] = 0
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What is the domain of f
frez [133]

Answer:

[ - 6 , 6 ]

Step-by-step explanation:

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Please mark for Brainliest!! :D Thanks!!

For more questions or more information, please comment below!

8 0
3 years ago
Just Need A Few Questions Answered To Finish My Quiz, Any Help Would Be Much Appreciated.
kvv77 [185]

Given the triangle

PQR

with points

P(8,0)

Q(6,2)

R(-2,-4)

And the triangle

P'Q'R'

with points

P'(4,0)

Q'(3,1)

R'(-1,-2)

Part A. Scale factor

Using the vertex

P( 8, 0)

P'(4,0)

the dilatation factor is given by

\frac{4}{8}=\frac{1}{2}

The triangle has a dilatation factor of 1/2

Part B:

P''Q''R'' after using P'Q'R' reflected about the y axis

to make a reflection over the y axis

coordinates (x,y) turn into coordinates (-x,y)

as follows

P^{\prime}(4,0)\rightarrow P^{\prime\prime}(-4,0)Q^{\prime}(3,1)\rightarrow Q^{\prime\prime}(-3,1)R^{\prime}(-1,-2)\rightarrow R^{\prime\prime}(1,-2)

Then triangle P''Q''R'' has coordinates

P''(-4,0)

Q''(-3,1)

R''(1,-2)

Part C:

PQR is congruent to P''Q''R''?

Congruent triangles are triangles that have the same size and shape. This means that the corresponding sides are equal and the corresponding angles are equal.

Then the triangles are not congruent

8 0
1 year ago
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