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stealth61 [152]
4 years ago
14

Write the equation of the line that passes through (2, 4) and has a slope of –1 in point-slope form.

Mathematics
2 answers:
wel4 years ago
6 0

The other person is correct i just did the test

klemol [59]4 years ago
5 0
Point slope form is just a point and the slope written in the form y-y1 =m (x-x1).
Where x1 and y1 (should be subscripts) are (x1,y1) the point you are given.
So in this case all you must do is plug in the given values:

( y-y1 ) = m ( x-x1 )
( y-4. ) = -1 ( x-2 )
This being your answer

I hope this helps!!
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Easy question:
ivolga24 [154]

Answer:

midpoint = (-2 , 6)

Step-by-step explanation:

The midpoint can be found with the formula {(x1 + x2 )/2, (y1 + y2.) P1( x1 , y1 ), and P2 (x2 , y2 ) are the coordinates of two points, and the midpoint is a point lying equidistant and between these two points.

6 0
3 years ago
How do you know when to use the Estimation strategy vs the Guess and Check strategy ?​
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The guessing strategy only works when you don’t u don’t understand a question so I advise you NOT to do it , only when it’s necessary but the Estimation strategy works when you don’t have a any but a answer where theirs decimals or a answer that is close to one another and can be rounded
6 0
3 years ago
Complete the equation and tell which property you used (18•2)•5=18•(2•?)
RUDIKE [14]

Answer:

https://brainly.com/

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Determine which of the following sets of three points constitute the vertices of a right triangle: (a) 3 + 5i,2 +2i,5i; (b)2i,3
Illusion [34]

Answer:

Option (c) is correct

Step-by-step explanation:

Case (a)

A = 3 + 5i = (3, 5)

B = 2 + 2i = (2, 2)

C = 5i = (0, 5)

Use the distance formula to find the distance between two points

AB = \sqrt{(2-3)^{2}+(2-5)^{2}}=\sqrt{10}

BC = \sqrt{(0-2)^{2}+(5-2)^{2}}=\sqrt{13}

CA = \sqrt{(0-3)^{2}+(5-5)^{2}}=\sqrt{9}

For the triangle to be right angles triangle

BC^{2}=AB^{2}+CA^{2}

Here, it is not valid, so these are not the points of a right angled triangle.

Case (b)

A = 2i = (0, 2)

B = 3 + 5i = (3, 5)

C = 4 + i = (4, 1)

Use the distance formula to find the distance between two points

AB = \sqrt{(3-0)^{2}+(5-2)^{2}}=\sqrt{18}

BC = \sqrt{(4-3)^{2}+(1-5)^{2}}=\sqrt{17}

CA = \sqrt{(4-0)^{2}+(1-2)^{2}}=\sqrt{17}

For the triangle to be right angles triangle

AB^{2}=BC^{2}+CA^{2}

Here, it is not valid, so these are not the points of a right angled triangle.

Case (c)

A = 6 + 4i = (6, 4)

B = 7 + 5i = (7, 5)

C = 8 + 4i = (8, 4)

Use the distance formula to find the distance between two points

AB = \sqrt{(7-6)^{2}+(5-4)^{2}}=\sqrt{2}

BC = \sqrt{(8-7)^{2}+(4-5)^{2}}=\sqrt{2}

CA = \sqrt{(8-6)^{2}+(4-4)^{2}}=\sqrt{4}

For the triangle to be right angles triangle

CA^{2}=BC^{2}+AB^{2}

Here, it is valid, so these are the points of a right angled triangle.

7 0
3 years ago
OD
MAXImum [283]
I need help tooooo cause at this point it’s frustrating me
4 0
3 years ago
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