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AlekseyPX
3 years ago
14

Complete the point-slope equation of the line through (3,6)(3,6) (3,6) left parenthesis, 3, comma, 6, right parenthesis and (5,−

8)(5,-8) (5,−8) left parenthesis, 5, comma, minus, 8, right parenthesis .
Mathematics
1 answer:
Elanso [62]3 years ago
3 0

Answer:

y = -7x + 27 is the point slope equation that passes through the two points

Step-by-step explanation:

Here, we want to write the equation of the line between (3,6) and (5,-8)

Mathematically, the equation of the line that passes through both points can be represented by ;

y = mx + c

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

Let’s find the slope m first;

Mathematically;

slope m = y2-y1/x2-x1

where (x1,y1) = (3,6) and (x2,y2) = (5,-8)

Substitute these values in the slope equation , we have the following;

m = (-8-6)/(5-3) = -14/2 = -7

So the equation becomes;

y = -7x + c

we still need the value of c

To get this, we can substitute any of the points in the equation, where x is the x coordinate of the point and y is the coordinate of the point.

Let’s use (3,6)

Thus we have;

6 = -7(3) + c

c = 6 + 21

c = 27

So the equation becomes;

y = -7x + 27

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What is 1 +1 pls HELP
ElenaW [278]

Answer:

2

Step-by-step explanation:

Subject: Re: Need the math proof for 1 + 1 = 2

The proof starts from the Peano Postulates, which define the natural

numbers N. N is the smallest set satisfying these postulates:

P1. 1 is in N.

P2. If x is in N, then its "successor" x' is in N.

P3. There is no x such that x' = 1.

P4. If x isn't 1, then there is a y in N such that y' = x.

P5. If S is a subset of N, 1 is in S, and the implication

(x in S => x' in S) holds, then S = N.

Then you have to define addition recursively:

Def: Let a and b be in N. If b = 1, then define a + b = a'

(using P1 and P2). If b isn't 1, then let c' = b, with c in N

(using P4), and define a + b = (a + c)'.

Then you have to define 2:

Def: 2 = 1'

2 is in N by P1, P2, and the definition of 2.

Theorem: 1 + 1 = 2

Proof: Use the first part of the definition of + with a = b = 1.

Then 1 + 1 = 1' = 2 Q.E.D.

Note: There is an alternate formulation of the Peano Postulates which

replaces 1 with 0 in P1, P3, P4, and P5. Then you have to change the

definition of addition to this:

Def: Let a and b be in N. If b = 0, then define a + b = a.

If b isn't 0, then let c' = b, with c in N, and define

a + b = (a + c)'.

You also have to define 1 = 0', and 2 = 1'. Then the proof of the

Theorem above is a little different:

Proof: Use the second part of the definition of + first:

1 + 1 = (1 + 0)'

Now use the first part of the definition of + on the sum in

parentheses: 1 + 1 = (1)' = 1' = 2 Q.E.D.

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