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KiRa [710]
3 years ago
8

What is the point-slope form and standard form of (3,1) and (4, 2)?

Mathematics
1 answer:
Brrunno [24]3 years ago
5 0

The point-slope form and standard form of (3,1) and (4, 2) are y – 1 = x – 3 and x – y = 2 respectively

<u>Solution:</u>

Given, two points are (3, 1) and (4, 2)

We have to find that a line that passes through the given two points.

First let us find the slope of the line that passes through given two points.

<em><u>Slope of line "m" is given as:</u></em>

\mathrm{m}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

\text { where, }\left(x_{1}, y_{1}\right) \text { and }\left(x_{2}, y_{2}\right) \text { are two points on line. }

\text { Here } x_{1}=3 \text { and } y_{1}=1 \text { and } x_{2}=4 \text { and } y_{2}=2

\mathrm{m}=\frac{2-1}{4-3}=\frac{1}{1}=1

<em><u>The point slope form is given as:</u></em>

y-y_{1}=m\left(x-x_{1}\right)

\text { where } m \text { is slope and }(x_1, y_1) \text { is point on the line. }

y – 1 = 1(x – 3)  

y - 1 = x - 3  

Line equation in point slope form is y – 1 = x – 3  -- eqn 1

Now, line equation in standard form i.e. ax + by = c is  found out by eqn 1

y – 1 = x – 3  

x – y = 3 – 1  

x – y = 2

Hence, the line equation in point slope form and standard forms are y – 1 = x – 3 and x – y = 2 respectively

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Help!! Find missing value for each quadratic function
Svetllana [295]

I'll do the first one to get you started. The answer is -3

=================================================

Explanation:

When x = -1, y = 0 as shown in the first column. The second column says that (x,y) = (0,1) and the third column says (x,y) = (1,0)

We'll use these three points to determine the quadratic function that goes through them all.

The general template we'll use is y = ax^2 + bx + c

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

Plug in (x,y) = (0,1). Simplify

y = ax^2 + bx + c

1 = a*0^2 + b*0 + c

1 = 0a + 0b + c

1 = c

c = 1

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

Plug in (x,y) = (-1,0) and c = 1

y = ax^2 + bx + c

0 = a*(-1)^2 + b(-1) + 1

0 = 1a - 1b + 1

a-b+1 = 0

a-b = -1

a = b-1

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

Plug in (x,y) = (1,0), c = 1, and a = b-1

y = ax^2 + bx + c

0 = a(1)^2 + b(1) + 1 ... replace x with 1, y with 0, c with 1

0 = a + b + 1

0 = b-1 + b + 1 ... replace 'a' with b-1

0 = 2b

2b = 0

b = 0/2

b = 0

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

If b = 0, then 'a' is...

a = b-1

a = 0-1

a = -1

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

In summary so far, we found: a = -1, b = 0, c = 1

Therefore, y = ax^2 + bx + c turns into y = -1x^2 + 0x + 1 which simplifies to y = -x^2 + 1

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

The last thing to do is plug x = 2 into this equation and simplify

y = -x^2 + 1

y = -2^2 + 1

y = -4 + 1

y = -3


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3 years ago
Simplify 7(exponent 6) multiplied by 7 (exponent 5)
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7(6) x 7(5) = 7(6+5) = 7(11) is the simplified answer.
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