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kicyunya [14]
2 years ago
13

How i can answer this question

Mathematics
1 answer:
Advocard [28]2 years ago
4 0

Answer:

A. 2(x+y+1). lmk if you need any more awnsers

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(x1,y1) = (-1,-1)

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m = (y2 - y1)/(x2 - x1)

m = (-2 + 1)/(2 + 1)

m = -1/3

Option → C

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Given the following roots build a polynomial function in standard form: -3 and +2
swat32

Answer:

  f(x) = x² +x -6

Step-by-step explanation:

The standard form will look like ...

  f(x) = x² +bx +c

where b is the opposite of the sum of the roots, and c is their product.

  f(x) = x² -(-3+2)x +(-3)(2)

  f(x) = x² +x -6

_____

<em>Additional comment</em>

In general, "standard form" is ax²+bx+c. In this case, the coefficient 'a' can be 1 since neither of the roots is expressed as a fraction. The sum of roots is (-b/a) and the product of roots is (c/a).

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Every 11 days my dog eats 3 cupcakes. How many does my dog eat in 22 days?
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6 0
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Read 2 more answers
What is the equation of the line perpendicular to 3x+y= -8that passes through -3,1? Write your answer in slope-intercept form. S
Gekata [30.6K]

Slope intercept form of a line perpendicular to 3x + y = -8, and passing through (-3,1) is y=\frac{1}{3} x+2

<u>Solution:</u>

Need to write equation of line perpendicular to 3x+y = -8 and passes through the point (-3,1).

Generic slope intercept form of a line is given by y = mx + c

where m = slope of the line.

Let's first find slope intercept form of 3x + y = -8

3x + y = -8

=> y = -3x - 8

On comparing above slope intercept form of given equation with generic slope intercept form y = mx + c , we can say that for line 3x + y = -8 , slope m = -3  

And as the line passing through (-3,1) and is  perpendicular to 3x + y = -8, product of slopes of two line will be -1  as lies are perpendicular.

Let required slope = x  

\begin{array}{l}{=x \times-3=-1} \\\\ {=>x=\frac{-1}{-3}=\frac{1}{3}}\end{array}

So we need to find the equation of a line whose slope is \frac{1}{3} and passing through (-3,1)

Equation of line passing through (x_1 , y_1) and having lope of m is given by

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

\text { In our case } x_{1}=-3 \text { and } y_{1}=1 \text { and } \mathrm{m}=\frac{1}{3}

Substituting the values we get,

\begin{array}{l}{(\mathrm{y}-1)=\frac{1}{3}(\mathrm{x}-(-3))} \\\\ {=>\mathrm{y}-1=\frac{1}{3} \mathrm{x}+1} \\\\ {=>\mathrm{y}=\frac{1}{3} \mathrm{x}+2}\end{array}

Hence the required equation of line is found using slope intercept form

4 0
3 years ago
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