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ArbitrLikvidat [17]
3 years ago
5

Determine the standard form of the equation of the line that passes through ( -5,0 ) ( 0 ,-9)

Mathematics
2 answers:
Alecsey [184]3 years ago
6 0

Answer:

B

Step-by-step explanation:

insens350 [35]3 years ago
5 0

The general form for a line through two points (a,b) and (c,d) is

(c-a)(y-b)=(d-b)(x-a)

This is better than the slope forms because it works in the no slope case, as does the standard form.

If you haven't seen it before, it works because when (x,y)=(a,b) we get (c-a)(b-b)=(d-b)(a-a), both sides zero, and when (x,y)=(c,d) we get (c-a)(d-b)=(d-b)(c-a), clearly equal sides.

Here we have

(0 - -5)(y - 0) = (-9 - 0)(x - - 5)

5y = -9(x+5)

5y = -9x - 45

9x + 5y = -45

Ironically there are two standards for standard form; one with the constant alone on the right and one with the whole thing equal to zero.  I like the constant alone.

Answer: 9x + 5y = -45

Check:

We check each point is on the line

(-5,0)

9(-5) + 5(0) = -45, good

(0, -9)

9(0) + 5(-9) = -45, good again


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Answer:

<em>The estimated sales were $260 million</em>

Step-by-step explanation:

Assume the endpoints of a segment are (x1,y1) and (x2,y2).

The midpoint (xm,ym) is calculated as follows:

\displaystyle x_m=\frac{x_1+x_2}{2}

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The sales ov Cars, Inc. were (2008,240 million) and (2010,280 million). We need to use the midpoint to estimate the sales in 2009:

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The equation y= 2(x+1)^2 has one real root and that is x=-1.

What is real roots of the equation?

    We are aware that when we resolve a linear or quadratic equation, we always arrive at the value variable of the equation, or, to put it another way, we always locate the equation's solution. This "solution" is what we refer to as the real roots. For instance, when the equation X^2-7x+12=0 is solved, the actual roots are 3 and 4.

Here given,

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Hence the given equation has one real root and that is x=-1.

To learn more about real roots refer the below link

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