How do you solve y=2.5x+5

2 answers:

7 0

$slope\ is\ for\ y=0\\\\
y=2,5x+5\\
0=2,5x+5\ \ \ | subtract\ 5\\
-5=2,5x \ \ \ | divide\ by\ 2,5\\
x=-2 ---> SLOPE\\\\\
intercept\ is\ for\ x=0\\\\
y=5----> INTERCEPT$

d1i1m1o1n [39]2 years ago

3 0

If you are solving for x.
y=2.5x+5
you subtract 5 from both sides.
-5=2.5x
then divide both sides with 2.5
-5/2.5 2.5/2.5
the 2.5s will cancel each other out.
x=-2

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Read 2 more answers

Find the coordinates of the midpoint of the segment with endpoints A(-5, 6) and B(3, 2). You must show all work in order to rece

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The point P(1,1/2) lies on the curve y=x/(1+x). (a) If Q is the point (x,x/(1+x)), find the slope of the secant line PQ correct

lukranit [14]

Answer:

See explanation

Step-by-step explanation:

You are given the equation of the curve

$y=\dfrac{x}{1+x}$

Point $P\left(1,\dfrac{1}{2}\right)$ lies on the curve.

Point $Q\left(x,\dfrac{x}{1+x}\right)$ is an arbitrary point on the curve.

The slope of the secant line PQ is

$\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{\frac{x}{1+x}-\frac{1}{2}}{x-1}=\dfrac{\frac{2x-(1+x)}{2(x+1)}}{x-1}=\dfrac{\frac{2x-1-x}{2(x+1)}}{x-1}=\\ \\=\dfrac{\frac{x-1}{2(x+1)}}{x-1}=\dfrac{x-1}{2(x+1)}\cdot \dfrac{1}{x-1}=\dfrac{1}{2(x+1)}\ [\text{When}\ x\neq 1]$