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2 years ago

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The slope-intercept form of a linear equation is y = mx + b, where x and y are coordinates of an ordered pair, m is the slope of

the line, and b is where the line crosses the y-axis. Which is an equivalent equation solved for the slope, m? m = yx + b m = m equals StartFraction y minus b Over x EndFraction. m = m equals StartFraction y Over x EndFraction minus b. – b m = y – m equals y minus StartFraction b Over x EndFraction.

1 answer:

anzhelika [568]2 years ago

7 0

We have been given that the slope-intercept form of a linear equation is $y = mx+b$ , where x and y are coordinates of an ordered pair, m is the slope of the line, and b is where the line crosses the y-axis. We are asked to solve the equation for slope.

First of all, we will subtract 'b'from both sides.

$y-b= mx+b-b$

$y-b= mx$

Let us switch the sides:

$mx=y-b$

Now we will divide both sides by x.

$\frac{mx}{x}=\frac{y-b}{x}$

$m=\frac{y-b}{x}$

Therefore, the value of m is $\frac{y-b}{x}$ .

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QUICK! 75 POINTS !!Select all that are part of the solution set of csc(x) > 1 and over 0 ≤ x ≤ 2π.

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Answer:

$\frac{\pi}{4}$

$\frac{5\pi}{6}$

Step-by-step explanation:

The answer uses the unit circle and that sine and cosecant are reciprocals.

The first choice doesn't even fit the criteria that $x$ is between $0$ and $2\pi$ (inclusive of both endpoints) because of the $x=\frac{-7\pi}{6}$ .

Let's check the second choice.

$\csc(\frac{\pi}{4})=\frac{2}{\sqrt{2}} \text{ since } \sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$ .

$\csc(\frac{\pi}{4})>1 \text{ since } \frac{2}{\sqrt{2}}>1$

$\csc(\frac{\pi}{2})=1 \text{ since } \sin(\frac{\pi}{2})=1$ which means $\csc(\frac{\pi}{2})=1$ which is not greater than 1.

So we can eliminate second choice.

Let's look at the third.

$\csc(\frac{5\pi}{6})=2 \text{ since } \sin(\frac{5\pi}{6})=\frac{1}{2}$ which means $\csc(\frac{5\pi}{6})>1$ .

$\csc(\pi)$ isn't defined because $\sin(\pi)=0$ .

So we are eliminating 3rd choice now.

Let's look at the fourth choice.

$\csc(\frac{7\pi}{6})=-2 \text{ since } \sin(\frac{7\pi}{6})=\frac{-1}{2}$ which means $\csc(\frac{7\pi}{6})$ and not greater than 1.

I was looking at the rows as if they were choices.

Let me break up my choices.

So we said $x=-\frac{7\pi}{6}$ doesn't work because it is not included in the inequality $0\le x \le 2\pi$ .

How about $x=0$ ? This leads to $\csc(0)$ which doesn't exist because $\sin(0)=0$ .

So neither of the first two choices on the first row.

Let's look at the second row again.

We said $\frac{\pi}{4}$ worked but not $\frac{\pi}{2}$

Let's look at the choices on the third row.

We said $\frac{5\pi}{6}$ worked but not $x=\pi$

Let's look at at the last choice.

We said it gave something less than 1 so this choice doesn't work.

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