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3 years ago

Someone knows how to do this?

Mathematics

1 answer:

Zarrin [17]3 years ago

6 0

Answer:

Most likely B

Step-by-step explanation:

Most likely B because it starts up around 5 which is the y int of B

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SOLVE THIS PLS JSJSXHHX

madam [21]

Step-by-step explanation:

y = ax + b

I see already the result (y = x/10 × 1/6), but let's go in formally.

we have multiple function points to use to officially calculate a and b.

1/6 = a×10 + b

2/3 = a×40 + b

5/6 = a×50 + b

1 2/3 = a×100 + b

let's e.g subtract equation 1 from equation 3.

4/6 = a×40 + 0

a = 4/40 / 6 = 4 / 240 = 1/60

1/6 = 10/60 + b

1/6 = 1/6 + b

b = 0

so, the function is

y = x/60

x = 25

y = 25/60 = 5/12 cups

5 0

3 years ago

HELLLPPPPPPP FASTTTTTTTT Jeremiah has to pay for his phone calls that are $.11 the minute if he has eight dollars how many minut

fenix001 [56]

72 minutes or an hour and 12 minutes

3 0

3 years ago

Read 2 more answers

Write the equation of the line that passes through (−3,1) and (2,−1) in slope-intercept form

Alex787 [66]

Answer:

$y=-\frac{2}{5}x-\frac{1}{5}$

Step-by-step explanation:

The equation of a line is y = mx + b

Where:

m is the slope

b is the y-intercept

First, let's find what m is, the slope of the line.

Let's call the first point you gave, (-3,1), point #1, so the x and y numbers given will be called x1 and y1.

Also, let's call the second point you gave, (2,-1), point #2, so the x and y numbers here will be called x2 and y2.

Now, just plug the numbers into the formula for m above, like this:

$m = -\frac{2}{5}$

So, we have the first piece to finding the equation of this line, and we can fill it into y=mx+b like this:

$y=-\frac{2}{5}x + b$

Now, what about b, the y-intercept?

To find b, think about what your (x,y) points mean:

(-3,1). When x of the line is -3, y of the line must be 1.
(2,-1). When x of the line is 2, y of the line must be -1.

Now, look at our line's equation so far: $y=-\frac{2}{5}x + b$ . $b$ is what we want, the - $-\frac{2}{5}$ is already set and x and y are just two 'free variables' sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the two points (-3,1) and (2,-1).

So, why not plug in for x and y from one of our (x,y) points that we know the line passes through? This will allow us to solve for b for the particular line that passes through the two points you gave!

You can use either (x,y) point you want. The answer will be the same:

(-3,1). y = mx + b or $1=-\frac{2}{5} * -3 + b$ , or solving for b: $b = 1-(-\frac{2}{5})(-3)$ . $b = -\frac{1}{5}$ .
(2,-1). y = mx + b or $-1=-\frac{2}{5} * 2 + b$ , or solving for b: $b = 1-(-\frac{2}{5})(2)$ . $b = -\frac{1}{5}$ .