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4 years ago

20 points for one question. You know you want to.

Mathematics

2 answers:

spayn [35]4 years ago

8 0

Answer:

The answer is A.

Step-by-step explanation:

We only have to test one coordinate to find the answer. The original coordinate for N is (-1, -1). Using the formula, N' is (0, -2).

DiKsa [7]4 years ago

5 0

You see, it's quite easy. All you need to do is find the coordinates of FNQ, then add the 1 to the x values and subtract 1 from the y values as stated in the translation rule.

However, in your case, you don't have numbers on your graph but you can easily draw some. Be mindful of the quadrants.

Let's try out point F just for gags.

F = (3, 3)

(Technically you only need one point to determine the answer.)

Now, apply rule: x + 1, y - 1

(3, 3) becomes (4, 2)

Now, which multiple choice answer contains that specific point for F?

Choice A.

You might be interested in

Find all the zeroes of the polynomial function f(x)=x^3-5x^2+6x-30 using synthetic division.

WITCHER [35]

Write the coeeficientes of the polynomial in order:

| 1 - 5 6 - 30
|
|
|
------------------------

After some trials you probe with 5

 | 1 - 5 6 - 30
 |
 |
5 | 5 0 30
-----------------------------
 1 0 6 0 <---- residue

Given that the residue is 0, 5 is a root.

The quotient is x^2 + 6 = 0, which does not have a real root.

Therefore, 5 is the only root. You can prove it by solving the polynomial x^2 + 6 = 0.

3 0

3 years ago

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: $b = 1-(-\frac{2}{5})(2)$ . $b = -\frac{1}{5}$ .