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ss7ja [257]
1 year ago
12

Write an equation in slope-intercept form for each line. (4,-1) and (-2,-1)

Mathematics
1 answer:
professor190 [17]1 year ago
3 0

With each line, a slope-intercept relationship (4,-1) and (-2,-1).

Y = 0 ,X = 5.

<h3>What is slope-intercept form?</h3>

The slope-intercept form is just a means of stating a line's equation so that both the slope (steepness) as well as y-intercept (where another line crosses this same vertical y-axis) are obvious. This expression is frequently referred to as y = mx + b.

<h3>According to the given information:</h3>

To begin, are using the slope formula to figure out the average slope between two points here.

Let:

x1 = 4

y1 = -1

x2 = -2

y2 = -1

m = (y2-y1)/(x2-x1)

= (-1 - (-1))/ (4-(-1))

= 0/5

= 0

So the slope is 0.

We must now calculate the y-intercept. This will be accomplished by converting a single of the points and also the slope together into a point-slope linear equation. y2-y1 = m(x2-x1).

Let’s plug in the point (5,0).

So we get y-(-0) = (0/5)(x-(-5)) ⇒  y+0

= (0/5)x +  0 ⇒  y

So,

Y = 0

X = 5

with each line, a slope-intercept relationship (4,-1) and (-2,-1).

Y = 0 ,X = 5

To know more about slope-intercept visit:

brainly.com/question/19824331

#SPJ4

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3 years ago
HELP PLEASE!!! I need help with 94 if you could show the steps that would be very helpful!
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n!/((n-r)!r!)

In this case, there could be 0, 1, 2, 3, 4, or all 5 cards discarded. There is only one possible combination each for 0 or 5 cards being discarded (either none of them or all of them). We will be the above equation to find the number of combination s for 1, 2, 3, and 4 discarded cards.

5!/((5-1)!1!) = 5!/(4!*1!) = (5*4*3*2*1)/(4*3*2*1*1) = 5

5!/((5-2)!2!) = 5!/(3!2!) = (5*4*3*2*1)/(3*2*1*2*1) = 10

5!/((5-3)!3!) = 5!/(2!3!) = (5*4*3*2*1)/(2*1*3*2*1) = 10

5!/((5-4)!4!) = 5!/(1!4!) = (5*4*3*2*1)/(1*4*3*2*1) = 5

Notice that discarding 1 or discarding 4 have the same number of combinations, as do discarding 2 or 3. This is being they are inverses of each other. That is, if we discard 2 cards there will be 3 left, or if we discard 3 there will be 2 left.

Now we add together the combinations

1 + 5 + 10 + 10 + 5 + 1 = 32 choices combinations to discard.

The answer is 32.

-------------------------------

Note: There is also an equation for permutations which is:

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Notice it is very similar to combinations. The only difference is that a permutation is an ORDERED arrangement while a combination is UNORDERED.

We used combinations rather than permutations because the order of the cards does not matter in this case. For example, we could discard the ace of spades followed by the jack of diamonds, or we could discard the jack or diamonds followed by the ace of spades. These two instances are the same combination of cards but a different permutation. We do not care about the order.

I hope this helps! If you have any questions, let me know :)








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