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Mazyrski [523]
3 years ago
10

What is the slope of the line through (2, -1) and (-2, -3)?

Mathematics
2 answers:
prohojiy [21]3 years ago
8 0

\text{Given that,}\\\\\\(x_1,y_1) = (2,-1)~~\text{and}~~ (x_2,y_2) = (-2,-3)\\\\\\\text{Slope,}~ m =\dfrac{y_2-y_1}{x_2 -x_1} = \dfrac{-3+1}{-2-2} = \dfrac{-2}{-4} = \dfrac 12

Nookie1986 [14]3 years ago
4 0

Answer:

Step-by-step explanation:

Using slope formula:

(y2-y1)/(x2-x1)

(-3 + 1) / (-2-2)

-2/-4

slope: 1/2

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The school paper charges $1.50 for an ad, plus 25 cents per word. Carl wants to spend only $12.00 on the ad.
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42 words

Step-by-step explanation:

To solve just solve 0.25x +1.50 = 12.00 for x

7 0
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Solving system by graphing<br> y= -5/3x+3<br> y=1/3x-3
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the point of intersection on the graph are (3,-2)
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When given an algebraic expression involving subtraction, why is it best to rewrite the expression using addition before identif
Ray Of Light [21]
Usually when you do expressions involving subtraction you can do Keep, Change ,Change
That means you keep the sign of the first number and change the other two signs.
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3 years ago
A graph of a system of two equations is shown. HELP!!!!!
s344n2d4d5 [400]

Answer:

8 and 2

Step-by-step explanation:

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4 0
3 years ago
How many different linear arrangements are there of the letters a, b,c, d, e for which: (a a is last in line? (b a is before d?
inna [77]
A) Since a is last in line, we can disregard a, and concentrate on the remaining letters.
Let's start by drawing out a representation:

_ _ _ _ a

Since the other letters don't matter, then the number of ways simply becomes 4! = 24 ways

b) Since a is before d, we need to account for all of the possible cases.

Case 1: a d _ _ _ 
Case 2: a _ d _ _
Case 3: a _ _ d _
Case 4: a _ _ _ d

Let's start with case 1.
Since there are four different arrangements they can make, we also need to account for the remaining 4 letters.
\text{Case 1: } 4 \cdot 4!

Now, for case 2:
Let's group the three terms together. They can appear in: 3 spaces.
\text{Case 2: } 3 \cdot 4!

Case 3:
Exactly, the same process. Account for how many times this can happen, and multiply by 4!, since there are no specifics for the remaining letters.
\text{Case 3: } 2 \cdot 4!

\text{Case 4: } 1 \cdot 4!

\text{Total arrangements}: 4 \cdot 4! + 3 \cdot 4! + 2 \cdot 4! + 1 \cdot 4! = 240

c) Let's start by dealing with the restrictions.
By visually representing it, then we can see some obvious patterns.

a b c _ _

We know that this isn't the only arrangement that they can make.
From the previous question, we know that they can also sit in these positions:

_ a b c _
_ _ a b c

So, we have three possible arrangements. Now, we can say:
a c b _ _ or c a b _ _
and they are together.

In fact, they can swap in 3! ways. Thus, we need to account for these extra 3! and 2! (since the d and e can swap as well).

\text{Total arrangements: } 3 \cdot 3! \cdot 2! = 36
7 0
3 years ago
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