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GenaCL600 [577]
3 years ago
14

PLEASE HELP! its due todayy :D

Mathematics
2 answers:
valkas [14]3 years ago
8 0

Since both points are positive, they are located in the I quadrant.

bagirrra123 [75]3 years ago
5 0

Answer:

quadrant one

Step-by-step explanation:

This is simple though the 8 is not on the graph the numbers are both positive so it's quadrant one :)

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  x - 6

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7 0
3 years ago
Identify the graph of 5-i
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Answer:

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Step-by-step explanation:

5 0
2 years ago
What is 2x + 4 = x + 40
d1i1m1o1n [39]

{\boxed{\boxed { ⎆ Answer :- }}} \

2x + 4 = x + 40 \\ 2x - x = 40 - 4 \\ x = 36

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8 0
3 years ago
Read 2 more answers
How many ways can four runners finish a race?<br> 4<br> 1<br> 10<br> 24
Whitepunk [10]
That's 4! (faculty), ie., 4x3x2x1 = 24.
one of four can win, then there are 3 left for second place, and so on...
5 0
3 years ago
Read 2 more answers
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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