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lana [24]
2 years ago
7

The numbered list gives instructions to bisect a line segment, but the steps are out of order.

Mathematics
1 answer:
Step2247 [10]2 years ago
3 0

Answer: 4, 2, 3, 5, 1

Step-by-step explanation:

First, you need to set your compass width or set the compass at one of the terminal points.

After, you can draw arcs above and below.

Then, repeat for the other terminal point.

Finally, connect the intersection of the arcs.

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If f(x)= 0, what is x?
crimeas [40]

Answer:

X is the input to the function ... it appears that it is unrestricted...

it can be anything from -∞ to + ∞ (the domain) ..

in every case the output (range) is "0"

Step-by-step explanation:

7 0
3 years ago
HELP ME PLEASEEEE I BEG YOUUU
Alex777 [14]

Answer:

1.a. E>D

b. E<F

2. Yes. There are multiple combinations on how to do number 1. The first question asks how for a, you need a letter that is bigger than another letter, and for b, you need a letter that is smaller than the other letter's value. There are many ways to go about that previous problem.

Step-by-step explanation:

Hello!

To solve this question, we first need to realize that this is a simple number line. A number line is just a line that contains multiple points from the left being negative, middle is zero, and right becoming increasingly positive.

In this case, D is zero, and A-C are deemed negative. E-F are positive.

For a, we need to find a point that is bigger than another point.

We have various answers for this. F is bigger than A-E, E is bigger than A-D, D is bigger than A-C, C is bigger than A-B, and B is bigger than A. As long as any of your answer goes along this format it is correct.

B is asking a number that is smaller than that said number. A is smaller to B-F, B is smaller to C-F, C is smaller to D-F, E is smaller to F. So again, any letter combination that falls within this category is correct.

For the last one, there are multiple combinations, meaning there is just not one answer that fulfills that question.

6 0
2 years ago
Use the recursive formula to answer the question.
Oliga [24]

Answer:

5

Step-by-step explanation:

7 0
2 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
The formula y equals 15 left parenthesis 1.26 right parenthesis to the power of x gives the number of cellular phone users y (in
Alexus [3.1K]

Answer:

2.3 billion

Step-by-step explanation:

7 0
3 years ago
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