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Alexus [3.1K]
3 years ago
11

What distance will the divider be set at the scribe a 02.8" circle? .75" or 1.4" or 2.8" or 5.6"

Mathematics
1 answer:
lbvjy [14]3 years ago
7 0

9514 1404 393

Answer:

  (b)  1.4"

Step-by-step explanation:

A 2.8" circle is one that has a <em>diameter</em> of 2.8". The divider will be set to the radius, which is half the diameter.

The divider is set to 1.4".

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A postulate isn't required to be proven
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What is the answer a or b to this question
galina1969 [7]
The answer would be A.

Explanation:

If they enter with one car, that’s a price of $3 for the car alone. Because there’s also a fee for each person who enters, and there is a family of 5 who enters, x is representative of the cost for each person who enters. Therefore, there would be 5 x’s and one 3 that when solved would add up to the total of $45 they spent. That would mean A is correct because the equation would be x + x + x + x + x + 3 = 45.
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2 years ago
What number is between 3/4 and 6/5​
Sladkaya [172]

Answer:

19.5/20

Step-by-step explanation:

3/4=15/20

6/5=24/20

--|-----|-----|----|-----|-----|------|----|-----|----|-----------

15 16 17 18 19 20 21 22 23 24

5 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
2 people walk 3 meters each opposite direction They both turn right &amp; walk 4 meters each What is distance between them now
Andrej [43]
I think it would be 8 meter
3 0
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