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Natali [406]
3 years ago
11

Frank says that for any whole number n, the value of 6n-1 is always prime. Is Frank correct? Explain your answer.​

Mathematics
1 answer:
Lady bird [3.3K]3 years ago
8 0

Answer:

Frank is wrong

Step-by-step explanation:

6(6) - 1 = 35

35 is a multiply by 5 so it is not a prime number.

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KiRa [710]

Answer:

Not clear of the question

Step-by-step explanation:

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3 0
2 years ago
Hi what is the answer for this?
saw5 [17]
This is a common factor problem.

Pencils come in a pack of 12
Erasers come in a pack of 10

First, break the number into their prime factors(the idea is that we will break the number down into its smallest multiples, which are prime numbers):
10 = 2 * 5
12 = 2 * 2 *3

So now we take the unique multiples of each number, and when we multiply them together, we will get the smallest number that both 10 and 12 can be divided into(this is what the problem is asking for)
We have (2*2*3) that comes from 12, and the only unique number that comes from the 10 is (5)
So now, we multiply:
2*2*3*5=60
However, this isn't exactly out answer. Now we have to divide our answer by the number of each this in the pack to know how many packs to buy.
60/12=5 packs of pencils
60/10= 6 packs of erasers

I hope this helps. Let me know if you have any questions!!
5 0
3 years ago
Read 2 more answers
which strategy do you prefer to use to multiply with multiples of 10: base ten blocks, a number line, or place value? Explain wh
ladessa [460]
Place value. when multiplying by 10 you are moving a decimal, having everything lined up like place value would save you a lot of time.
7 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
Please help ASAP!<br> 3(T + 4) - 2(2T + 3) = -4 <br> Thank You!
kakasveta [241]
The answer would be t=10
5 0
3 years ago
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