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Oduvanchick [21]
2 years ago
12

The numbers 1, 2, 3, 4, 5, 6, 7, and 8 are arranged to form four two-digit numbers. What is the largest amount of primes that co

uld possibly be among these four numbers?
Mathematics
1 answer:
artcher [175]2 years ago
4 0
<h3>Answer:  3 primes</h3>

==========================================================

Reason:

The only even prime number is 2. The other primes are odd.

The list of the first few primes are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43

A prime is any number that has factors of 1 and itself, and nothing else.

For instance, 13 is prime because 1 and 13 are the only factors.

-----------

As you can see, if we want a two digit prime number, then the units digit must be odd. Otherwise, 2 is a factor making it not prime (aka composite).

So far we see that the units digit is 1, 3, 5, 7, or 9. But wait, if 5 is the units digit then 5 is a factor. Eg: 5 is a factor of 35 since 5*7 = 35. Refer to the divisibility by 5 rule.

So we reduce the list of choices to 1, 3, 7, 9 for the units digit.

Unfortunately 9 is not in the list of original numbers given, so we can't use it. We really have 1, 3, or 7 as our choices to form the units digit of the two-digit number.

-----------

Let's try to build some primes.

Pick the smallest odd number 1 as the units digit. Then pick 2 as the tens digit. The number 21 is composite because 21 = 7*3, so we rule it out.

On the other hand, 31 is prime since only 1 and 31 are factors.

The problem is that now "3" is tied up and cannot be used for another prime. The good news is that 41 is prime and that's what I'll go with.

Cross 1 and 4 off the list.

The next odd number is 3 which is our units digit. The value 23 is prime.

So far we have 41 and 23 as our two primes.

Like mentioned earlier, we cannot use 5 as the units digit. The number 57 is not prime because 57 = 19*3. So we'll skip over 5.

The number 67 is prime for similar reasoning mentioned earlier.

------------

We have these primes: 41, 23, 67

The next number either 58 or 85 isn't prime

This is one way to show an example of why we're only able to get 3 primes out of this.

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Pls help me I'm in neeeeed :(​
-BARSIC- [3]

Answer:

x = 30

Step-by-step explanation:

x+2x= 90-------> 3x=90

Divide by 3

x= 30

Hope this helps! Pls mark brainliest if correct:)

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2 years ago
What is the radius of the circle?
natima [27]

Answer: The circumference of a circle is equal to Pi π times the diameter d . Since the diameter d is equal to 2 times the radius r , the formula for circumference using radius is 2πr 2 π r .

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Please show the steps​
notka56 [123]

Answer:

  x = -3, 0, or 7

Step-by-step explanation:

After removing common factors, the remaining quadratic can be factored by comparison to the factored form of a quadratic.

__

<h3>Step 1</h3>

Write the equation in standard form.

  4x³ -16x² -84x = 0

<h3>Step 2</h3>

Factor out the common factor from all terms.

  = 4x(x² -4x -21) = 0

<h3>Step 3</h3>

Compare to the factored form of a quadratic:

  (x +a)(x +b) = x² +(a+b)x +ab

This tells you the constants 'a' and 'b' in the factors can be found by considering ...

  (a+b) = -4 . . . . the coefficient of the x term of the quadratic

  ab = -21 . . . . . the constant term of the quadratic

It is often helpful to list factor pairs of the constant:

  -21 = (-21)(1) = (-7)(3) . . . . integer pairs that have a negative sum

The sums of these pairs are -20 and -4. We are interested in the latter. We can choose ...

  a = -7, b = 3

<h3>Step 4</h3>

Put it all together.

  4x³ -16x² -84 = (4x)(x -7)(x +3) = 0 . . . . . factored form of the equation

<h3>Step 5</h3>

Apply the zero product rule. This rule tells you the product of factors will be zero when one or more of the factors is zero:

  4x = 0   ⇒   x = 0

  x -7 = 0   ⇒   x = 7

  x +3 = 0   ⇒   x = -3

Solutions to the equation are x ∈ {-3, 0, 7}.

_____

<em>Additional comment</em>

What we did in Step 3 is sometimes referred to as the X-method of factoring a quadratic. The constant (ab product) is put at the top of the X, and the sum (a+b) is put at the bottom. The sides of the X are filled in with values that match the product and sum: -7 and 3. The method is modified slightly if the coefficient of x² is not 1.

A graphing calculator often provides a quick and easy method of finding the real zeros of a polynomial.

3 0
2 years ago
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arlik [135]
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3 years ago
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7x to the 2nd power when x=6
topjm [15]
Your answer would be 252 Hope this helps
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