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3 years ago

Can pie be written as a natural number

1 answer:

7 0

For example, 5 can be written as 5/1. The natural numbers, whole numbers, and integers are all subsets of rational numbers. In other words, an irrational number is a number that can not be written as one integer over another. It is a non-repeating, non-terminating decimal.

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25 POINTS! WILL GIVE BRAINLIEST!

iVinArrow [24]

Answer:

I think the answer is A.

Step-by-step explanation:

A minute is 60 seconds so you would have to see how many times 5 sec and 6 sec goes into 1 minute in order to find out how many times they pass each other in 1 minute.

-I hope this helps!-

-Good luck!-

-Please mark as brainliest!-

-Please let me know if I got it correct or incorrect!-

5 0

3 years ago

Peter states that the roots of a quadratic equation are 6 and -3. He believes its factored form is (x + 6)(x - 3). Is Peter righ

Nat2105 [25]

Answer:

Step-by-step explanation:

If the factored form is (x + 6)(x - 3), then the two roots should be -6 and 3. Whatever the factored numbers are, the opposite sign should be the root in this situation, and vice versa.

Hope this helps!

(please mark brainliest)

7 0

3 years ago

What is the vertex of the quadratic function f(x) = (x-8)(x - 2)?

Sergio [31]

Answer:

The vertex is (5, -9).

Step-by-step explanation:

One way of approaching this problem consists of finding the roots. Seet the quadratic equal to zero and solve for x: {2, 8}.

The axis of symmetry is exactly halfway between 2 and 8: x = 5.

Now find the y-component of the vertex by evaluating f(5):

f(5) = (5 - 8)(5 - 2) = -9

The vertex is (5, -9). This is the minimum value of the function. The graph of this quadratic opens up.

8 0

2 years ago

18.

sdas [7]

p=2(l+b) =2(3x-3) =6x=48=x=8

8 0

3 years ago

Read 2 more answers

If A ⊂ B, then A ∩ B = A ∪ B. always sometimes or never

Lisa [10]

Sometimes. Proof:

Assume $A\subset B$ .

Let $a\in A\cap B$ , so that $a\in A$ and $a\in B$ . Then of course $a\in A\cup B$ , so $A\cap B\subseteq A\cup B$ .

Now let $a\in A\cup B$ , so that $a\in A$ *or* $a\in B$ . Since $A\subset B$ , if $a\in A$ then $a\in B$ , so that $a\in A\cap B$ . But if $a\in B$ , then $a\not\in A$ , so it's not always the case that $a\in A\cap B$ , and therefore in general $A\cup B\not\subseteq A\cap B$ .

6 0

3 years ago