All categories

3 years ago

Between each whole number does √199

Mathematics

1 answer:

Gala2k [10]3 years ago

7 0

between 14 and 15

because

14^2 = 196

15^2 = 225

You might be interested in

Is pi/pi irrational

dedylja [7]

Answer:

No, $\frac{\pi}{\pi}$ is 1 which can be written as $\frac{1}{1}$ . The definition of rationals is any number that can be written as a ratio of an integer to an integer where the bottom integer is not 0.

Step-by-step explanation:

$\frac{\pi}{\pi}=1$ since $1(\pi)=\pi$ .

Rational numbers included anything that can be written as a fraction where the top and bottom are integers (bottom integer is not 0).

1 can be written many different ways. It can be written as $\frac{1}{1}$ .

Both the numerator and the denominator of the rewrite of one that I did are integers.

The integers include the following numbers:

{...,-4,-3,-2,-1,0,1,2,3,4,...}.

The integers are the whole numbers and the opposite of the whole numbers.

The whole numbers are {0,1,2,3,4,...}.

The opposite of the whole whole numbers are {...,-4,-3,-2,-1,0}.

3 0

3 years ago

3T= _______lbs

sveticcg [70]

Answer:

Step-by-step explanation:

Hope it is helpful...

3 0

3 years ago

What is the vaule of 0.77

padilas [110]

<span>seventy-seven hundredths.</span>

3 0

3 years ago

(1/2)x x 3/43=<br><br><br> Pls im so close for finishing 20 POINTS

Digiron [165]

3/86 I believe because; 1/2 x 3/43 = 1x3 and 2x43 as denominator = 3/86 MULTIPLY both numerators and denominators.

3 0

3 years ago

Read 2 more answers

Suppose x is a real number and epsilon > 0. Prove that (x - epsilon, x epsilon) is a neighborhood of each of its members; in

kaheart [24]

Answer:

See proof below

Step-by-step explanation:

We will use properties of inequalities during the proof.

Let $y\in (x-\epsilon,x+\epsilon)$ . then we have that $x-\epsilon$ . Hence, it makes sense to define the positive number delta as $\delta=\min\{x+\epsilon-y,y-(x-\epsilon)\}$ (the inequality guarantees that these numbers are positive).

Intuitively, delta is the shortest distance from y to the endpoints of the interval. Now, we claim that $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$ , and if we prove this, we are done. To prove it, let $z\in (y-\delta,y+\delta)$ , then $y-\delta$ . First, $\delta \leq y-(x-\epsilon)$ then $-\delta \geq -y+x-\epsilon$ hence $z>y-\delta \geq x-\epsilon$

On the other hand, $\delta \leq x+\epsilon-y$ then $z$ hence $z$ . Combining the inequalities, we have that $x-\epsilon$ , therefore $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$ as required.

3 0

3 years ago