All categories

3 years ago

Is .3333333333333 a rational number??

Mathematics

2 answers:

natali 33 [55]3 years ago

8 0

Answer: No

Step-by-step explanation: Because it is.

loris [4]3 years ago

7 0

Answer:

No .3333333333333 is a rational number

Explanation:

In mathematics, a rational number is any number that can be expressed as the quotient or fraction p/q of two integers, a numerator p and a non-zero denominator q. Since q may be equal to 1, every integer is a rational number.

You might be interested in

Simplify the expression

Paladinen [302]

$\log_4(16^x)=\log_4(4^{2x})=2x.$

4 0

3 years ago

The answer is 48 I just don’t know how to get there. Please solve using proportions and show me

OleMash [197]

Answer:

x=48

Step-by-step explanation:

What number out of 75 gives you 64%

What number out of 75 gives you 64 out of 100

x 64

------------ = -----------

75 100

Using cross product

100x = 64*75

100x =4800

Divide each side by 100

100x/100 = 4800/100

x=48

3 0

3 years ago

If for two non-zero vectors a,b : ∣a+b∣=∣a−b∣ then angle between them is:______a. 0b. 45c. 60d. 90

Eddi Din [679]

Answer:

$90^\circ$

Step-by-step explanation:

Given two non zero vectors, $\vec a, \vec b$ .

Let the angle between the two vectors = $\theta$

Given that:

$|\vec a+\vec b|=|\vec a-\vec b|$

Let us have a look at the formula for magnitude of addition of two vectors:

$|\vec x+\vec y|=\sqrt{x^2+y^2+2xycos\theta}$

Where $\theta$ is the angle between the two vectors.

formula for magnitude of subtraction of two vectors:

$|\vec x-\vec y|=\sqrt{x^2+y^2-2xycos\theta}$

As per the given condition:

$\sqrt{a^2+b^2+2abcos\theta}=\sqrt{a^2+b^2-2abcos\theta}$

Squaring both sides:

$a^2+b^2+2abcos\theta=a^2+b^2-2abcos\theta\\\Rightarrow 2abcos\theta=-2abcos\theta\\\Rightarrow cos\theta = -cos\theta\\\Rightarrow 2cos\theta = 0\\\Rightarrow cos\theta = 0\\\Rightarrow \theta = 90^\circ$