Which number below is an integer but not a natural number?

Find the locus of a point which moves such that its distance from (1,2) is always 5 units

Rina8888 [55]

Answer:

TbH my brain is to tired lol

Step-by-step explanation:

sorry

8 0

2 years ago

Read 2 more answers

6500 is what percent of 32,500?

blagie [28]

i belive its 6500 is 20% of 32500

3 0

3 years ago

Read 2 more answers

The measure of angle θ is . The measure of its reference angle is _°, and tan θ is _?

defon

Answer:

0000

Step-by-step explanation:

5 0

3 years ago

What form is 4.6 X 10' written in?

ziro4ka [17]

Answer: Standard Form

4 0

3 years ago

Please answer the two questions!

shutvik [7]

Given:

The expressions are

(c) $\left\{\left(\dfrac{2^4\times 3^6}{12^2}\right)^0\right\}^3$

(d) $\dfrac{13^3\times 7^0}{\{(65\times 49)^2\}^1}$

To find:

The simplified form of the given expression.

Solution:

(c)

We have,

$\left\{\left(\dfrac{2^4\times 3^6}{12^2}\right)^0\right\}^3$

We know that, zero to the power of a non-zero number is always 1. So, $\left(\dfrac{2^4\times 3^6}{12^2}\right)^0=1$

$\left\{\left(\dfrac{2^4\times 3^6}{12^2}\right)^0\right\}^3=(1)^3$

$\left\{\left(\dfrac{2^4\times 3^6}{12^2}\right)^0\right\}^3=1$

Therefore, the value of the given expression is 1.

(d)

We have,

$\dfrac{13^3\times 7^0}{\{(65\times 49)^2\}^1}$

It can be written as

$\dfrac{13^3\times 7^0}{\{(65\times 49)^2\}^1}=\dfrac{13^3\times 1}{(65\times 49)^2}$

$\dfrac{13^3\times 7^0}{\{(65\times 49)^2\}^1}=\dfrac{13\times 13\times 13}{(65\times 49)(65\times 49)}$

$\dfrac{13^3\times 7^0}{\{(65\times 49)^2\}^1}=\dfrac{13}{(5\times 49)(5\times 49)}$

$\dfrac{13^3\times 7^0}{\{(65\times 49)^2\}^1}=\dfrac{13}{60025}$

Therefore, the value of given expression is $\dfrac{13}{60025}$ .

7 0

3 years ago