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3 years ago

14

What is the smallest integer greater than the square root of 68?

2 answers:

Talja [164]3 years ago

7 0

This question is essentially just determining whether you understand the definition of integers and square roots. The square root of a quantity is a number that when multiplied with itself gives that quantity. The square root of 68 is expressed as:

√68 = 8.25

The square root of 68 is roughly 8.25. Now, an integer is simply a whole number, which is a number that is not a fraction. Therefore, the smallest integer that is greater than 8.25 is simply the next whole number, which is 9.

The smallest integer greater than the square root of 68 is 9.

const2013 [10]3 years ago

5 0

What is the smallest integer greater than the square root of 68?

The answer is 9.

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3 years ago

Find the distance between the complex numbers. z1 = 9 - 9i z2 = 10 - 9i

steposvetlana [31]

Answer:

The distance between the two given complex numbers = 9

Step-by-step explanation:

<u><em>Explanation</em></u>:-

<u><em>Step(i):</em></u>-

Given Z₁ = 9 - 9 i and Z₂ = 10 -9 i

Let A and B represent complex numbers Z₁ and Z₂ respectively on the argand plane

⇒ A = Z₁ = x₁ +i y₁ = 9 - 9 i and

B = Z₂ = x₂+ i y₂ = 10 -9 i

Let (x₁ , y₁) = ( 9, -9)

(x₂, y₂) = (10, -9)

<u>Step(ii)</u>:-

<em>The distance between the two points are </em>

A B = $\sqrt{(x_{2} - x_{1})^{2}+(y_{2} - y_{1})^{2} }$

A B = $\sqrt{(10 - 1)^{2}+(-9 - (-9))^{2} }$

AB = $\sqrt{(9)^{2} +(-9+9)^{2} }$

<em> AB = √81 = 9</em>

<u><em>Conclusion:-</em></u>

The distance between the two given complex numbers = 9

<u><em></em></u>

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3 years ago