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3 years ago

Find the distance between 2-4i and d 6+i

Mathematics

1 answer:

liq [111]3 years ago

8 0

Answer:

7.211

Step-by-step explanation:

-For two points in the complex plane, the distance between the points is the modulus of the of the difference of the two complex numbers.

-Point 2-4i has the coordinates (2,-4)

-Point 6+i has the coordinates (6,1)

#We must find the distance between the two coordinates (2,-4) and (6,1):

$=\sqrt{(6-2)^2+(1--4)^2}\\\\=\sqrt{4^2+6^2}\\\\=\sqrt{52}\\\\=7.211$

Hence, the distance between the two points is 7.211

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I need help with this .

vaieri [72.5K]

I would be glad to help you! I will solve through 3 to get you started.

This is a very important concept, so it would be helpful if you learn it on your own.

C= 2* pi* r

1. The radius is 5 cm. 5 cm represents r in the equation.
C= 2* 3.14 (or pi)* 5
Multiply it by using a calculator.
My answer was 31.4.

Let's try the second one.
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3 years ago

Can you see an easy way to work out the value of <img src="https://tex.z-dn.net/?f=%5Csqrt%7B%201%5E3%20%2B%202%5E3%20%2B%203%5E

Lina20 [59]

We have the sum of cubes identity

$a^3 + b^3 = (a + b) (a^2 - ab + b^2)$

and observing that 1 + 4 = 2 + 3, we have

$1^3 + 4^3 = (1 + 4) (1^2 - 4 + 4^2) = 5 \times 13$

and

$2^3 + 3^3 = (2 + 3) (2^2 - 6 + 3^2) = 5 \times 7$

Then

$\sqrt{1^3+2^3+3^3+4^3+5^3} = \sqrt{5\times13 + 5\times7 + 5\times5^2} = \sqrt{5 \times (20 + 25)} \\\\ ~~~~~~~~ = \sqrt{5^2 \times (4 + 5)} = \sqrt{5^2\times9} = \sqrt{5^2\times3^2} = 5\times3 = \boxed{15}$

Alternatively, we have the well-known sum of cubes formula

$\displaystyle \sum_{i=1}^n i^3 = \frac{n^2(n+1)^2}4$

The sum under the square root is this sum with $n=5$ . Then

$1^3+2^3+3^3+4^3+5^3 = \dfrac{5^2\times6^2}4 = 225 = 15^2$

and so the square root again reduces to 15.

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2 years ago