Question

Pls help with year 8 maths

Answer 1

Answer:

see explanation

Step-by-step explanation:

Given a point (x, y ) then its distance from the origin is

d = $\sqrt{x^2+y^2}$

(a) (2, 3 ) , then

d = $\sqrt{2^2+3^2}$

   = $\sqrt{4 + 9}$

   = $\sqrt{13}$ ≈ 3.6 units ( to 1 dec. place )

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(b) ( - 5, 7 ) , then

d = $\sqrt{(-5)^2+7^2}$

   = $\sqrt{25+49}$

    = $\sqrt{74}$ ≈ 8.6 units ( to 1 dec. place )

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(c) (- 1, 4 )

d = $\sqrt{(-1)^2+4^2}$

   = $\sqrt{1+16}$

    = $\sqrt{17}$ ≈ 4.1 units ( to 1 dec. place )

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(d) (- 4, - 5 ) , then

d = $\sqrt{(-4)^2+(-5)^2}$

   = $\sqrt{16+25}$

   = $\sqrt{41}$ ≈ 6.4 units ( to 1 dec. place )

Answer 2

Answer:

a) 1

b)11

c)4

d)1

I hope it helps you...