The distance between two (4, 6) and (5, 0) is 6.08 units.
Solution:
The given points are (4, 6) and (5, 0).
![x_1=4, y_1=6, x_2=5, y_2=0](https://tex.z-dn.net/?f=x_1%3D4%2C%20y_1%3D6%2C%20x_2%3D5%2C%20y_2%3D0)
<u>To find the distance between two points:</u>
Distance formula:
![d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
![d=\sqrt{(5-4)^2+(0-6)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%285-4%29%5E2%2B%280-6%29%5E2%7D)
![d=\sqrt{1^2+(-6)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B1%5E2%2B%28-6%29%5E2%7D)
The value of 1² = 1
The value of (–6)² = 36
![d=\sqrt{1+36}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B1%2B36%7D)
![d=\sqrt{37}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B37%7D)
d = 6.08
The distance between two (4, 6) and (5, 0) is 6.08 units.
It depreciates at 15%, which means it would be valued at 85% of the previous value.
After one year the value would be 9000 x 0.85 = £ 7650
Vas happenin!
1.74 x 10^6
Hope this helps
-Zayn Malik
Answer:
Step-by-step explanation:
y ∝ x^2
Introducing the proportionality constant, we have
y = kx^2
Given : y= 18 when x = 3
substitute the given values in order to get the constant
i.e 18 = k x 3^2
18 = 9k
k = 2
therefore the formula connecting x and y
⇒ y = 2x^2
To find y if x is 4, just substitute x = 4 into the formula connecting x and y
i.e y = 2 x 4^2
= 2 x 16
= 32