Let the length of unknown side be x
So, hypotenuse = 2x + 1
By pythagorean theorem ie , a² = b² + c²
(2x + 1)² = x² + 15²
4x² + 1 + 4x = x² + 225
3x² + 4x -224
(3x +28)(x-8)
x = -9.33 , 8
Since, length can't be negative
x = 8
∴Unknown side length = 8 m
Length of hypotenuse = 2 x 8 + 1 = 17 m
Hope it helps now.
I think B not 100%shure ,you could prob go ask this on yahoo and get a bit more help ;) sorry I couldent help more
Answer:
(0,0)
Step-by-step explanation:
y value divided by x value for slope, then use desmos
Answer:
(i) (f - g)(x) = x² + 2·x + 1
(ii) (f + g)(x) = x² + 4·x + 3
(iii) (f·g)(x) = x³ + 4·x² + 5·x + 2
Step-by-step explanation:
The given functions are;
f(x) = x² + 3·x + 2
g(x) = x + 1
(i) (f - g)(x) = f(x) - g(x)
∴ (f - g)(x) = x² + 3·x + 2 - (x + 1) = x² + 3·x + 2 - x - 1 = x² + 2·x + 1
(f - g)(x) = x² + 2·x + 1
(ii) (f + g)(x) = f(x) + g(x)
∴ (f + g)(x) = x² + 3·x + 2 + (x + 1) = x² + 3·x + 2 + x + 1 = x² + 4·x + 3
(f + g)(x) = x² + 4·x + 3
(iii) (f·g)(x) = f(x) × g(x)
∴ (f·g)(x) = (x² + 3·x + 2) × (x + 1) = x³ + 3·x² + 2·x + x² + 3·x + 2 = x³ + 4·x² + 5·x + 2
(f·g)(x) = x³ + 4·x² + 5·x + 2
