Answer:

Step-by-step explanation:
The formula for the length of a vector/line in your case.
![L = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = \sqrt{[4 - (-1)]^2 + [2 -(-3)]^2} = \sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2}](https://tex.z-dn.net/?f=L%20%3D%20%5Csqrt%7B%28x_2-x_1%29%5E2%20%2B%20%28y_2-y_1%29%5E2%7D%20%3D%20%5Csqrt%7B%5B4%20-%20%28-1%29%5D%5E2%20%2B%20%5B2%20-%28-3%29%5D%5E2%7D%20%3D%20%5Csqrt%7B5%5E2%20%2B%205%5E2%7D%20%3D%20%5Csqrt%7B50%7D%20%3D%205%5Csqrt%7B2%7D)
Answer:
a) 5.83 cm
b) 34.45°
Step-by-step explanation:
a) From Pythagoras theorem of right triangles, given right triangle ABC:
AB² + BC² = AC²
Therefore:
AC² = 5² + 3²
AC² = 25 + 9 = 34
AC = √34
AC = 5.83 cm
b) From triangle ACD, AC = 5.83 cm, AD = 4 cm and ∠A = 90°.
From Pythagoras theorem of right triangles, given right triangle ACD:
AD² + AC² = DC²
Therefore:
DC² = 5.83² + 4²
DC² = 34 + 16 = 50
DC = √50
DC = 7.07 cm
Let ∠ACD be x. Therefore using sine rule:

Answer:
Step-by-step explanation:
-3x^4 -9x^3+ 15x^2
-3x^2 ( x^2 + 3x -5)
if you use quadratic formula (see image)
~~~~~~~~~~~~~~~~~~~~~~
10x^3 -35x^2 +6x -21
5x2(2x-7) +3(2x-7)
(5
+ 3) (2x-7)
10LW=V
L/10W= V/10W
L= V/10W