R = 230 m
v = 17 m/s
Formula: |Ac| = v^2 / r
Point A
|Ac| = v^2 / r = (17m/s)^2 / (230m) = 1.257 m/s^2
Components:
cos(25) = - [ x-component / |Ac| ] => x-component = -|Ac|*cos(25)
sin(25) = - [y-component / |Ac| ] => y-component = - |Ac|*sin(25)
x-component = - 1.256 * cos(25) m/s^2 = - 1.139 m/s^2
y-component = -1.256 * sin (25) m/s^2 = - 0.531 m/s^2
Point B
|Ac| = 1.256
x-component = - 1.256 cos(58) = - 0.666 m/s^2
y-component = + 1.256 sin(58) = + 1.065 m/s^2
Answer:
Step-by-step explanation:
In the triangle ABC, m∠ABC=90°. Hence, it is right angle triangle.
--------1
In the triangle ABD, m∠ABD=90°. Hence, it is right angle triangle.
As 
, so
---------2
From equation 1 and 2,
Answer:
101 degrees
Step-by-step explanation:
both AB and CD are 13.2 in length, so the values of their arcs are equal as well
Answer:
x = 2
As x = 12/6 = 2
Step-by-step explanation:
3x-4+3x-8
3x+3x-4-8
= 6x-12
FACTORISING for check and set to zero
6(x-2)
6x -12 = 0
Cancel out left by adding 12 to both sides then applying /6 to both sides to cancel 6 left side.
6x--12 + 12 = 0 + 12
6x / 6 = 12 / 6
x = 12/6
x = 2
or balance as grouping by separation = multiply at end by -1
3x+3x = --4 - 8
6x = - 12
x = - 12/6
x = - 2
x = - 2 ( -1 )
x = 2
Although I am not clear as to whether the radius is r—1 or otherwise, but if i work with the values i see then:
Volume — T. S. A.
= πr²h — (2πr²+2πrh)
So after the simplification, I substitute the values given,
r= r—1 and h=h—1
And the answer is...
π(r—1)(rh—3r—3h+5)!
