The distance are
a. 42.42
b. 15. 52
<h3>How to solve for the distance</h3>
80 = 65 + <ABC
<ABC = 80 - 65
= 15
Through the use of sine rule we would have
By using sine rule,
Sine 15/a = sine 135/b =
sin 30/30.
Next we have to solve b by cross multiplying
b = (30× sin135) / sin 30
= 21.21/0.5
b = 42.42 km
Also, the value of a will be:
= (42.42 × sin 15) / sin 135
= 10.98/0.7071
= 15.52
Read more on bearing here:
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What we have so far:
Kinetic energy = 0. The reason behind that is because the beam is not moving at a height of 40m.
Gavity, g = 9.8m/s²
Height = 40m
Potential energy = mgh; this is equal to 0 because m, stands for mass and in this problem, we do not have a value for the mass of the beam. Hence, 0 x 9.8m/s² x 40m = 0. Potential energy = 0.
Solution:
We will use the equation of Total energy:
TE = potential energy + kinetic energy
TE = 0 + 0
∴ TE = 0
The answer is: Assuming no air resistance, the total energy of the beam as it hits the ground is 0.
Answer:
The equation is y = 2x + 11.
Step-by-step explanation:
It is given that the gradient of the equation is 2. Using slope-form formula, y = mx+b where m is gradient and b is y-intercept. In order to find b, you have to substitute x-coordinate and y-coordinate into the equation :
y = mx + b
m = 2
At(-4,3),
3 = 2(-4) + b
b = 3 - 2(-4)
= 11
F(x)=x^2+3x+5
f(3+h)=(3+h)^2+3(3+h)+5
f(3+h)=9+6h+h^2+9+3h+5
f(3+h)=23+9h+h^2
