Answer:
6
Step-by-step explanation:
We can use the geometric mean theorem:
The altitude on the hypotenuse is the geometric mean of the two segments it creates.
In your triangle, the altitude is the radius CM and the segments are AC and BC.
Answer:
-19k ≠ 0
No solutions
Step-by-step explanation:
k/4 - 5k + 1 = 1
4 (k/4 - 5k + 1 = 1)
k-20k + 4 = 4
-19k ≠ 0
You have a 2/6 chance of rolling <span>a number greater than 4 or less than 3 which reduces to 1/3
2/6 = 1/3</span>
Answer:
Step-by-step explanation:
From the picture attached,
a). Triangle in the figure is ΔBCF
b). Since, 
and 
are the parallel lines and m is a transversal line,
m∠FBC = m∠BFG [Alternate interior angles]
Since, and are the parallel lines and n is a transversal line,
m∠BCF = m∠CFE [Alternate interior angles]
By triangle sum theorem in ΔBCF
m∠FBC + m∠BCF + m∠BFC = 180°
From the properties given above,
m∠BFG + m∠CFE + m∠BFC = 180°
m∠GFE = 180°
Therefore, angle GFE is the straight angle that will be useful in proving that the sum of the measures of the interior angles of the triangle is 180°.
<span>The maxima of a differential equation can be obtained by
getting the 1st derivate dx/dy and equating it to 0.</span>
<span>Given the equation h = - 2 t^2 + 12 t , taking the 1st derivative
result in:</span>
dh = - 4 t dt + 12 dt
<span>dh / dt = 0 = - 4 t + 12 calculating
for t:</span>
t = -12 / - 4
t = 3
s
Therefore the maximum height obtained is calculated by
plugging in the value of t in the given equation.
h = -2 (3)^2 + 12 (3)
h =
18 m
This problem can also be solved graphically by plotting t
(x-axis) against h (y-axis). Then assigning values to t and calculate for h and
plot it in the graph to see the point in which the peak is obtained. Therefore
the answer to this is:
<span>The ball reaches a maximum height of 18
meters. The maximum of h(t) can be found both graphically or algebraically, and
lies at (3,18). The x-coordinate, 3, is the time in seconds it takes the ball
to reach maximum height, and the y-coordinate, 18, is the max height in meters.</span>
