Answer: x=5
Step-by-step explanation:
Answer:
0.28cm/min
Step-by-step explanation:
Given the horizontal trough whose ends are isosceles trapezoid
Volume of the Trough =Base Area X Height
=Area of the Trapezoid X Height of the Trough (H)
The length of the base of the trough is constant but as water leaves the trough, the length of the top of the trough at any height h is 4+2x (See the Diagram)
The Volume of water in the trough at any time
=8h(8+2x)
V=64h+16hx
We are not given a value for x, however we can express x in terms of h from Figure 3 using Similar Triangles
x/h=1/4
4x=h
x=h/4
Substituting x=h/4 into the Volume, V
h=3m,
dV/dt=25cm/min=0.25 m/min
=0.002841m/min =0.28cm/min
The rate is the water being drawn from the trough is 0.28cm/min.
Try graphing (or looking up) the tangent function. The graph of this function passes through the origin; the segment of the graph that does so has domain
(-pi/2, pi/2).
The equation of that function is simply y = tan x.
If we translate the entire graph upward by 2 units, we get the function
y = tan x + 2.
Y=mx+c
M=y1-y2/x1-x2
=26-79/45-40
=-1.3
Y=-1.3x+c
Replace any pair of x and y values in equation to get c
79=-1.3(5)+c
79+1.3=c
c=80.3
Y=-1.3x+80.3
Id choose the first option even tho the answer i did isnt up there but im guessing c is different for each point, but the slope is the same