Height of the woman = 5 ft
Rate at which the woman is walking = 7.5 ft/sec
Let us assume the length of the shadow = s
Le us assume the <span>distance of the woman's feet from the base of the streetlight = x
</span>Then
s/5 = (s + x)/12
12s = 5s + 5x
7s = 5x
s = (5/7)x
Now let us differentiate with respect to t
ds/dt = (5/7)(dx/dt)
We already know that dx/dt = 7/2 ft/sec
Then
ds/dt = (5/7) * (7/2)
= (5/2)
= 2.5 ft/sec
From the above deduction, it can be easily concluded that the rate at which the tip of her shadow is moving is 2.5 ft/sec.
For this case we can make the following rule of three:
9 1/2 m --------------> 28 1/2 rupees
26 1/3 m -------------> x
From here, we clear the value of x.
We have then:
Rewriting we have:
Answer:
samruth should pay 79 rupees to the shopkeeper
Answer: 60 yd
Step-by-step explanation:
Here: subdivided into 3 smaller pens (2w + w + w) extra widths for separation purposes (3 smaller pens)
180 yards of fencing material
180yd = 2L + 4w
180dy = 2*60 + 4w
60dy = 4w
w = 15yd, how wide can the rectangle can be with L = 60yd
Answer:
-40
Step-by-step explanation:
f(a) = a³ - a² + a - 1
f(-3) = (-3)³ - (-3)² + (-3) - 1
f(-3) = (-27) - (9) + (-3) -1
f(-3) = -27 - 9 -3 - 1
f(-3) = -40
