Answer:
- 60<em><u>÷</u></em><em><u>15</u></em><em><u> </u></em><em><u>=</u></em><em><u>4</u></em><em><u> </u></em>
<em><u>Therefore</u></em><em><u> </u></em><em><u>Martin</u></em><em><u> </u></em><em><u>uses</u></em><em><u> </u></em><em><u>his</u></em><em><u> </u></em><em><u>power</u></em><em><u> </u></em><em><u>saw</u></em><em><u> </u></em><em><u>4</u></em><em><u> </u></em><em><u>times</u></em><em><u> </u></em><em><u>in</u></em><em><u> </u></em><em><u>one</u></em><em><u> </u></em><em><u>hour</u></em><em><u> </u></em>
Multiplying ,you get 18x^2+21x-24x-28. You simplify to get:
18x^2-3x-28.
1/6 for the first question and 1/3 for the second
Answer:
x = 7; y = 3
Step-by-step explanation:
I used the Trial and Error Method and found that x = 7 and y = 3
=> 7 + 3 = 10
=> 10 = 10. ✔
=> 7 - 3 = 4
=> 4 = 4. ✔
Therefore, x = 7 and y = 3
Hoped this helped.
Answer:
Dh/dt = 0.082 ft/min
Step-by-step explanation:
As a perpendicular cross section of the trough is in the shape of an isosceles triangle the trough has a circular cone shape wit base of 1 feet and height h = 2 feet.
The volume of a circular cone is:
V(c) = 1/3 * π*r²*h
Then differentiating on both sides of the equation we get:
DV(c)/dt = 1/3* π*r² * Dh/dt (1)
We know that DV(c) / dt is 1 ft³ / 5 min or 1/5 ft³/min
and we are were asked how fast is the water rising when the water is 1/2 foot deep. We need to know what is the value of r at that moment
By proportion we know
r/h ( at the top of the cone 0,5/ 2) is equal to r/0.5 when water is 1/2 foot deep
Then r/h = 0,5/2 = r/0.5
r = (0,5)*( 0.5) / 2 ⇒ r = 0,125 ft
Then in equation (1) we got
(1/5) / 1/3* π*r² = Dh/dt
Dh/dt = 1/ 5*0.01635
Dh/dt = 0.082 ft/min
