Answer:
13t² + 2t + 9
Step-by-step explanation:
(9t² - 3t + 5) - (-4t² + 5t + 4)
Solving like terms
-(-4t²) becomes + 4t²
(9t² + 4t²) + (-3t + 5t) + (5+4) =
13t² + 2t + 9
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-Chetan K
Answer:
angle NMP = 63°
angle LMP = 74°
Step-by-step explanation:
Let angle NMP be x° . It's given that angle LMP is 11° more than angle NMP. So, angle LMP = x° + 11°
But angle NML = 137°.
So,
angle NMP + angle LMP = 137°
=> x° + x° + 11° = 137°
=> 2x° + 11° = 137°
=> 2x° = 137° - 11°
= 126°
=> x° = 126/2 = 63°
angle NMP = 63°
angle LMP = 63 + 11 = 74°
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
Answer:
25
Step-by-step explanation:
Answer:
-6
Explanation: use a calculator lol
