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Rainbow [258]
1 year ago
14

Determine whether the statement is true or false. If f '(c) = 0, then f has a local maximum or minimum at c.

Physics
1 answer:
ankoles [38]1 year ago
4 0

It is False that, If f '(c) = 0, then f has a local maximum or minimum at c.

Local maximum and minimum points are very distinctive on the graph of a function and are thus, helpful in grasping the shape of the graph. Either a local minimum or a local maximum can be considered a local extremum.

A counterexample can be used for:

f(x) = x³

f'(x) = 2 × x²

and,

f'(0) = 2×0² = 0

However, the assertion is false because x = 0 is actually an inflection point rather than a maximum or minimum in f(x) = x³.

Know more about f(x) here: brainly.com/question/28058540

#SPJ4

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Please help
Allisa [31]

<u>We are given:</u>

Mass of the rocket = 10 kg

Weight of the Rocket = 100 N

Upward thrust applied by the rocket = 400 N

<u>Net upward force on the rocket:</u>

We are given that gravity pulls the rocket with a force of 100 N

Also, the rocket applied a force of 400N against gravity

Net upward force = Upward thrust - Force applied by gravity

Net upward force = 400 - 100

Net upward force = 300 N

<u>Upward Acceleration of the Rocket:</u>

From newton's second law:

F = ma

<em>replacing the variables</em>

300 = 10 * a

a = 30 m/s²

5 0
3 years ago
What force must be used to do 224 Joules of work on an object over a distance of 32 meters?
cestrela7 [59]
7.625 Newtons

work = force× distance
Newtons is an accepted value for force

so take the total 224 joules and decide by distance 32 meters to find force in Newtons
4 0
3 years ago
Two identical balls are thrown vertically upward. the second ball is thrown with an initial speed that is twice that of the firs
Temka [501]
The motion of the ball on the vertical axis is an accelerated motion, with acceleration 
a=g=-9.81 m/s^2
The following relationship holds for an uniformly accelerated motion:
2aS=v_f^2 - v_i^2
where S is the distance covered, vf the final velocity and vi the initial velocity.

If we take the moment the ball reaches the maximum height (let's call this height h), then at this point of the motion the vertical velocity is zero:
v_f =0
So we can rewrite the equation as
2(-9.81 m/s^2) h=-v_i^2
from which we can isolate h
h= \frac{v_i^2}{19.62} (1)

Now let's assume that v_i is the initial velocity of the first ball. The second ball has an initial velocity that is twice the one of the first ball: 2v_i. So the maximum height of the second ball is
h= \frac{(2v_i)^2}{19.62}= \frac{4v_i^2}{19.62} (2)

Which is 4 times the height we found in (1). Therefore, the maximum height of ball 2 is 4 times the maximum height of ball 1.
8 0
3 years ago
What type of specialized training is necessary to learn the skills needed to work in this career field
Nonamiya [84]

Answer:

DONTKNOW NEED CAREER FILD FIRST TO ANWSER

Explanation:

5 0
3 years ago
Whos the best character in assassination classroom
lutik1710 [3]
Karma, and sensei. Maybe nagisa
6 0
3 years ago
Read 2 more answers
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