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Anestetic [448]
3 years ago
9

Find the rate of change for x3. You need to work out the change in f(x)=x3 when x is increased by a small number h to x+h. So yo

u will work out f(x+h)-f(x). Then do some algebra to simplify this. Then divide this by h to get the average rate of change of f(x) between x and x+h.
Physics
1 answer:
Marat540 [252]3 years ago
4 0

Answer:

Explanation:

Given

F(x)=x^3

Rate of change of F(x) is given by

F'(x)=\lim_{h\rightarrow 0}\\\frac{F(x+h)-F(x)}{x+h-x}

F'(x)=\lim_{h\rightarrow 0}\\\frac{(x+h)^3-x^3}{h}

F'(x)=\lim_{h\rightarrow 0}\\\frac{x^3+h^3+3x^2h+3xh^2-x^3}{h}

F'(x)=\lim_{h\rightarrow 0}\\\frac{h^3+3x^2h+3xh^2}{h}

F'(x)=\lim_{h\rightarrow 0}\\h^2+3x^2+3xh

Putting limits

F'(x)=3x^2

                     

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northen henisphere,southern hemisphere, Eastern hemisphere, Western hemisphere.

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3 years ago
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Two masses are attracted by a gravitational force of 7 N.
damaskus [11]

Answer:

The answer to your question is: F  = 0.4375 N. The force will be 16 times lower than with the first conditions.

Explanation:

Data

F = 7 N

F = ?  if the masses is quartered

Formula

F = \frac{Km1m2}{r2}

Process

Normal conditions F = Km₁m₂/r²  = 7              

When masses quartered        F = K(m₁/4)(m₂/4)/r²  = ?

                                                F = K(m₁m₂/16)/r²

                                                F = K(m₁m₂/16r²      = 7/16  = 0.4375 N

3 0
3 years ago
A lady walks 10 m to the north, then she turns and continues walking 30 m due east.
ANEK [815]

Answer:

The distance covered is 40 m and the displacement is 31,6m.

Explanation:

The distance covered is the sum of the two distances (10+30). The displacement is equal to the distance of the hipotenusa of the triangle that the two distances (10 m to north and 30m to east) create. Using the Pythagoras theorem the displacent is equal to the Square root of (30^2 +10^2) .

4 0
3 years ago
What is the only continent with land in all four hemispheres?
dimaraw [331]
It’s either Asia or the America
7 0
3 years ago
At the instant a traffic light turns green, a car starts with a constant acceleration of 1.3 m/s2. At the same instant a truck,
Bas_tet [7]

Answer:

(a) Distance traveled = 75.3846 m

(b) Velocity of car at that instant will be 14 m/sec

Explanation:

We have given acceleration of the car a=1.3m/sec^2

Initial velocity of the cart u = 0 m/sec

(a) According to second equation of motion we know that s=ut+\frac{1}{2}at^2

So distance traveled by car s_c=0\times t+\frac{1}{2}\times 1.3t^2=0.65t^2

As the truck is moving with constant speed

So distance traveled by truck s_t=ut=7t

As the truck overtakes the car

So s_c=s_t

0.65t^2=7t

t=10.769sec

So distance traveled s_c=s_t=7\times 10.769=75.3846m

(b) From second equation of motion we know that v = u+at

So v = 0+1.3×10.769 = 14 m /sec

7 0
3 years ago
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