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ololo11 [35]
3 years ago
6

If f(1)=1 and f(n)=(n-2)^2 +n then find the value of f(3)

Mathematics
1 answer:
Blababa [14]3 years ago
4 0

Answer:

f(3) =4

Step-by-step explanation:

f(n)=(n-2)^2 +n

We want to find f(3)

Put in n=3

f(3) = (3-2) ^2 +3

f(3) = 1^2 +3

f(3) =1+3

f(3) =4

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What is the standard error of the sampling distribution of sample proportion?.
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Hope this helps you! Feel free to give feedback

5 0
1 year ago
33. Suppose you were on a planet where the
tatyana61 [14]

<u>Answer:</u>

a) 3.675 m  

b) 3.67m

<u>Explanation:</u>

We are given acceleration due to gravity on earth =9.8ms^-2

And on planet given = 2.0ms^-2

A) <u>Since the maximum</u><u> jump height</u><u> is given by the formula  </u>

\mathrm{H}=\frac{\left(\mathrm{v} 0^{2} \times \sin 2 \emptyset\right)}{2 \mathrm{g}}

Where H = max jump height,  

v0 = velocity of jump,  

Ø = angle of jump and  

g = acceleration due to gravity

Considering velocity and angle in both cases  

\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 2}{\mathrm{g} 1}

Where H1 = jump height on given planet,

H2 = jump height on earth = 0.75m (given)  

g1 = 2.0ms^-2 and  

g2 = 9.8ms^-2

Substituting these values we get H1 = 3.675m which is the required answer

B)<u> Formula to </u><u>find height</u><u> of ball thrown is given by  </u>

 \mathrm{h}=(\mathrm{v} 0 * \mathrm{t})+\frac{\mathrm{a} *\left(t^{2}\right)}{2}

which is due to projectile motion of ball  

Now h = max height,

v0 = initial velocity = 0,

t = time of motion,  

a = acceleration = g = acceleration due to gravity

Considering t = same on both places we can write  

\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 1}{\mathrm{g} 2}

where h1 and h2 are max heights ball reaches on planet and earth respectively and g1 and g2 are respective accelerations

substituting h2 = 18m, g1 = 2.0ms^-2  and g2 = 9.8ms^-2

We get h1 = 3.67m which is the required height

6 0
3 years ago
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