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makvit [3.9K]
3 years ago
12

two sets of 3 consecutive integers have a property that the product of the larger two is one less than seven times the smallest

set up and solve an equation that can be used to find both sets of intigers
Physics
1 answer:
Alchen [17]3 years ago
7 0
Answer:
First set is 3,4 and 5
Second set is 1,2 and 3

Explanation:
Assume that the smallest number is n.
Since the three numbers are consecutive, this means that the other two numbers would be n+1 and n+2

Now, we know that the product of the larger two numbers is 1 less tan than 7 times the smallest.
This means that: 
(n+1)(n+2) = 7n - 1 .........> equation used to find the sets
n² + 2n + n + 2 = 7n - 1
n² + 3n + 2 = 7n - 1
n² + 3n + 2 - 7n + 1 = 0
n² - 4n + 3 = 0
(n-3)(n-1) = 0
This means that:
either n-3=0 ..........> n = 3
or n-1=0 ..........> n = 1

At n = 3:
The three numbers would be 3,4 and 5

At n = 1:
The three numbers would be 1,2 and 3

Hope this helps :)

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a hawk flies in a horizontal arc of radius 10.3 m at a constant speed of 4.8 m/s. find its centripetal acceleration. answer in u
n200080 [17]

The hawk’s centripetal acceleration is 2.23 m/s²

The magnitude of the acceleration under new conditions is 2.316 m/s²

radius of the horizontal arc = 10.3 m

the initial constant speed = 4.8 m/s

we know that the centripetal acceleration is given by

    a_{c}  = \frac{v^{2} }{r}

   a_{c}  = 23.04/10.3

    a_{c}  = 2.23 m/s²

It continues to fly but now with some tangential acceleration

a_{t} = 0.63 m/s²

therefore the net value of acceleration is given by the resultant of the centripetal acceleration and the tangential acceleration

so

a_{net}  =  \sqrt{a_{c} ^{2} +a_{t} ^{2}   }

a_{net}  =  \sqrt{4.97 + 0.396}

a_{net}  =  2.316 m/s²

So the magnitude of  net acceleration will become 2.316 m/s².

learn more about acceleration here :

brainly.com/question/11560829

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8 0
1 year ago
A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1 2 and t = 1 s? Use G
Romashka-Z-Leto [24]

There is one mistake in the question.The Correct question is here

A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1/2 and t = 1 s? Use Galileo's formula v(t) = −9.8t m/s.

Answer:

y(1s) - y(1/2s) =  - 3.675 m  

The cat falls 3.675 m between time 1/2 s and 1 s.

Explanation:

Given data

time=1/2 sec to 1 sec

v(t)=-9.8t m/s

To find

Distance

Solution

As the acceleration as first derivative of velocity with respect to time  

So

acceleration(-g)=  dv/dt

Solve it

dv  =  a dt

dv =  -g dt

v - v₀  =  -gt

v=  dy/dt

dy  =  v dt

dy =  ( v₀ - gt ) dt

y(1s) - y(1/2s)  =  ( v₀ ) ( 1 - 1/2 ) - ( g/2 )[ ( t1)² -( t1/2s )² ]

y(1s) - y(1/2s)  = ( - 9.8/2 ) [ ( 1 )² - ( 1/2 )² ]

y1s - y1/2s  = ( - 4.9 m/s² ) ( 3/4 s² )

y(1s) - y(1/2s) =  - 3.675 m  

The cat falls 3.675 m between time 1/2 s and 1 s.

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Explanation:

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