The stiffness of one short spring is 13500 N/m
<em><u>Solution:</u></em>
We are given a chain with number of spring N = 30 and linked end to end ( in series) and stiffness of this chain is 450 N/m
We have to find the stiffness of one short spring
The springs are identical, which means they have same stiffness
<em><u>The stiffness of one spring in series is given as:</u></em>
Where,
N is the number of springs
is the stiffness of one spring
is the stiffness of this chain
Substituting,
Therefore,
Thus the stiffness of one short spring is 13500 N/m
The linear equation that gives the height on the x-th month is:
y = 2.5*x + 74.5
<h3>
Which rule gives the height of the tree for the n-th month?</h3>
This will be modeled with a linear equation of the form:
y = a*x + b
Where a is the slope and b is the y-intercept.
Remember that for a line with the points (x₁, y₁) and (x₂, y₂), the slope is given by:
Here we have the two points (3, 82) and (7, 92). Where the first value is the number of months and the second value is the height.
So the slope is:
Then the linear equation is:
y = 2.5*x + b
To get the value of b, we use the fact that when x = 3, we have y = 82,
82 = 2.5*3 + b
82 = 7.5 + b
82 - 7.5 = b = 74.5
Then the linear equation is:
y = 2.5*x + 74.5
If you want to learn more about linear equations:
brainly.com/question/1884491
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(D) -3w2<span> - 9w - 4
I hope this helped :)
</span>
Answer:
im proably wrong but, (2,-4)
Step-by-step explanation:
