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aksik [14]
3 years ago
13

If the sides of square are increased by 2 meters, the area becomes 64 square meters. Find the lenght of a side of the original s

quare
Mathematics
2 answers:
weeeeeb [17]3 years ago
8 0

Answer:

The length of a side of the original square is 6 meters long.

Step-by-step explanation:

First, you need to know that finding the area of a square means multiplying two sides together. You also need to know that a square has four equal sides, meaning that the area is a square of a number. This means that since the area of the new square is 64 square meters, then finding a side length would mean finding the square root of 64. Using a calculator, you see that the square root of 64 is eight. This means that the second square has a side length of 8 meters.

Next, the problem tells you the sides of the new square were increased by 2 meters. All you would have to do is subtract 8 by two, which equals six. That being said, the length of a side of the original square would be six meters.

+The area of the original square would be 36 meters, because 6 times 6 (side times side) equals 36.

zepelin [54]3 years ago
8 0
The length of one side Is 6 meter long
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Multiplying the whole function by -1 reflects the function across the x axis
so f(x) to -f(x) would be a reflection across the x axis


multiplying the whole function by a fraction is a vertical transformation, if you multiply by a value x, such that 0<x<1, then it is a vertical shrink, if x>1, then it is a vertical stretch
so like f(x) to 2f(x) is a vertical stretch by a factor of 2

adding a value to the whole function moves it up by that number



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so f(x)=(-1)(\frac{1}{2})(x^2)+2
we have multiplied the whole thing by -1 and 1/2 and then added 2 to the whole function

that is a reflection across the x axis, a vertical shrink by a factor of 1/2, and translated up by 2 units in that order

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The displacement of a particle from a reference point at any instant is given by s = 3 t 2 - 4 t + 5 , where s is in metres and
lina2011 [118]

The average velocity of the particle in the time interval between 3s and 5s is 20 ms⁻¹ and its instantaneous velocity at 4s is 20 ms⁻¹.

How to determine average velocity and instantaneous velocity?
Average velocity is defined as the body's overall displacement divided by its time of motion. While instantaneous velocity is defined as a body's speed at a certain instant in time, or its displacement at that instant. When the velocity is constant, average and instantaneous velocities will equalize at just one condition.
The definition of instantaneous velocity is the rate of change of position over a relatively brief time period (almost zero). Simply said, the speed of an object at that precise moment. The definition of instantaneous velocity is "The velocity of an item in motion at a certain point in time." The instantaneous velocity of an object may be equal to its standard velocity if it has uniform velocity.
Mathematically, average velocity = [s(t₂) - s(t₁)]/[t₂ - t₁]
Instantaneous velocity at time, t is = (ds/dt) at time = t

Given, the displacement for the particle is given by s = 3t² - 4t + 5
Time interval, t₁ = 5s and t₂ = 3s;
Using formula in literature, average velocity of the particle in the time interval between 3s and 5s is:
Average velocity = (s(5) - s(3))/(5 - 3) = (60 - 20)/2 = 20 ms⁻¹
Instantaneous velocity at t = 4 is ds/dt at that time-frame:
Now, v = ds/dt = 6t -4
Now, v(4) = 6(4) - 4 = 20 ms⁻¹
The average velocity of the particle in the time interval between 3s and 5s is 20 ms⁻¹ and its instantaneous velocity at 4s is 20 ms⁻¹.

To learn more about this, tap on the link below:
brainly.com/question/2234298

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