Answer:
Solving for y gives:

Step-by-step explanation:
Linear equations are the equations that are degree one equations. When it is asked to solve an equation for a specific variable that means that the specified variable has to be isolated on one side of the equation i.e. in the given question solving for y means isolating y on one side of the equation.
Given equation is:

Multiplication property of equality(Multiplying both sides by q)

Division property of equality(Dividing both sides by 3)

Hence,
Solving for y gives:

Answer:
29.9 to the 6th power for number 2
0.1 to the -9 th power
i think lol
its better than ?
Step-by-step explanation:
Using the perimeter concept, it is found that the perimeter of the polygon is of 24.1 units.
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- The perimeter of a polygon is the sum of the lengths of all its sides.
- One side has length 1 unit.
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- The second side is the hypotenuse of a right triangle with sides of length 3 and 4, thus, applying the Pytagorean Theorem:




- One of the sides has length of 5 units.
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- The third side is the hypotenuse of a right triangle of sides with lengths 5 and 7, thus:




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- The final side is the hypotenuse of a right triangle with sides of lengths 3 and 9, thus:




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- The lengths of the sides are: 1, 5, 8.6 and 9.5.
- Thus, the perimeter is of:

The perimeter of the polygon is of 24.1 units.
A similar problem is given at brainly.com/question/6139098
Correct!
Your notes on the right side are correct, too.
<em>In</em>dependent -- <em>not </em>dependent on another variable
Dependent -- dependent on another variable
The height of the plant varies because/due to of the quality/quantity of fertilizer.
So, the height is changed by the fertilizer.
Thus: height is dependent and fertilizer amount is independent.