My sandbox is a square each side is 6 feet long what is the area of the sandbox
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Point (-2 , 4) lies in the 2nd quadrant
Point (3 , -1) lies in the 4th quadrant
Point (-1 , 0) lies on the negative part of the x-axis
Point (1 , 2) lies in the 1st quadrant
Point (-3 , -5) lies in the 3rd quadrant
Step-by-step explanation:
Let us revise the signs of the coordinates in each quadrant
- x and y coordinates are positive in the 1st quadrant
- x-coordinate is negative and y-coordinate is positive in the 2nd quadrant
- x and y coordinates are negative in the 3rd quadrant
- x-coordinate is positive and y-coordinate is negative in the 4th quadrant
- The general point on the positive part of x-axis is (x , 0), and on the negative part of x-axis is (-x , 0)
- The general point on the positive part of y-axis is (0 , y), and on the negative part of y-axis is (0 , -y)
Now let us check the points
Point (-2 , 4) ⇒ red point
∵ x = -2 and y = 4
∵ x-coordinate is negative and y-coordinate is positive
∴ Point (-2 , 4) lies in the 2nd quadrant
Point (3 , -1) ⇒ blue point
∵ x = 3 and y = -1
∵ x-coordinate is positive and y-coordinate is negative
∴ Point (3 , -1) lies in the 4th quadrant
Point (-1 , 0) ⇒ green point
∵ x = -1 and y = 0
∵ x-coordinate is negative and y-coordinate is zero
∴ Point (-1 , 0) lies on the negative part of the x-axis
Point (1 , 2) ⇒ purple point
∵ x = 1 and y =2
∵ x and y coordinates are positive
∴ Point (1 , 2) lies in the 1st quadrant
Point (-3 , -5) ⇒ black point
∵ x = -3 and y = -5
∵ x and y coordinates are negative
∴ Point (-3 , -5) lies in the 3rd quadrant
<em>Look to the attached graph for more understand</em>
Learn more:
You can learn more about the four quadrant in brainly.com/question/9381523
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It's evident that the first four terms are 4, 4/3, 4/9, and 4/27. So the fourth partial sum of the series is
It's as easy as adding up the fractions, but I bet this is supposed to be an exercise in taking advantage of the fact that the series is geometric and use the well-known formula for computing such a sum.
Multiply the sum by 1/3 and you have
Now subtracting this from
gives
That is, all the matching terms will cancel. Now solving for, you
have
It's as easy as adding up the fractions, but I bet this is supposed to be an exercise in taking advantage of the fact that the series is geometric and use the well-known formula for computing such a sum.
Multiply the sum by 1/3 and you have
Now subtracting this from
That is, all the matching terms will cancel. Now solving for
have