The equation of the circle is given as (x-h)² + (y-k)² = r². Then the value of y will be ± √5/3.
<h3>What is an equation of a circle?</h3>
A circle can be characterized by its center's location and its radius's length.
Let the center of the considered circle be at (h, k) coordinate.
Let the radius of the circle be 'r' units.
Then, the equation of that circle would be:
(x-h)² + (y-k)² = r²
The equation of the circle has the center at the origin and the radius is one unit. Then we have
x² + y² = 1
The point P = (-2/3, y) lies on the unit circle. Then the value of y will be
(-2/3)² + y² = 1
y² = 1 - 4/9
y² = 5/9
y = ± √5 / 9
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Answer:
2.2
Step-by-step explanation:
41/9 is 2.15789473684
since you need it to the nearest tenth, just get rid of the other numbers past the five, you will then have 2.15
since you are looking for the nearest tenth, you need to look at the 2nd number after the decimal place to determine whether it rounds up or down. ( a helpful rule is 5 or above, give it a shove. 4 or below, leave it alone. because the 2nd digit is 5 you need to give the 1 a shove and turn it into a 2. Since you gave the 5 a shove, you need to get rid of that 2nd digit in your answer.
FINAL ANSWER: 2.2
Answer:
36
Step-by-step explanation:
Answer:
Increased by 5%
Step-by-step explanation:
Let at the beginning the student's score be
points.
At first, it increased by 40%. So, the new score is
![s+0.4s=1.4s](https://tex.z-dn.net/?f=s%2B0.4s%3D1.4s)
Then it decreased by 25%, so the new score is
![1/4s-0.25\cdot 1.4s=1.4s-0.35s=1.05s](https://tex.z-dn.net/?f=1%2F4s-0.25%5Ccdot%201.4s%3D1.4s-0.35s%3D1.05s)
At the beginning of a month the score was
and at the end of that month the score was
. Since
the score increased by ![1.05-1=0.05=5\%](https://tex.z-dn.net/?f=1.05-1%3D0.05%3D5%5C%25)