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Alenkinab [10]
2 years ago
9

Find the coordinates of S after a rotation of the triangle 90 degrees counterclockwise about the origin

Mathematics
1 answer:
Neko [114]2 years ago
4 0

We first start by finding the actual coordinates of the pre - image's point S.

Assuming the x and y axis count by 1s, as there is no labels, we can conclude that:

The point's x coordiate is 4 to the left from the origin, or -4.

The point's y coordinate is 4 up from the origin, or 4.

So, the coordinates of Point S is (-4, 4).

Next, we apply the translation rule for a 90 degree counterclockwise rotation to the point's coordinates. The rule is (x, y) --> (-y, x). Using this rule on our coordinates, we get:

(-4, 4) --> (-4, -4)

So, the new coordinates of point S prime after a 90 degree counterclockwise rotation are (-4, -4).

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<em>15.4 </em>

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-11/20

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    2      8   1

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       3   9      5

2      -8

Divide  —  by  ——

        3      9

To divide fractions, write the divison as multiplication by the reciprocal of the divisor :

2     -8       2      9

—  ÷  ——   =   —  •  ——

3     9        3     -8

(0--3/4)+1/5

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3 0
3 years ago
Read 2 more answers
What is the solution to the trigonometric inequality 2sin(x)+3&gt;sin ^2(x) over the interval
navik [9.2K]

The intervals that satisfy the given trigonometric Inequality are; 0 ≤ x < 3π/2 and 3π/2 < x ≤ 2π

<h3>How to solve trigonometric inequality?</h3>

We are given the trigonometric Inequality;

2 sin(x) + 3 > sin²(x)

Rearranging gives us;

sin²(x) - 2 sin(x) - 3 < 0

Factorizing this gives us;

(sin(x) - 3)(sin(x) + 1) < 0

Thus;

sin(x) - 3 = 0 or sin(x) + 1 = 0

sin(x) = 3 or sin(x) = -1

sin(x) = 3 is not possible because sin(x) ≤ 1.

Thus, we will work with;

sin(x) = -1 for the interval 0 ≤ x ≤ 2π radians.

Then, x = sin⁻¹(-1)

x = 3π/2.

Now, if we split up the solution domain into two intervals, we have;

from 0 ≤ x < 3π/2, at x = 0. Then;

sin²(0) - 2 sin(0) - 3

= 0² - 0 - 3

= -3 < 0

Thus, the interval 0 ≤ x < 3π/2 is true.

From 3π/2 < x ≤ 2π, take x = 2π. Then;

sin²(2π) - 2 sin(2π) - 3

= 0² - 0 - 3

= -3 < 0

Thus, the interval 3π/2 < x ≤ 2π is also true.

Read more about trigonometric inequality at; brainly.com/question/27862380

#SPJ1

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