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The line AB touching the circle at point B in the considered diagram is not tangent to the circle O.

What is Pythagoras Theorem?

If ABC is a triangle with AC as the hypotenuse and angle B with 90 degrees then we have:

where |AB| = length of line segment AB. (AB and BC are rest of the two sides of that triangle ABC, AC being the hypotenuse).

How are radius and tangent to a circle related?

There is a theorem in mathematics that:

If there is a circle O with tangent line L intersecting the circle at point A, then the radius OA is perpendicular to the line L.

So, if AB is a tangent, then ∠ABO = 90° and therefore satisfies Pythagoras theorem.

Assuming AB is tangent, then ABO is right angled we should get:

This statement is false, and therefore, so as our assumption is false that ABis tangent to circle O. Thus, AB is not tangent to circle O.

(so it might be that even if AB looks like touching at one point the circle O, but AB might be intersecting the circle at two points, or not touching it at all)

Thus, the line AB touching the circle at point B in the considered diagram is not tangent to the circle O.

Learn more about tangent to a circle here:

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Answer:

not tangent

Step-by-step explanation:

two reasons, first

Triangle AOB is not a right triangle

line AB intersects the circle O at two points.

The value of cos 7pi/6 = -√3/2.

What is the value of cos?

cos The value of 1 degree is 0.9998476. .. .. Cos 1 degree radians are written as cos (1 ° x π / 180 °). H. cos (0.017453. .. .. ). This article describes how to find the value of cos1 using an example. Cos1 °: 0.9998476.

Trigonometric functions, fields of mathematics related to specific trigonometric functions, and their computational applications. There are six trigonometric functions commonly used in trigonometric functions. Their names and abbreviations are sine (sin), cosine (cos), tangent (tan), cotangent (cot), second (sec), and cosecant (CSC).

A function of arc or angle (sine, cosine, tangent, cotangent, second, cotangent, etc.) is most easily expressed by the ratio of the pair of sides of a right triangle.

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Answer: -√3/2

Step-by-step explanation: cos 7 /6= cos ( + /6) = cos (/6) = -√3/2

Answer:

Step-by-step explanation:

Since the lines are parallel to each other that means they have the same gradient. So use the gradient formula y=mx+c where m is the gradient rearrange 3x-2y=10 using the formula to become -2y=-3x +10

divide both sides by -2 so that m=1.5. It passed through the point (2,2) use the equation of a line formula to find the the answer.

[{y - 2}/{x -2}=1.5

cross multiply to get 1.5x-y=1

hope that helps

Final answer:

The equation of the line parallel to 3x - 2y = 10 and passes through the point (2,2) is y = 1.5x -1 in slope-intercept form, and 3x - 2y = -2 in standard form.

Explanation:

First, we must find the slope of the line parallel to the given line 3x - 2y = 10. The slope of this line, written in the form ax + by = c, is -a/b, so the slope is -3/(-2) = 1.5. The equation of the line we want to find has this same slope. This is because parallel lines have the same slope.

To find our line, we use the point-slope form of an equation y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point the line passes through. Plugging in given point (2,2) and the slope 1.5, the equation becomes y - 2 = 1.5(x - 2). Simplifying it, we get y - 2 = 1.5x - 3. Adding 2 to both sides gives our line in slope-intercept form: y = 1.5x -1.

To express the equation in standard form, we manipulate the equation to be in the form Ax + By = C. Multiplying the equation by 2 to remove fractions gives us y * 2 = x * 3 - 2, so the line in standard form is 3x - 2y = -2.

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Answer:

Just answered the question the answer was -1.3 :)

Step-by-step explanation: