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nekit [7.7K]
2 years ago
12

A square OABC with the length of a side 6 cm and circle k(O) with a radius 5 cm are given. Which of the lines OA , AB , BC , or

AC are secants to the circle k(O)?
Mathematics
1 answer:
Pavel [41]2 years ago
5 0

Answer:

AC and OA

Step-by-step explanation:

-A secant is a  line connecting two points on the circle.

-Given the square OABC of sides 6cm and a circle of r=5cm and  the center of the circle as O, and that the radius of the circle is less than the side of the square:

-The circle passes through OA and OC, but doesn't pass through AB and BC.

Hence, AC and OA are the circle's secants.

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One factor of the polynomial x^3 − 7x^2 + 13x − 3 is (x − 3). What is the other factor of the polynomial? (Note: Use long or syn
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Answer:

Option (B)

Step-by-step explanation:

Given polynomial in the question is,

x³ - 7x² + 13x - 3

One factor of the polynomial is (x - 3).

That means x = 3 is a zero factor of the polynomial.

For the other factor,

By using synthetic division,

3 |   1         -7           13           -3

 

   <u>               3           -12          3   </u>

     1          -4            1            0

= (x² - 4x + 1)(x - 3)

Therefore, the other factor of the polynomial is (x² - 4x + 1).

Option (B) will be the answer.

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2 years ago
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The steps below show the work of a student used to calculate the number of yards in 6,436 meters.
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The conversion factor should be multiplied in Step 2 instead of being divided.

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A slide has an angle of elevation of 25°. It is 60 feet from the end of the slide to the stairway beneath the top of the slide.
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If the median of the score distribution on this pop quiz is more than 170, the teacher will drop the next pop quiz. The pop quiz
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The median of a set of data is obtained by arranging the data from lowest to highest. In this pop quiz scores, the arrangements is as follows:

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There are 22 items in the list and the median is located in the middle position. In this case, it is the average of 174 (11th) and 175 (12th). Thus, the median is 174.5. 
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