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Dmitriy789 [7]
3 years ago
5

Any time you inscribe a square in a circle or circumscribe a right triangle by a circle, which type of line or segment will alwa

ys contain two of the vertices of the square or right triangle?
Mathematics
1 answer:
pogonyaev3 years ago
5 0

Answer:

Step-by-step explanation:

A diameter of the circle.

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0,3,5,7,8 because it would make it higher pls heart and good luck !
7 0
3 years ago
Please need help with son's math question!!!!!!!! If any one could help with the following :
myrzilka [38]

L = lees free throws

L = 2/3 Brians free throws

multiply by 3 on each side

3L = 2B

ratio is 3:2

if brian makes 15

3L = 2 (15)

3L = 30

divide by 3

L = 30/3 = 10

Lee makes 10

Answer:  ratio 3:2

lee makes 10


6 0
4 years ago
Read 2 more answers
I need help with problem 1 with a through explanation and solution please <br><br>​
ki77a [65]

Explanation:

The cubic ...

  f(x) = ax³ +bx² +cx +d

has derivatives ...

  f'(x) = 3ax² +2bx +c

  f''(x) = 6ax +2b

<h3>a)</h3>

By definition, there will be a point of inflection where the second derivative is zero (changes sign). The second derivative is a linear equation in x, so can only have one zero. Since it is given that a≠0, we are assured that the line described by f''(x) will cross the x-axis at ...

  f''(x) = 0 = 6ax +2b   ⇒   x = -b/(3a)

The single point of inflection is at x = -b/(3a).

__

<h3>b)</h3>

The cubic will have a local extreme where the first derivative is zero and the second derivative is not zero. These will only occur when the discriminant of the first derivative quadratic is positive. Their location can be found by applying the quadratic formula to the first derivative.

  x=\dfrac{-2b\pm\sqrt{(2b)^2-4(3a)(c)}}{2(3a)} = \dfrac{-2b\pm\sqrt{4b^2-12ac}}{6a}\\\\x=\dfrac{-b\pm\sqrt{b^2-3ac}}{3a}\qquad\text{extreme point locations when $b^2>3ac$}

There will be zero or two local extremes. A local extreme cannot occur at the point of inflection, which is where the formula would tell you it is when there is only one.

__

<h3>c)</h3>

Part A tells you the point of inflection is at x= -b/(3a).

Part B tells you the midpoint of the local extremes is x = -b/(3a). (This is half the sum of the x-values of the extreme points.) You will notice these are the same point.

The extreme points are located symmetrically about their midpoint, so are located symmetrically about the point of inflection.

_____

Additional comment

There are other interesting features of cubics with two local extremes. The points where the horizontal tangents meet the graph, together with the point of inflection, have equally-spaced x-coordinates. The point of inflection is the midpoint, both horizontally and vertically, between the local extreme points.

6 0
3 years ago
-10 |h+5| -3= -83<br> what is the answer to this the topic is absolute value.
Andru [333]

Answer:

the absolute value is 3,-13

6 0
3 years ago
Select the correct answer from each drop-down menu.The center of the circle is at O. Segment AB is tangent to circle O at point
IRISSAK [1]

Based on the tangent theorem, m∡ADO = 90°.

Based on the angles of intersecting chords theorem, m∡CGF = 1/2(mCF + mDE).

What is the Angles of Intersecting Chords Theorem?

The angles of intersecting chords theorem states that when two chords in a circle intersect, the angle formed inside the circle at the point of intersection has a measure that is half of the sum of the measures of the intercepted arcs that are formed by the angle and its vertical angle.

<h3>What is the Tangent Theorem?</h3>

According to the tangent theorem, if a line is tangent to a circle, the segment forms a right angle at the point of tangency with the radius of the circle.

Since line AB is tangent to circle O at point D, therefore, based on the tangent theorem:

The measures of angle ADO = 90°

Chord CE and DF intersect in the circle, therefore, based on the angles of intersecting chords theorem:

The measure of CGF = 1/2(mCF + mDE).

Learn more about the tangent theorem on:

brainly.com/question/9892082

#SPJ1

4 0
1 year ago
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