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3 years ago

True or false the opposite angles of a quadrilateral in a circumscribed circle are always complementary

Mathematics

2 answers:

ella [17]3 years ago

7 0

Answer: This statement is false.

Step-by-step explanation:

False, the opposite angles of a quadrilateral in a circumscribed circle is not complementary.

As we know that the sum of opposite angles of a cyclic quadrilateral ( quadrilateral circumscribed circle ) is always supplementary.

So, Sum of opposite angles is 180° .

Hence, this statement is false.

sukhopar [10]3 years ago

7 0

Answer:

False

Step-by-step explanation:

We are given that

Opposite angles of a quadrilateral in a circumscribed circle are always complementary.

We have to find that this statement is true or not.

We know that

When a quadrilateral in a circumscribed circle is called cyclic quadrilateral.

We know that

Sum of opposite angles of cyclic quadrilateral is always 180 degrees.

When sum of two angle is equal to 180 degrees then , the angles are called supplementary.

Hence, the sum of opposite angles of cyclic quadrilateral is always supplementary.

Therefore, the given statement is false.

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For f(x)=4sin(x2) between x=0 and x=3, find the coordinates of all intercepts, critical points, and inflection points to two dec

telo118 [61]

Answer:

Intercepts:

x = 0, y = 0

x = 1.77, y = 0

x = 2.51, y = 0

Critical points:

x = 1.25, y = 4

x = 2.17 , y = -4

x = 2.8, y = 4

Inflection points:

x = 0.81, y = 2.44

x = 1.81, y = -0.54

x = 2.52, y = 0.27

Step-by-step explanation:

We can find the intercept by setting f(x) = 0

$4sin(x^2) = 0$

$sin(x^2) = 0$

$x^2 = n\pi$ where n = 0, 1, 2,3, 4, 5,...

$x = \sqrt(n\pi)$

Since we are restricting x between 0 and 3 we can stop at n = 2

So the function f(x) intercepts at y = 0 and x:

x = 0

x = 1.77

x = 2.51

The critical points occur at the first derivative = 0

$f^{'}(x) = 4cos(x^2)2x = 8xcos(x^2) = 0$

$xcos(x^2) = 0$

$x = 0$ or

$cos(x^2) = 0$

$x^2 = \frac{\pi}{2} + n\pi$ where n = 0, 1, 2, 3

$x = \sqrt{\pi(n+1/2)}$

Since we are restricting x between 0 and 3 we can stop at n = 2

So our critical points are at

x = 1.25, $y = f(1.25) = 4sin(1.25^2) = 4$

x = 2.17 , $y = f(2.17) = 4sin(2.17^2) = -4$

x = 2.8, $y = f(2.8) = 4sin(2.8^2) = 4$

For the inflection point, we can take the 2nd derivative and set it to 0

$f^[''}(x) = 8(cos(x^2) - xsin(x^2)2x) = 8cos(x^2) - 16x^2sin(x^2) = 0$

$cos(x^2) = 2x^2sin(x^2)$

$tan(x^2) = \frac{1}{2x^2}$

We can solve this numerically to get the inflection points are at

x = 0.81, $y = f(0.81) = 4sin(0.81^2) = 2.44$

x = 1.81, $y = f(1.81) = 4sin(1.81^2) = -0.54$

x = 2.52, $y = f(2.52) = 4sin(2.52^2) = 0.27$

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3 years ago

Explain why the product of a nonzero rational number and an irrational number is irrational

spayn [35]

Answer:

If you multiply any irrational number by the rational number zero, the result will be zero, which is rational. Any other situation, however, of a rational times an irrational will be irrational.

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3 years ago

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Wewaii [24]

Answer:

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Step-by-step explanation:

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erik [133]

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So the answer is 32%
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</span>

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2 years ago

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