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3 years ago

What is the value of b ?

Mathematics

1 answer:

Reptile [31]3 years ago

4 0

Answer: 120

Step-by-step explanation:

Since Angle formed at the center is equal to the angle measure of its opposite arc

⇒ ∠AOD = 104°

Also, The angle formed in the segment is half the angle formed at the center of the circle

⇒ ∠DOB = 2 × ∠BAD

⇒ ∠DOB = 2 × 68°

⇒ ∠DOB = 136°

Now, ∠AOB = 360 - ∠BOD + ∠AOD

⇒ ∠AOB = 360 - 136 - 104

⇒ ∠AOB = 120°

Now, Angle formed at the center is equal to the angle measure of its opposite arc

⇒ b = ∠AOB

⇒ b = 101°

Hence, the correct option is 120°

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Read 2 more answers

Need help and explain please!!

lukranit [14]

Answer:

$x=-4\text{ and } x=3$

Step-by-step explanation:

We are given the second derivative:

$g''(x)=(x-3)^2(x+4)(x-6)$

And we want to find its inflection points.

To do so, we will first determine possible inflection points. These occur whenever g''(x) = 0 or is undefined.

Next, we will test values for the intervals. Inflection points occur if and only if the sign changes before and after the point.

So first, finding the zeros, we see that:

$0=(x-3)^2(x+4)(x-6)\Rightarrow x=-4, 3, 6$

So, we can draw the following number-line:

<----(-4)--------------(3)----(6)---->

Now, we will test values for the intervals x < -4, -4 < x < 3, 3 < x < 6, and x > 6.

Testing for x < -4, we can use -5. So:

$g^\prime^\prime(-5)=(-5-3)^2(-5+4)(-5-6)=704>0$

Since we acquired a positive result, g(x) is concave up for x < -4.

For -4 < x < 3, we can use 0. So:

$g^\prime^\prime(0)=(0-3)^2(0+4)(0-6)=-216$

Since we acquired a negative result, g(x) is concave down for -4 < x < 3.

And since the sign changed before and after the possible inflection point at x = -4, x = -4 is indeed an inflection point.

For 3 < x < 6, we can use 4. So:

$g^\prime^\prime(4)=(4-3)^2(4+4)(4-6)=-16$

Since we acquired a negative result, g(x) is concave down for 3 < x < 6.

Since the sign didn't change before and after the possible inflection point at x = 3 (it stayed negative both times), x = -3 is not a inflection point.

And finally, for x > 6, we can use 7. So:

$g^\prime^\prime(7)=(7-3)^2(7+4)(7-6)=176>0$