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3 years ago

AD is a diameter of a circle and AB is a chord. If AD=34cm, AB=30cm, the distance of AB from the centre of the circle is :

Mathematics

1 answer:

e-lub [12.9K]3 years ago

7 0

Answer:

The distance is 8 cm

Step-by-step explanation:

The chord and the diameter form one leg and the hypotenuse of a right triangle. The other leg, BD, has length ...

BD² +AB² = AD²

BD² = AD² -AB² = 34² -30² = 256

BD = √256 = 16

The segment from the center of the circle to the midpoint of the chord is the midline of triangle ABD, so is half the length of BD.

distance from AB to the center = 16/2 = 8 . . . cm

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Find a polynomial for the sum of the shaded areas of the figure. A = 6, B = 4

soldi70 [24.7K]

Answer:

The polynomial for the sum of the shaded $\pi$ r² - 20 $\pi$

Step-by-step explanation:

Given as :

The figure is shown which is of concentric circle with radius B , A , r

The radius B = 4 unit

The radius A = 6 unit

Let The sum of shaded portion = x unit

Now, The circumference of circle = 2 $\pi$ R , where R is the radius

So, for circle with radius B.

The circumference = 2 $\pi$ R = 2 $\pi$ B

Or, The circumference = 2 $\pi$ × 4 = 8 $\pi$

Similarly

For circle with radius A.

The circumference = 2 $\pi$ R = 2 $\pi$ A

Or, The circumference = 2 $\pi$ × 6 = 12 $\pi$

Now, The area of circle with radius r is

Area = $\pi$ ×radius × radius

Or, Area = $\pi$ r²

Now,

The sum of shaded region area = The area of circle with radius r - ( The circumference with radius B + The circumference with radius A )

Or, The sum of shaded region area = $\pi$ r² - ( 8 $\pi$ + 12 $\pi$ )

Or, The sum of shaded region area = $\pi$ r² - 20 $\pi$

Hence The polynomial for the sum of the shaded area is $\pi$ r² - 20 $\pi$ Answer

5 0

3 years ago

Marvin collects trading cards.He stores them in boxes that hold 235 cards each.If Marvin has 4 boxes full of cards, how many car

Taya2010 [7]

235 * 4 = 940 your answer is A

5 0

2 years ago

Read 2 more answers

Suppose a fair 6-sided die is tossed five times. What is the probability that we see at least one three in the five tosses?

Marta_Voda [28]

Answer:

C. 0.1667

Step-by-step explanation:

1/6 +0/6 + -1/6

The closest answer is C. 0.1667

Based on 2nd throw of dice shows zero then 0.1667 of the answer is shown on the 3rd throw in decimal form of -1/6

3 0

2 years ago

Whitch three capital letters have rotational symmetry but do not have line symmetry

deff fn [24]

First you want to figure out what exactly it is you are looking for. We are looking for "capital letters that have rotational symmetry but do not have line symmetry"

So:
1. Must have rotational symmetry.
This means that if we rotate the capital letter 180 deg, either clockwise or counterclockwise, it will still look the same

2. Must not have line symmetry.
If an object has line symmetry, it means that if you draw a line down the middle (in any way), it will be symmetrical on both sides. We need capital letters that do not fit that condition.

Now we look at all capital letters.
We find that H, I, N,O, S, X, and Z are all rotationally symmetrical. Think about it. If you rotate them, they still look the same.

But, we have to make sure they do not have line symmetry. If we draw a line right down the middle of H, I, O and X (**note, the have multiple lines of symmetry), they are symmetrical on both sides of the line.

Now we are left with N, S, and Z

4 0

3 years ago

Read 2 more answers

Solve for zzz. 42=-7(z-3)42=−7(z−3)

xxTIMURxx [149]

Answer:

False

Step-by-step explanation:

Well to solve for z we single it out, distribute, use the communicative property, and combine like terms.

42 = -7(z - 3)42 = -7(z - 3)

Distribute

42 = -7z + 21*42 = -7z + 21

42 = -7z + 882 = -7z + 21

-42 to both sides

-7z + 840 = -7z + 21

-21 to both sides

-7z + 819 = -7z

+7z to both sides

819 = 0

Thus,

the given equation is false.

I hope this helps :)

5 0

3 years ago