All categories

2 years ago

Please help I’ll give you brainliest

Mathematics

1 answer:

dem82 [27]2 years ago

5 0

Answer: 226.9 square in.

Step-by-step explanation: Remember that the formula for the area of a circle is (pi)r^2. Since the diameter is 17, you divide that by 2 to get your radius: 8.5. Now, put it into the formula to get your answer. (226.865 is the original, 226.9 is rounded to tenths)

You might be interested in

For x, y ∈ R we write x ∼ y if x − y is an integer. a) Show that ∼ is an equivalence relation on R. b) Show that the set [0, 1)

vodomira [7]

Answer:

A. It is an equivalence relation on R

B. In fact, the set [0,1) is a set of representatives

Step-by-step explanation:

A. The definition of an equivalence relation demands 3 things:

The relation being reflexive (∀a∈R, a∼a)
The relation being symmetric (∀a,b∈R, a∼b⇒b∼a)
The relation being transitive (∀a,b,c∈R, a∼b^b∼c⇒a∼c)

And the relation ∼ fills every condition.

∼ is Reflexive:

Let a ∈ R

it´s known that a-a=0 and because 0 is an integer

a∼a, ∀a ∈ R.

∼ is Reflexive by definition

∼ is Symmetric:

Let a,b ∈ R and suppose a∼b

a∼b ⇒ a-b=k, k ∈ Z

b-a=-k, -k ∈ Z

b∼a, ∀a,b ∈ R

∼ is Symmetric by definition

∼ is Transitive:

Let a,b,c ∈ R and suppose a∼b and b∼c

a-b=k and b-c=l, with k,l ∈ Z

(a-b)+(b-c)=k+l

a-c=k+l with k+l ∈ Z

a∼c, ∀a,b,c ∈ R

∼ is Transitive by definition

We´ve shown that ∼ is an equivalence relation on R.

B. Now we have to show that there´s a bijection from [0,1) to the set of all equivalence classes (C) in the relation ∼.

Let F: [0,1) ⇒ C a function that goes as follows: F(x)=[x] where [x] is the class of x.

Now we have to prove that this function F is injective (∀x,y∈[0,1), F(x)=F(y) ⇒ x=y) and surjective (∀b∈C, Exist x such that F(x)=b):

F is injective:

let x,y ∈ [0,1) and suppose F(x)=F(y)

[x]=[y]

x ∈ [y]

x-y=k, k ∈ Z

x=k+y

because x,y ∈ [0,1), then k must be 0. If it isn´t, then x ∉ [0,1) and then we would have a contradiction

x=y, ∀x,y ∈ [0,1)

F is injective by definition

F is surjective:

Let b ∈ R, let´s find x such as x ∈ [0,1) and F(x)=[b]

Let c=║b║, in other words the whole part of b (c ∈ Z)

Set r as b-c (let r be the decimal part of b)

r=b-c and r ∈ [0,1)

Let´s show that r∼b

r=b-c ⇒ c=b-r and because c ∈ Z

r∼b

[r]=[b]

F(r)=[b]

∼ is surjective

Then F maps [0,1) into C, i.e [0,1) is a set of representatives for the set of the equivalence classes.

4 0

2 years ago

What is the probability of being dealt exactly three of a kind (like three kings or three 7’s, etc.) in a five card hand from a

Drupady [299]

Answer:

P=0.00564

Step-by-step explanation:

From Exercise we have 52 cards.

We calculate the number of combinations to draw 5 cards from a deck of 52 cards. We get

{52}_C_{5}=\frac{52!}{5!(52-5)!}=2598960

We now count the number of favorable combinations:

{13}_C_{1} · {48}_C_{2}= 13 · \frac{48!}{2!(48-2)!}=14664

Therefore, the probabilitiy is

14664/2598960=0.00564

P=0.00564

6 0

3 years ago

Picture Question please help!!!! Thank you so much!

ra1l [238]

The answer is 33. This is because there are 3 colours and the probability that the spinner will land on each colour is 1/3 (one third). To work out the number of times it will land on blue you have to do the probability of it landing on blue (1/3) multiplied by the number of times spun (100) which gives approximately 33.

6 0

3 years ago

Read 2 more answers

What are the zeros of the function f(x)=x^2-2x-15?

valkas [14]

To find the zeros of this function, we must first set the entire function equal to 0

f(x) = x² - 2x - 15 = 0

Since this is a quadratic function, we must use the quadratic formula, which is:

$\frac{-b +/- \sqrt{b^{2} - 4(a)(c) } }{2a}$

Let's assign a, b, and c using our first function
x² means a = 1 (because it could be written as 1x²)
-2x means b = -2
-15 means c = -15

Now let's plug those in:

$\frac{-(-2) +/- \sqrt{(-2)^{2} - 4(1)(-15) } }{2(1)}$

which simplifies to:

$\frac{2 +/- \sqrt{(4 + 60} }{2}$

Simplified further:

$\frac{2 +/- \sqrt{(64} }{2}$
$\frac{2 +/- 8 }{2}$
And divide it by the 2 on the bottom gives us:

$2 +/- 4$

2+4 = 6
2-4 = -2

So the zeros of this function are -2 and 6

6 0

3 years ago

Read 2 more answers

Write the equation of the circle centered at (4, - 9) with diameter 16.<br> HELP A GIRL OUT

iragen [17]

Answer:

16= (x-4)^2 +(y+9)^2

Step-by-step explanation:

hey! The formula for the equation of a circle is r^2= (x-h)^2 + (y-k)^2

r= the radius (in your case you have the diameter so you do not need to square it)

(h,k)= is the center coordinates, so your center coordinate would be (4, -9). You now have your h (h=4) and your k (k=-9) so you can plug it in the formula! :)

a nice rating would help out

4 0

2 years ago