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Alja [10]
3 years ago
6

Find the diameter of each circle.

Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
5 0

Answer:

<em>Diameter Length: ( About ) 5.4 km; Option B</em>

Step-by-step explanation:

~ Let us apply the Area of the Circle formula πr^2, where r ⇒ radius of the circle ~

1. We are given that the area of the circle is 22.9 km^2, so let us substitute that value into the area of the circle formula, solving for r ( radius ) ⇒ 22.9 = π * r^2 ⇒ r^2 = 22.9/π ⇒ r^2 = 7.28929639361.... ⇒

<em>radius = ( About ) 2.7</em>

2. The diameter would thus be 2 times that of the radius by definition, and thus is: 2.7 * 2 ⇒ ( About ) 5.4 km

<em>Diameter Length: ( About ) 5.4 km</em>

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Tangent is positive in quadrants I and IV only.<br>True or False
zhannawk [14.2K]

Answer:

The statement is false

Step-by-step explanation:

we know that

tan(x)=\frac{sin(x)}{cos(x)}

The tangent function will be positive when the sine function and the cosine function have the same sign

so

In the first quadrant the tangent function is positive

In the third quadrant the tangent function is positive

so

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6 0
3 years ago
Read 2 more answers
Please help me !!! need help will mark brainly
Ronch [10]

Answer:

1. is 3 2/3

2. is 3 1/4

3. is 2 1/4

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3 years ago
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7 0
3 years ago
The nth term of a sequence is 3n^2- 1
Thepotemich [5.8K]

Answer: The number that belongs to both sequences is 26.

Step-by-step explanation:

We have two sequences, let's call one as A and the other as B.

The n-th term of sequence A is written as:

aₙ = 3*n^2 - 1

the nth term of sequence B is written as:

bₙ = 30 - n^2

We want to find a term that belongs to both sequences, (it can be for different integers, we can use n for sequence A and x for sequence B)

Then we want to find:

aₙ = bₓ

where n and x are integer numbers.

Then we will heave:

3*n^2 - 1 = 30 - x^2

To find the pair, we could isolate one of the variables, then input different integers in the other variable and see if the outcome is also an integer.

Let's isolate n.

3*n^2 = 30 - x^2 + 1

3*n^2 = 31 - x^2

n^2 = (31 - x^2)/3

n = √(  (31 - x^2)/3)

Now let's input different values for x, and see if the outcome is also an integer, notice that x is in a negative term inside a square root, then we have only a few values of x such that the equation can be true.

Then let's start with x = 1.

n(1) = √(  (31 - 1^2)/3) = √(30/3) = √10

We know that √10 is not an integer.

now with x = 2,

n(2) = √(  (31 - 2^2)/3)  = √( (31 - 4)/3) = √(27/3) = √9 = 3

then if x = 2, we have n = 3.

Both of them are integers, then we get:

a₂ = 3*(3)^2 - 1 = 27 - 1 = 26

b₃ = 30 - 2^2 = 30 - 4 = 26

The number that belongs to both sequences is 26.

5 0
2 years ago
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3 years ago
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