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kozerog [31]
3 years ago
10

What does radius mean

Mathematics
2 answers:
Katyanochek1 [597]3 years ago
8 0
Radius is half of a circle
dlinn [17]3 years ago
8 0
Radius is a half of a circle.
You might be interested in
Convert 2000m to centimeters​
iragen [17]

Answer:

200,000cm

Step-by-step explanation:

1m = 100cm

2000m = ?

2000 × 100 = 200,000cm

5 0
3 years ago
Read 2 more answers
2 Points
mart [117]
The distance around the circle is called the A. circumference
8 0
3 years ago
Write (8,3)m=-7 in point slope and slope intercept form
Lerok [7]

Answer:

The slope intercept form is y=-7x+59

Step-by-step explanation:

y-y=m(x-x1)

y-3=-7(x-8)

y-3=-7x+56

+3 +3

y=-7x+59

8 0
3 years ago
the hypotenuse of the right angled triangle is 6cm more than twice the shortest side. if the third side is 2 cm less than the hy
Mumz [18]

Answer:

Let the base be p

Hypotenuse = 2p +6

Perpendicular = 2p + 4

By Pythagoras theoram

(2p+6)^2 = (2p+4)^2 +p^2

=> 4p^2 +36 + 24p = 4p^2 + 16 +16p +p^2

=> 36+ 24p = p^2 + 16p + 16

=> p^2 - 8p - 20 = 0

=> p^2 - 10p +2p - 20 = 0

=> p(p-10) +2(p-10) = 0

=> (p-10)(p+2) = 0

p = 10 and - 2

Length can't be negative

So,

p = 10

Base = 10

Perpendicular = 24

Hypotenuse = 26

6 0
3 years ago
Does anyone know how to do this?? Help please!!!!
Doss [256]

Answer:

When we have a rational function like:

r(x) = \frac{x + 1}{x^2 + 3}

The domain will be the set of all real numbers, such that the denominator is different than zero.

So the first step is to find the values of x such that the denominator (x^2 + 3) is equal to zero.

Then we need to solve:

x^2 + 3 = 0

x^2 = -3

x = √(-3)

This is the square root of a negative number, then this is a complex number.

This means that there is no real number such that x^2 + 3 is equal to zero, then if x can only be a real number, we will never have the denominator equal to zero, so the domain will be the set of all real numbers.

D: x ∈ R.

b) we want to find two different numbers x such that:

r(x) = 1/4

Then we need to solve:

\frac{1}{4} = \frac{x + 1}{x^2 + 3}

We can multiply both sides by (x^2 + 3)

\frac{1}{4}*(x^2 + 3) = \frac{x + 1}{x^2 + 3}*(x^2 + 3)

\frac{x^2 + 3}{4} = x + 1

Now we can multiply both sides by 4:

\frac{x^2 + 3}{4}*4 = (x + 1)*4

x^2 + 3 = 4*x + 4

Now we only need to solve the quadratic equation:

x^2 + 3 - 4*x - 4 = 0

x^2 - 4*x - 1 = 0

We can use the Bhaskara's formula to solve this, remember that for an equation like:

a*x^2 + b*x + c = 0

the solutions are:

x = \frac{-b +- \sqrt{b^2 - 4*a*c} }{2*a}

here we have:

a = 1

b = -4

c = -1

Then in this case the solutions are:

x = \frac{-(-4) +- \sqrt{(-4)^2 - 4*1*(-1)} }{2*(1)} = \frac{4 +- 4.47}{2}

x = (4 + 4.47)/2 = 4.235

x = (4 - 4.47)/2 = -0.235

5 0
3 years ago
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