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Kryger [21]
3 years ago
9

The endpoints of a diameter of a circle are located at (5,9) and (11,17) which is an equation of the circle?

Mathematics
1 answer:
lord [1]3 years ago
8 0

Answer:

(x - 8)² + (y - 13)² = 25

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

The centre is located at the midpoint of the endpoints of the diameter.

Use the midpoint formula to find the centre

[\frac{x_{1}+x_{2}  }{2}, \frac{y_{1}+y_{2}  }{2} ]

with (x₁, y₁ ) = (5, 9) and (x₂, y₂ ) = (11,17)

centre = ( \frac{5+11}{2}, \frac{9+17}{2} ) = (8, 13)

The radius is the distance from the centre to either end of the diameter

Calculate r using the distance formula

r = √ (x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = (8, 13) and (x₂, y₂ ) = (5, 9)

r = \sqrt{(5-8)^2+(9-13)^2}

  = \sqrt{(-3)^2+(-4)^2}

  = \sqrt{9+16} = \sqrt{25} = 5 ⇒ r² = 25

Hence

(x - 8)² + (y - 13)² = 25

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Hunter-Best [27]

(a) If

<em>f(x)</em> = <em>a</em>₀ + <em>a</em>₁ <em>x </em>+<em> a</em>₂ <em>x</em> ² + <em>a</em>₃ <em>x</em> ³

then from the given conditions we get the system of equations,

<em>f</em> (-1) = <em>a</em>₀ - <em>a</em>₁<em> </em>+<em> a</em>₂ - <em>a</em>₃ = -1

<em>f</em> (1) = <em>a</em>₀ + <em>a</em>₁<em> </em>+<em> a</em>₂ + <em>a</em>₃ = 2

<em>f</em> (2) = <em>a</em>₀ + 2<em>a</em>₁<em> </em>+ 4<em>a</em>₂ + 8<em>a</em>₃ = 1

<em>f</em> (3) = <em>a</em>₀ + 3<em>a</em>₁<em> </em>+<em> </em>9<em>a</em>₂ + 27<em>x</em> ³ = 5

(b) Similarly, if

<em>f(x)</em> = <em>a</em>₀ + <em>a</em>₁ <em>x </em>+<em> a</em>₂ <em>x</em> ² + <em>a</em>₃ <em>x</em> ³

then

<em>f'(x)</em> = <em>a</em>₁<em> </em>+<em> </em>2<em>a</em>₂ <em>x</em> + 3<em>a</em>₃ <em>x</em> ²

so that the given conditions yield the system,

<em>f</em> (1) = <em>a</em>₀ + <em>a</em>₁<em> </em>+<em> a</em>₂ + <em>a</em>₃ = 0

<em>f'</em> (1) = <em>a</em>₁<em> </em>+<em> </em>2<em>a</em>₂ + 3<em>a</em>₃ = 2

<em>f</em> (2) = <em>a</em>₀ + 2<em>a</em>₁<em> </em>+<em> </em>4<em>a</em>₂ + 27<em>a</em>₃ = 3

<em>f'</em> (2) = <em>a</em>₁<em> </em>+<em> </em>4<em>a</em>₂ + 12<em>a</em>₃ = -1

7 0
2 years ago
4. Assume that the chances of a basketball player hitting a 3-pointer shot is 0.4 and the probability of hitting a free-throw is
vovikov84 [41]

Answer:

0.0334 = 3.34% probability that the player will make exactly 3 3-pointers and 5-free throws.

Step-by-step explanation:

For each 3-pointer shot, there are only two possible outcomes. Either the player makes it, or the player does not. The same is valid for free throws. This means that both the number of 3-pointers and free throws made are given by binomial distributions.

Since 3-pointers and free throws are independent, first we find the probability of making exactly 3 3-pointers out of 10, then the probability of making exactly 5 free throws out of 10, and then the probability that the player will make exactly 3 3-pointers and 5-free throws is the multiplication of these probabilities.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Probability of making 3 3-pointers out of 10:

The chances of a basketball player hitting a 3-pointer shot is 0.4, which means that p = 0.4. So this is P(X = 3) when n = 10.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 3) = C_{10,3}.(0.4)^{3}.(0.6)^{7} = 0.21499

Probability of making 5 free throws out of 10:

The probability of hitting a free-throw is 0.65, which means that p = 0.65. The probability is P(X = 5) when n = 10.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 3) = C_{10,5}.(0.65)^{5}.(0.35)^{5} = 0.15357

Calculate the probability that the player will make exactly 3 3-pointers and 5-free throws.

0.21499*0.15537 = 0.0334

0.0334 = 3.34% probability that the player will make exactly 3 3-pointers and 5-free throws.

5 0
3 years ago
Classify each pair of angels listed below as adjacent or vertical.
Stolb23 [73]

Answer:

ntersecting lines DA and CE.

To find:

Each pair of adjacent angles and vertical angles.

Solution:

Adjacent angles are in the same straight line.

Pair of adjacent angles:

(1) ∠EBD and ∠DBC

(2) ∠DBC and ∠CBA

(3) ∠CBA and ∠ABE

(4) ∠ABE and ∠EBD

Vertical angles are opposite angles in the same vertex.

Pair of vertical angles:

(1) ∠EBD and ∠CBA

(2) ∠DBC and ∠EBA

5 0
3 years ago
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daser333 [38]

Answer:

Option B (35°).

Step-by-step explanation:

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3 years ago
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