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3 years ago

Write the equation of the parabola when the vertex is (0,5) and it passes through (2,9)

Mathematics

1 answer:

Alika [10]3 years ago

7 0

We know that
the equation of a vertical parabola in vertex form
y=a*(x-h)²+k
(h,k)------> (0,5)
y=a*(x-0)²+5
y=a*x²+5
substitute the point (2,9) in the equation
9=a*(2)²+5------> 9=4*a+5-------> 4*a=9-5-----> 4*a=4-----> a=1

the equation of the vertical parabola is
y=x²+5

the equation of a horizontal parabola in vertex form
x=a*(y-k)²+h
(h,k)------> (0,5)
x=a*(y-5)²+0
x=a*(y-5)²
substitute the point (2,9) in the equation
2=a*(9-5)²------> 2=16*a------> a=1/8

the equation of the horizontal parabola is
x=(1/8)*(y-5)²

the answer is
the equation of the vertical parabola is y=x²+5
the equation of the horizontal parabola is x=(1/8)*(y-5)²

see the attached figure

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Find the probability of the following events , when a dice is thrown once:

Fudgin [204]

Answer:

Step-by-step explanation:

s a die is rolled once, therefore there are six possible outcomes, i.e., 1,2,3,4,5,6.

(a) Let A be an event ''getting a prime number''.

Favourable cases for a prime number are 2,3,5,

i.e., n(A)=3

Hence P(A)=n(A)n(S)=36=12

(b) Let A be an event ''getting a number between 3 and 6''.

Favourable cases for events A are 4 or 5.

i.e., n(A)=2

P(A)=n(A)n(S)=26=13

Favourable cases of events A are 5, 6.

i.e., n(A)=2

P(A)=n(A)n(S)=26=13

(d) Let A be the event of getting a number at most 4.

∴ A={1,2,3} ⇒ n(A)=4,n(S)=6

∴ Required probability =n(A)n(S)=42=23

(e) Let A be the event of getting a factor of 6.

∴ A={1,2,36} ⇒ n(A)=4,n(A)=6

∴ Required probability =46=23

(ii) Since, a pair of dice is thrown once, so there are 36 possible outcomes. i.e.,

(a) Let A be an event ''a total 6''. Favourable cases for a total of 6 are (2,4), (4,2), (3,3), (5,1), (1,5).

i.e., n(A)=5

Hence P(A)=n(A)n(S)=536

(b) Let A be an event ''a total of 10n. Favourable cases for total of 10 are (6,4), (4,6), (5,5).

i.e., n(A)=5

P(A)=n(A)n(S)=336=112

(c) Let A be an event ''the same number of the both the dice''. Favourable cases for same number on both dice are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).

i.e., n(A)=6

P(A)=n(A)n(S)=636=16

(d) Let A be an event ''of getting a total of 9''. Favourable cases for a total of 9 are (3,6), (6,3), (4,5), (5,4).

i.e., n(A)=4

P(A)=n(A)n(S)=436=19

(iii) We have, n(S) = 36

(a) Let A be an event ''a sum less than 7'' i.e., 2,3,4,5,6.

Favourable cases for a sum less than 7 ar

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