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2 years ago

Which of these are rational? (Select all that apply)

Mathematics

1 answer:

Anon25 [30]2 years ago

6 0

ABCEG. Not D because non terminating numbers are irrational, same goes for Pi

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divide 2 into each digit of the larger number

Step-by-step explanation:

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Write the point where the linear equation 3x + 4y = 12 cuts the x-axis.

Pani-rosa [81]

Answer:

When the eqn cuts the x-axis , y=0

3x=12

x=12/3

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1 year ago

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Write an equation for a function that has vertical asymptotes at x=3 & x=-10

stepan [7]

Answer:

$\frac{5}{ {x}^{2} + 7x - 30}$

Step-by-step explanation:

We can write a rational function.

We need to make sure our denominator both have zeroes at 3 and 10.

Set an equation equal to zero to find the function

$0 - x = 3$

$0 - x = - 10$

$0 - ( - 3) = = 3$

$0 - 10 = 10$

So we would represent that's as

$x - 3$

and

$x + 10$

Multiply the two binomial together.

$(x - 3)(x + 10) = {x}^{2} + 7x - 30$

Let our numbetator be any interger.

Use any equation as long as the quadratic is the denominator and the interger is the numerator.

$\frac{5}{ {x}^{2} + 7x - 30 }$

7 0

2 years ago

3. Two balls are drawn in succession without replacement from an urn containing 4 red balls and 3 blue balls. Find the probabili

choli [55]

Answer:

X ____ 0 ______ 1 _______ 2

P(x) __ 1/7 _____ 4/7 _____ 2/7

Step-by-step explanation:

Given that :

Number of red balls = 4

Number of Blue balls = 3

Total number of balls = (4 + 3) = 7

Picking without replacement :

Number of picks = 2

1st pick is red :

Probability = (required outcome / Total possible outcomes)

USing combination formula :

nCr = n! / (n-r)! r!

Total possible outcome :

7C2 = 7! / 5!2! = 7*6 / 2 = 42 / 2 = 21

X = number of red

For red = 0;

P(x =0) = required outcome / Total possible outcomes

P(x = 0)

Required outcome : [4C0 * 3C2)]

P(x = 0) = [(4C0 * 3C2)] / 21 = (1*3)/21 = 3/21 = 1/7

P(x = 1)

Required outcome : [4C1 * 3C1)]

P(x = 1) = [(4C1 * 3C1)] / 21 = (4*3)/21 = 12/21 = 4/7

P(x = 2)

Required outcome : [4C2 * 3C0)]

P(x = 2) = [(4C2 * 3C0)] / 21 = (6*1)/21 = 6/21 = 2/7

Hence ;

X ____ 0 ______ 1 _______ 2

P(x) __ 1/7 _____ 4/7 _____ 2/7

8 0

2 years ago