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Vadim26 [7]
3 years ago
11

Which graph represents the function f (x) = StartFraction 1 Over x + 3 EndFraction minus 2? On a coordinate plane, a hyperbola i

s shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = negative 3, and the horizontal asymptote is at y = negative 2. On a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrants 1 and 4, and the other curve opens down and to the left in quadrants 2 and 3. A vertical asymptote is at x = 3, and the horizontal asymptote is at y = negative 2. On a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrants 1 and 4, and the other curve opens down and to the left in quadrants 2 and 3. A vertical asymptote is at x = negative 3, and the horizontal asymptote is at y = 2. On a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1 , and the other curve opens down and to the left in quadrants 2, 1, and 4. A vertical asymptote is at x = 3, and the horizontal asymptote is at y = 2.
Mathematics
2 answers:
Zinaida [17]3 years ago
6 0

Answer:

D on edge 2020

Step-by-step explanation:

Ede4ka [16]3 years ago
5 0

Answer:

its A

Step-by-step explanation:

i just did it swag

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(B) x=\sqrt{30}

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3 ft<br> 10 ft<br> Perimeter:<br> Area:
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The area = 3 × 10 = 30

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Nicole has a collection of dolls that weigh 2/2 gram. How much would 2 dolls weigh?
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3 years ago
If you roll two fair dice repeatedly, what is the probability that you will get a sum of 4 before you get a sum of 5 ? (a) (b) (
AlexFokin [52]

Answer:

The probability that you will get a sum of 4 before you get a sum of 5 is \frac{1}{12}

Step-by-step explanation:

Given : If you roll two fair dice repeatedly.

To find : What is the probability that you will get a sum of 4 before you get a sum of 5 ?

Solution :

When two dice are rolled the outcomes are

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Total number of outcomes = 36

Favorable outcome get a sum of 4 before you get a sum of 5 is (1,3) ,(2,2) and (3,1) = 3

The probability that you will get a sum of 4 before you get a sum of 5 is

P=\frac{3}{36}

P=\frac{1}{12}

Therefore, The probability that you will get a sum of 4 before you get a sum of 5 is \frac{1}{12}

5 0
3 years ago
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