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Nadusha1986 [10]

2 years ago

14

Heeellllppp~ I beg you, I'm crying~ because I have been waiting for a long time now and I have asked this question multiple time

s and I don't get an answer~ help me

1 answer:

sp2606 [1]2 years ago

6 0

Answer:

5969593829191937373

Step-by-step explanation:

Sister I gotcha b

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Give an example of a rational function that has a horizontal asymptote at y = 0 and a vertical asymptote at

Vesnalui [34]

F(x)=x^2/(x-2)(x-1), the degree of the numerator and the degree of the denominator equal and the quotient of their coefficient equal to 1(horizontal asymptote means that the lines approach a certain number when the input approaches positive or negative infinity)
and when plugging in 2 or 1 as input, the output is undefined.

8 0

3 years ago

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If u = 8 + 4i and v = 3i + 6, then what is u – v?

lozanna [386]

$u=8+4i\ and\ v=31+6\\\\u-v=(8+4i)-(3i+6)=8+4i-3i-6=(8-6)+(4i-3i)\\\\\huge\boxed{=2+i}\leftarrow \boxed{d}$

7 0

3 years ago

A standard six-sided die is rolled twice and the top faces are observed. What is the probability that the sum of the numbers on

melisa1 [442]

<u>Answer: </u>

A standard six-sided die is rolled twice and the top faces are observed and the probability that the sum of the numbers on the top faces is 10 is $\frac{1}{12}$

<u>Solution: </u>

The sample space of two dice outcomes is given as follows:

{(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 events n(s) = 36

Expected Outcomes = {(4, 6), (5, 5), (6, 4)}

Number of favorable outcomes n(E) = 3

Probability (sum of numbers on the top faces is 10) = $\frac{n(E)}{n(s)}$ = $\frac{3}{36}$ = $\frac{1}{12}$

Therefore the probability that the sum of the numbers on the top faces is 10 is $\frac{1}{12}$

3 0

3 years ago

HELPPPP<br> please help me answer this question-

erik [133]

Answer:

$\displaystyle GH = 2 \sqrt{30}$

$\displaystyle FH = 2 \sqrt{10}$

$\displaystyle m \angle G = {30}^{ \circ}$

Step-by-step explanation:

<h3>QUESTION-2:</h3>

we are given a right angle triangle

it's a 30-60-90 triangle of which FH is the shortest side

remember that,in case of 30-60-90 triangle the the longest side is twice as much as the shortest side thus

our equation is

$\displaystyle 2FH=4\sqrt{10}$

divide both sides by 2

$\displaystyle\frac{2FH}{2}= \frac{ 4 \sqrt{10} }{2}$

$\displaystyle FH =2 \sqrt{10}$

<h3>Question-1:</h3>

in order to figure out GH we can use Trigonometry because the given triangle is a right angle triangle

as we want to figure out GH we'll use sin function

remember that,

$\displaystyle \sin( \theta) = \frac{opp}{hypo}$

let our opp, hypo and $\theta$ be GH, 4√10 and 60° respectively

thus substitute:

$\displaystyle \sin( {60}^{ \circ} ) = \frac{GH}{4 \sqrt{10} }$

recall unit circle:

$\displaystyle \frac{ \sqrt{3} }{2} = \frac{GH}{4 \sqrt{10} }$

cross multiplication:

$\displaystyle 2 GH = 4 \sqrt{10} \times \sqrt{3}$

simplify multiplication:

$\displaystyle 2 GH = 4 \sqrt{30}$

divide both sides by 2:

$\displaystyle GH = 2 \sqrt{30}$

<h3>QUESTION-3:</h3>

Recall that, the sum of the interior angles of a triangle is 180°

therefore,

$\displaystyle m \angle G + {60}^{ \circ} + {90}^{ \circ} = {180}^{ \circ}$

simplify addition:

$\displaystyle m \angle G + {150}^{ \circ} = {180}^{ \circ}$

cancel 150° from both sides

$\displaystyle m \angle G = {30}^{ \circ}$

6 0

3 years ago

Read 2 more answers

There are 9 marbles representing 3 different colors. Write a problem where 2 marbles are selected at random without replcement a

Nata [24]

There are 9 marbles in the bag. We pick 2 without replacement and get a probability of 1/6.

Each draw of a marble has a probability associated with it. Multiplying these gives 1/6 so let us assume the probabilities are (1/3) and (1/2).

In order for the first draw to have a probability of 1/3 we need to draw a color that has (1/3)(9)=3 marbles. So let's say there are 3 red marbles. The P(a red marble is drawn) = 1/3.

Now that a marble has been drawn there are 8 marbles left. In order for the second draw to have a probability of 1/2 we must draw a color that has (1/2)(8) = 4 marbles. So let's say there are 4 blue marbles out of the 8.

Since there are 9 marbles to start and we have 3 red marbles and 4 blue marbles, the remaining 2 marbles must be a different color. Let us say they are green.

The problem is: There are 3 red marbles, 4 blue marbles and 2 green marbles in a jar. A marble is picked at random, it's color is noted and the marble is not replaced. A second marble is drawn at random and its color noted. What is the probability that the first marble is red and the second blue?

8 0

3 years ago