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3 years ago

For each of the regions D in R^2 below, fill in the blank so that the statements are true:

Mathematics

2 answers:

Dmitrij [34]3 years ago

8 0

Answer:

b. might or might not

Step-by-step explanation:

For each of the regions D in R^2 below, fill in the blank so that the statements are true: might or might not

andrey2020 [161]3 years ago

7 0

Answer:

b. might or might not

Step-by-step explanation:

Theorem:

If f is a continuous function in a bounded interval, it must have an absolute maximum.

In this question:

The interval is 2x^4+6y^2 >= 152, which goes to infinite, meaning it is not bounded, and so the function might or might not have an absolute maximum, and the correct answer is given by option b.

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4) The path of a satellite orbiting the earth causes it to pass directly over two

Naily [24]

Answers:

Satellite is approximately 2446.43 km from station A.
Satellite is approximately 2441.61 km above the ground.

=========================================================

Explanation:

I'm assuming tracking stations A and B are at the same elevation and are on flat ground. In reality, this is likely not the case; however, for the sake of simplicity, we'll assume this is the case.

The diagram is shown below. Points A and B describe the two stations, while point C is the satellite's location. Point D is on the ground directly below the satellite. We have these lengths

AB = 60 km
AD = x
CD = h

Focusing on triangle ACD, we can apply the tangent rule to isolate h.

tan(angle) = opposite/adjacent

tan(A) = CD/AD

tan(86.4) = h/x

x*tan(86.4) = h

h = x*tan(86.4)

We'll use this later in the substitution below.

--------------------

Now move onto triangle BCD. For the reference angle B = 85, we can use the tangent rule to say

tan(angle) = opposite/adjacent

tan(B) = CD/DB

tan(B) = CD/(DA+AB)

tan(85) = h/(x+60)

tan(85)*(x+60) = h

tan(85)*(x+60) = x*tan(86.4) ............. apply substitution; isolate x

x*tan(85)+60*tan(85) = x*tan(86.4)

60*tan(85) = x*tan(86.4)-x*tan(85)

60*tan(85) = x*(tan(86.4)-tan(85))

x*(tan(86.4)-tan(85)) = 60*tan(85)

x = 60*tan(85)/(tan(86.4)-tan(85))

x = 153.612786190499

--------------------

We'll use this approximate x value to find h

h = x*tan(86.4)

h = 153.612786190499*tan(86.4)

h = 2441.60531869599

h = 2441.61 km is how high the satellite is above the ground.

Return to triangle ACD. We'll use the cosine rule to determine the length of the hypotenuse AC

cos(angle) = adjacent/hypotenuse

cos(A) = AD/AC

cos(86.4) = x/AC

cos(86.4) = 153.612786190499/AC

AC*cos(86.4) = 153.612786190499

AC = 153.612786190499/cos(86.4)

AC = 2446.43279498247

AC = 2446.43 km is the distance from the satellite to station A.

6 0

3 years ago

Daddy chill 6969696969

iren2701 [21]

Answer:

I've never been to waffle house

8 0

3 years ago

Read 2 more answers

Where on the graph of 2x - 6y = 7 is the x-coordinate twice the y-coordinate?

Nikolay [14]

Answer:

The required point is P(-7, $\frac{-7}{2}$ ) where, x-coordinate twice the y-coordinate on line 2x - 6y = 7

Step-by-step explanation:

Given equation of line is 2x-6y=7

To find point on graph whose x-coordinate twice the y-coordinate:

Let y be y-coordinate of point

Hence, x-coordinate of point will be 2y

The required point is P(2y,y) on equation of line 2x-6y=7

Now,

2x-6y=7

2(2y)-6y=7

-2y=7

y= $\frac{-7}{2}$

Thus,

The required point is P(-7, $\frac{-7}{2}$ )

Note: Figure show equation of line red line and point P as blue dot.

5 0

3 years ago

The ice cream shop has 80 different possible sundaes consisting of 1 flavor of ice cream, 1 syrup, and 1 candy topping. If the i

adoni [48]

4: 80*5*4 
= 1600 diff ice creams flavors. yum.

4 0

3 years ago

In a football or soccer game, you have 22 players, from both teams, in the field. what is the probability of having at least any

Tamiku [17]

We can solve this problem using complementary events. Two events are said to be complementary if one negates the other, i.e. E and F are complementary if

$E \cap F = \emptyset,\quad E \cup F = \Omega$

where $\Omega$ is the whole sample space.

This implies that

$P(E) + P(F) = P(\Omega)=1 \implies P(E) = 1-P(F)$

So, let's compute the probability that all 22 footballer were born on different days.

The first footballer can be born on any day, since we have no restrictions so far. Since we're using numbers from 1 to 365 to represent days, let's say that the first footballer was born on the day $d_1$ .

The second footballer can be born on any other day, so he has 364 possible birthdays:

$d_2 \in \{1,2,3,\ldots 365\} \setminus \{d_1\}$

the probability for the first two footballers to be born on two different days is thus

$1 \cdot \dfrac{364}{365} = \dfrac{364}{365}$

Similarly, the third footballer can be born on any day, except $d_1$ and $d_2$ :

$d_3 \in \{1,2,3,\ldots 365\} \setminus \{d_1,d_2\}$

so, the probability for the first three footballers to be born on three different days is

$1 \cdot \dfrac{364}{365} \cdot \dfrac{363}{365}$

And so on. With each new footballer we include, we have less and less options out of the 365 days, since more and more days will be already occupied by another footballer, and we can't have two players born on the same day.

The probability of all 22 footballers being born on 22 different days is thus

$\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}$

So, the probability that at least two footballers are born on the same day is

$1-\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}$

since the two events are complementary.

8 0

3 years ago