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4 years ago

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Two planes, A and B, start from the same place and move in different directions, making an angle of 50º between them. The speed

of plane A is 200 miles per hour, and the speed of plane B is 300 miles per hour. The two planes are how many miles apart after one hour?

2 answers:

cluponka [151]4 years ago

6 0

Answer: The distance between them is 230.65 miles.

Step-by-step explanation:

Since we have given that

Speed of plane A (b) = 200 miles per hour

Speed of plane B (c) = 300 miles per hour

Angle between them = 50°

So,

We will use the "Cosine formula", we get

$a^2=b^2+c^2-2bc\cos A\\\\a^2=200^2+300^2-2\times 200\times 300\cos 50\textdegree\\\\a^2=40000+90000-120000\cos 50\textdegree\\\\a^2=130000-120000\times 0.64\\\\a^2=130000-76800\\\\a^2=53200\\\\a=\sqrt{53200}\\\\a=230.65\ miles\ per\ hour$

So, distance between the two planes after one hour will be

$Distance=\frac{Speed}{time}=\frac{230.65}{1}=230.65\ miles$

So, the distance between them is 230.65 miles.

Darina [25.2K]4 years ago

4 0

I believe the the plains are 250 miles apart, i think :3

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Answer:

a)109 of the surveyed households only own a hand gun and do not support the ban on hand guns.

b)328 of the surveyed households do not own a gun and support the ban on hand guns

c )112 of the surveyed households do not own a gun and do not support the ban on handguns.

Step-by-step explanation:

To solve this problem, we must build the Venn's Diagram of this set.

I am going to say that:

-The set A represents those who own a hand gun.

-The set B represents those who own a rifle

-The set C represents those who support the ban of handguns.

-D represents those who do not own a gun and do not support the ban on hand guns.

We have that:

$A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)$

In which $a$ are those that only own a hand gun, $A \cap B$ are those who own both a handgun and a rifle, $A \cap C$ are those who own a hand gun and support the ban on handguns, and $A \cap B \cap C$ are those who own both a hand gun and a rifle, and also support the ban on hand guns.

By the same logic, we have:

$B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)$

In which $b$ are those who only own a rifle, and $B \cap C$ are those who own a rifle and support the ban on handguns.

$C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

In which $c$ are those who support the ban on hand guns and do not own any handguns.

This diagram has the following subsets:

$a,b,c,D(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C)$

There were 1000 households surveyed, so:

$a + b + c + D + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 1000$

We start finding the values from the intersection of three sets.

Solution:

48 own both a hand gun and a rifle and also support the ban on hand guns.

$A \cap B \cap C = 48$

70 own a hand gun and support the ban on hand guns:

$(A \cap C) + (A \cap B \cap C) = 70$

$A \cap C = 70 - 48$

$A \cap C = 22$

206 own a rifle but no hand gun and do not support the ban on hand guns.

$b = 206$ .

136 own both a hand gun and a rifle:

$(A \cap B) + (A \cap B \cap C) = 136$

$A \cap B = 136 - 48$

$A \cap B = 88$

493 supported the ban on hand guns.

$C = 493$

437 own a rifle:

$B = 437$

267 own a hand gun:

$A = 267$

a) How many of the surveyed households only own a hand gun and do not support the ban on hand guns?

This is the value of $a$ , that we can find from the following equation:

$A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)$

$267 = a + 88 + 22 + 48$

$a = 267 - 158$

$a = 109$

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b) How many of the surveyed households do not own a gun and support the ban on hand guns?

This is the value of $c$ , that we can find from the following equation:

$C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

$493 = c + 22 + (B \cap C) + 48$

Here, we also have to find $B \cap C$ , that are those who own a rifle and support the ban on hand guns. We can find this from the following equation:

$B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)$

$437 = 206 + (B \cap C) + 88 + 48$

$B \cap C = 437 - 342$

$B \cap C = 95$

So

$493 = c + 22 + (B \cap C) + 48$

$493 = c + 22 + 95 + 48$

$c = 493 - 165$

$c = 328$

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$a + b + c + D + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 1000$

$109 + 206 + 328 + D + 88 + 22 + 95 + 40 = 1000$

$D = 1000 - 888$

$D = 112$

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