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2 years ago

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Gandalf the grey started in the forest of mirkwood at a point pp with coordinates (−2,1)(−2,1) and arrived in the iron hills at

the point qq with coordinates (-1, 6). if he began walking in the direction of the vector v=3i+2jv=3i+2j and changes direction only once, when he turns at a right angle, what are the coordinates of the point where he makes the turn?

1 answer:

AlladinOne [14]2 years ago

8 0

Travel direction 3i + 2j, slope = 2/3.
We have slope and a point (-2,1) this gives us a line equation:
LINE BEFORE THE TURN, EQUATION: y = (2/3)x + 1

The point of intersection of the two lines:
1) through (-2,1) with slope 2/3 and;
2) through (-1,6) with slope 3/2.

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Let’s throw some numbers in here just so it becomes clear what we are asked for.

Let’s say the teacher hands out 12 notebooks that came in packs with 4 notebooks in each pack. If you are asked for the number of packs here you would see that it’s 3. That is 3 packs of 4 give 12.

So you divide the number of notebooks handed out (m) by the number of notebooks per pack (c) to get the number of packs.

The number of packs is m/c

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1 year ago

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Now evaluate f(x) = 2x4 - 4x3-11x2+3x-6 for x=-2 f (-2) =

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2 years ago

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450x2y5 3,000x4y3 What is the least common multiple (LCM) of these monomials?

ivolga24 [154]

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2 years ago

M∠1=45° and m∠3=65°

7nadin3 [17]

Answer:

In Right Δ ABC with right angle B,

∠A=(3 x -8)°, ∠B=90°, ∠C=(x-2)°

∠A+∠B+∠C=180°[∠ sum property of triangle]

3 x-8+90 + x-2=180

Adding and subtracting like terms

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2 years ago

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Suppose that only 20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions whe

Lina20 [59]

Answer:

a) 91.33% probability that at most 6 will come to a complete stop

b) 10.91% probability that exactly 6 will come to a complete stop.

c) 19.58% probability that at least 6 will come to a complete stop

d) 4 of the next 20 drivers do you expect to come to a complete stop

Step-by-step explanation:

For each driver, there are only two possible outcomes. Either they will come to a complete stop, or they will not. The probability of a driver coming to a complete stop is independent of other drivers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions when no other cars are visible.

This means that $p = 0.2$

20 drivers

This means that $n = 20$

a. at most 6 will come to a complete stop?

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{20,0}.(0.2)^{0}.(0.8)^{20} = 0.0115$

$P(X = 1) = C_{20,1}.(0.2)^{1}.(0.8)^{19} = 0.0576$

$P(X = 2) = C_{20,2}.(0.2)^{2}.(0.8)^{18} = 0.1369$

$P(X = 3) = C_{20,3}.(0.2)^{3}.(0.8)^{17} = 0.2054$

$P(X = 4) = C_{20,4}.(0.2)^{4}.(0.8)^{16} = 0.2182$

$P(X = 5) = C_{20,5}.(0.2)^{5}.(0.8)^{15} = 0.1746$

$P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091$

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 + 0.1091 = 0.9133$

91.33% probability that at most 6 will come to a complete stop

b. Exactly 6 will come to a complete stop?

$P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091$

10.91% probability that exactly 6 will come to a complete stop.

c. At least 6 will come to a complete stop?

Either less than 6 will come to a complete stop, or at least 6 will. The sum of the probabilities of these events is decimal 1. So

$P(X < 6) + P(X \geq 6) = 1$

We want $P(X \geq 6)$ . So

$P(X \geq 6) = 1 - P(X < 6)$

In which

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 = 0.8042$

$P(X \geq 6) = 1 - P(X < 6) = 1 - 0.8042 = 0.1958$

19.58% probability that at least 6 will come to a complete stop

d. How many of the next 20 drivers do you expect to come to a complete stop?

The expected value of the binomial distribution is

$E(X) = np = 20*0.2 = 4$

4 of the next 20 drivers do you expect to come to a complete stop

4 0

2 years ago