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Setler [38]
3 years ago
6

If a car travels 10 ft due north and then 15 feet due west, how far is the car from it starting point? This is for homework and

it’s on the Pythagorean theorem
Mathematics
1 answer:
krok68 [10]3 years ago
5 0
Answer:

Explanation:

The car travel 10ft north then 15ft due west. In order to find the distance from starting point. We need to find the hypotenuse where the distance to north and west are the adjacent side.

We can use Pythagorean’s theorem according to that:

Let c be the distance from starting point.
And a and b be the distance travel to north and west.

a^2 + b^2 = c^2
10^2 + 15^2 = c^2
100 + 225 = c^2
325 = c^2
Sqrt(325) = c
5sqrt(13) = c

Therefore, the distance from starting point is 5sqrt(13) ft
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Henry wants to earn more than $68 trimming trees. He charges $6 per hour and pays $4 in equipment fees. What are the possible nu
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there are n counters in a bag. 4 of the counters are red and the rest are blue. Ross takes a counter from the bag at random and
Ainat [17]

Answer: n = 10.

Step-by-step explanation:

In the bag, we have n counters.

4 of the counters are red.

the rest are blue, then we have:

(n - 4) blue counters.

Now, the probability that Ross takes a blue counter from the bag is equal to the quotient between the number of blue counters (n - 4) and the total number of counters, n

Then the probability is:

p1 = (n - 4)/n

Now he draws another, and it must be blue again, then we can calculate the probability in the same way as above, but he already take a blue counter, so the number of blue counters is (n - 5) and the total number of counters is (n - 1)

The probability of this event is:

p2 = (n - 5)/(n - 1)

The joint probability (the probability that Ross takes two blue counters) is equal to the product of the individual probabilities, and we know that this is equal to 1/3, then we have the equation:

1/3 = ( (n - 4)/n)*((n - 5)/(n - 1))

Now let's solve this for n.

n*(n - 1)/3 = (n - 4)*(n - 5)

(n^2 - n)/3 = n^2 - 4*n - 5*n + 20

n^2 - n = 3*(n^2 - 9*n + 20)

n^2 - n = 3*n^2 - 27*n + 60

0 = (3*n^2 - n^2) - 27*n + n + 60

0 = 2*n^2 - 26*n + 60

The two solutions of this equation can be found with Bhaskara's equation:

n = \frac{-(-26) +- \sqrt{(-26)^2 - 4*2*60} }{2*2} = \frac{26+- 14}{4}

Then the two solutions are:

n = (26 - 14)/4 = 3

This is not an option, because we know for sure that we have 4 red counters, then this option can be discarded.

The other solution is:

n = (26 + 14)/4 = 40/4 = 10

Then we have n = 10, 10 counters in total.

4 0
3 years ago
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