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grandymaker [24]
2 years ago
6

Line m passes through points (1, 4) and (10, 9). Line n is parallel to line m. What is the slope

Mathematics
1 answer:
Assoli18 [71]2 years ago
3 0
We be back in school I don’t want
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What is the average velocity between 12.00 and 15.00 seconds
bekas [8.4K]

The average between 12.00 and 15.00 seconds
can be defined in a few different ways.

Way #1:  (speed)

     Average speed =
         (1/2) (the speed at 12 seconds  +  the speed at 15 seconds) .


Way #2:  (speed)

   Average speed =
       (1/3) (the distance covered between 12.00 and 15.00 seconds)


Way #3:  (velocity)

   Average velocity =
     (1/3) (distance between location at 12 sec and location at 15 sec)
       in the direction from location at 12 sec to location at 15 sec.


That's the best we can do with the combination of given and
not given information in the question.

3 0
3 years ago
Read 2 more answers
Simplify: 2⁶X5⁴X3¹²/2²x5²x3⁴​
Sholpan [36]

Step-by-step explanation:

2⁶X5⁴X3¹²/2²x5²x3⁴

2⁴ × 5²×3⁸

4 0
2 years ago
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I need help with number 7 and 8
devlian [24]
7.

The point-slope form of the equation of a line is

y - y1 = m(x - x1)

where m is the slope, and (x1, y1) is a point on the line.

We are given two points, so we can find the slope.

slope = m = (y2 - y1)/(x2 - x1)

The slope of this line is

m = (-3 - 7)/(5 - (-15)) = -10/20 = -1/2

Since every choice has a 7 and a 15, we now use point (-15, 7) and slope -1/2.

y - y1 = m(x - x1)

y - 7 = -1/2(x - (-15))

y - 7 = -1/2(x + 15)
5 0
3 years ago
Niall owes $187 to his cousin. However, he doesn't have any money right now. He hopes to earn some money by painting a house. He
Alla [95]
The equation would be 34 x 2x-187= y. 
8 0
3 years ago
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Somebody please assist me here
Anettt [7]

The base case of n=1 is trivially true, since

\displaystyle P\left(\bigcup_{i=1}^1 E_i\right) = P(E_1) = \sum_{i=1}^1 P(E_i)

but I think the case of n=2 may be a bit more convincing in this role. We have by the inclusion/exclusion principle

\displaystyle P\left(\bigcup_{i=1}^2 E_i\right) = P(E_1 \cup E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) = P(E_1) + P(E_2) - P(E_1 \cap E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) \le P(E_1) + P(E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) \le \sum_{i=1}^2 P(E_i)

with equality if E_1\cap E_2=\emptyset.

Now assume the case of n=k is true, that

\displaystyle P\left(\bigcup_{i=1}^k E_i\right) \le \sum_{i=1}^k P(E_i)

We want to use this to prove the claim for n=k+1, that

\displaystyle P\left(\bigcup_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^{k+1} P(E_i)

The I/EP tells us

\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cup E_{k+1}\right) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) - P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cap E_{k+1}\right)

and by the same argument as in the n=2 case, this leads to

\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) - P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cap E_{k+1}\right) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1})

By the induction hypothesis, we have an upper bound for the probability of the union of the E_1 through E_k. The result follows.

\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^k P(E_i) + P(E_{k+1}) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^{k+1} P(E_i)

5 0
2 years ago
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