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Sati [7]
4 years ago
8

How would you define the distance between parallel lines in a plane?

Mathematics
1 answer:
Verdich [7]4 years ago
3 0
This article is about a mathematical relationship between lines. For other uses, see Parallel (disambiguation). "Parallel lines" redirects here. For other uses, see Parallel lines (disambiguation). Line art drawing of parallel lines and curves.

In geometry, parallel lines are lines in a plane which do not meet; that is, two lines in a plane that do not intersect or touch each other at any point are said to be parallel. By extension, a line and a plane, or two planes, in three-dimensional Euclidean space that do not share a point are said to be parallel. However, two lines in three-dimensional space which do not meet must be in a common plane to be considered parallel; otherwise they are called skew lines. Parallel planes are planes in the same three-dimensional space that never meet.

Parallel lines are the subject of Euclid's parallel postulate.[1] Parallelism is primarily a property of affine geometries and Euclidean geometry is a special instance of this type of geometry. In some other geometries, such as hyperbolic geometry, lines can have analogous properties that are referred to as parallelism.


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Which expression is equivalent to7(a+2)
Vanyuwa [196]

Answer:

7a + 14

just multiply 7 by both a and 2

6 0
3 years ago
Read 2 more answers
Help me pleaseeee just a yes or no question
oee [108]
The points have to have the coordinate y=4 to be on the y=4 line

only the (1, 4) point matches this, so yes it is a solution, all others not and are therefore no
3 0
3 years ago
Not sure if any of this is correct, but it’s what I got so far
Irina18 [472]

Problem 1 is correct. You use the pythagorean theorem to find the hypotenuse.

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

Problem 2 has the correct answer, but one part of the steps is a bit strange. I agree with the 132 ft/sec portion; however, I'm not sure why you wrote \frac{1 \text{ sec}}{132 \text{ ft}}=\frac{0.59\overline{09}}{78 \text{ ft}}*127 \text{ ft}

I would write it as \frac{1\text{ sec}}{132 \text{ ft}}*127 \text{ ft} = \frac{127}{132} \text{ sec} \approx 0.96 \text{ sec}

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

For problem 3, we first need to convert the runner's speed from mph to feet per second.

17.5 \text{ mph} = \frac{17.5 \text{ mi}}{1 \text{ hr}}*\frac{1 \text{ hr}}{60 \text{ min}}*\frac{1 \text{ min}}{60 \text{ sec}}*\frac{5280 \text{ ft}}{1 \text{ mi}} \approx 25.667 \text{ ft per sec}

Since the runner needs to travel 90-12 = 78 ft, this means\text{time} = \frac{\text{distance}}{\text{speed}} \approx \frac{78 \text{ ft}}{25.667 \text{ ft per sec}} \approx 3.039 \text{ sec}

So the runner needs about 3.039 seconds. In problem 2, you calculated that it takes about 0.96 seconds for the ball to go from home to second base. The runner will not beat the throw. The ball gets where it needs to go well before the runner arrives there too.

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

The question is now: how much of a lead does the runner need in order to beat the throw?

Well the runner needs to get to second base in under 0.96 seconds.

Let's calculate the distance based on that, and based on the speed we calculated earlier above.

\text{distance} = \text{rate}*\text{time} \approx (25.667 \text{ ft per sec})*(0.96 \text{ sec}) \approx 24.64032 \text{ ft}

This is the distance the runner can travel if the runner only has 0.96 seconds. So the lead needed is 90-24.64032 = 65.35968 feet

This is probably not reasonable considering it's well over halfway (because 65.35968/90 = 0.726 = 72.6%). If the runner is leading over halfway, then the runner is probably already in the running motion and not being stationary.

As you can see, the runner is very unlikely to steal second base. Though of course such events do happen in real life. What may explain this is the reaction time of the catcher may add on just enough time for the runner to steal second base. For this problem however, we aren't considering the reaction time. Also, not all catchers can throw the ball at 90 mph which is quite fast. According to quick research, the MLB says the average catcher speed is about 81.8 mph. This slower throwing speed may account for why stealing second base isn't literally impossible, although it's still fairly difficult.

5 0
3 years ago
Evaluate the expression.<br> 38 +16 - 12 - 2 - (302)<br> Enter your answer in the box.
masha68 [24]

Answer:

-262

Step-by-step explanation:

Evaluate Subtraction & Addition from right to left [or left to right, as long as you know what you are doing]:

38 + 16 - 12 - 2 - 302

|_______||____________|

44 - 316 =  -262

I am joyous to assist you anytime.

5 0
3 years ago
What value of m makes the equation true? m + 2m - 6 = -12 + 2m.
denpristay [2]

the answer is 6 i think

4 0
4 years ago
Read 2 more answers
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