Hey or range for each

Mathematics

1 answer:

topjm [15]3 years ago

7 0

Answer : 100 meters 2:35/mile or 1:36/km 2:49/mile or 1:45/km

200 meters 2:35/mile or 1:36/km 2:52/mile or 1:47/km

400 meters 2:54/mile or 1:48/km 3:12/mile or 1:59/km

800 meters 3:23/mile or 2:06/km 3:48/mile or 2:21/km

1,500 meters 3:41/mile or 2:17/km 4:07/mile or 2:34/km

1 mile 3:43/mile or 2:19/km 4:13/mile or 2:37/km

5K 4:04/mile or 2:31/km 4:34/mile or 2:50/km

10K 4:14/mile or 2:38/km 4:45/mile or 2:57/km

Half Marathon

(13.11 miles / 21.098 km) 4:27/mile or 2:46/km 4:58/mile or 3:05/km

Marathon

(26.22 miles / 42.195 km) 4:41/mile or 2:55/km 5:10/mile or 3:13/km

Step-by-step explanation: sorry if this is wromg :/ ( but if it is please add me as the brainist answer )

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22 less than 5 times a number f

olya-2409 [2.1K]

Answer:

5f - 22

Step-by-step explanation:

pls igve brainliest

8 0

2 years ago

Read 2 more answers

I need help with #18,20,22 and 24. Please show your work and I'm really thankful for your help as soon as possible!

Artist 52 [7]

I'll start 18 and 22 for you, and you should then be able to do the rest on your own!

For 18, what we literally do is apply the distance formula for all the points and add them up. For B to C, we get the distance between them to be
sqrt((x1-x2)^2+(y1-y2)^2)=sqrt((0-4)^2+(3-(-1))^2)=sqrt((-4)^2+4^2)=sqrt(16+16)=sqrt(32). Repeat the process for C to E, E and F, and F to B then add the results up to get your answer!

For 22, since the area of a rectangle is length*width (we know given the right angles and that the opposite sides are equal in how long they are), we can multiply 2 perpendicular lines, for example, BC and CE to get sqrt(32)*sqrt(8)=16 as the area

4 0

2 years ago

The eccentricity e of an ellipse is defined as the number c/a, where a is the distance of a vertex from the center and c is the

Anna71 [15]

Answer:

Check below, please.

Step-by-step explanation:

Hi, there!

Since we can describe eccentricity as $e=\frac{c}{a}$

a) Eccentricity close to 0

An ellipsis with eccentricity whose value is 0, is in fact, a degenerate one almost a circle. An ellipse whose value is close to zero is almost a degenerate circle. The closer the eccentricity comes to zero, the more rounded gets the ellipse just like a circle. (Check picture, please)

$\frac{x^2}{a^2} +\frac{y^2}{b^2} =1 \:(Ellipse \:formula)\\a^2=b^2+c^2 \: (Pythagorean\: Theorem)\:a=longer \:axis.\:b=shorter \:axis)\\a^2=b^2+(0)^2 \:(c\:is \:the\: distance \: the\: Foci)\\\\a^2=b^2 \\a=b\: (the \:halves \:of \:each\:axes \:measure \:the \:same)$

b) Eccentricity =5

$5=\frac{c}{a} \:c=5a$

An eccentricity equal to 5 implies that the distance between the Foci has to be five (5) times larger than the half of its longer axis! In this case, there can't be an ellipse since the eccentricity must be between 0 and 1 in other words:

$If\:e=\frac{c}{a} \:then\:c>0 , and\: c>0 \:then \:1>e>0$

c) Eccentricity close to 1

In this case, the eccentricity close or equal to 1 We must conceive an ellipse whose measure for the half of the longer axis a and the distance between the Foci 'c' they both have the same size.

$a=c\\\\a^2=b^2+c^2\:(In \:the\:Pythagorean\:Theorem\: we \:should\:conceive \:b=0)$