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muminat
3 years ago
5

A walker on a trail travels along the path described by the parametric equations

Mathematics
1 answer:
lana [24]3 years ago
6 0

Answer:

no,the pathway of the walker and cub don't intersect

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Answer:A,B,E

Step-by-step explanation:

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6. A graph with a<br> slope decreases from left to right.
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Read 2 more answers
Both the ranges and the interquartile ranges for the data sets are the same.
aleksley [76]

Answer:

to find range of interquartile sets of data

you see 5 vertical lines 1 is start point 2 is start of 2nd quartile 3 is start of 3rd quartile and also represents the median.  the 4th start of 4th quartile and 5th is very end point.

Therefore the interquartile is from start line 2 to start line 4

apply to each by if it says 10 and 34 then we show workjings 34-10 = 24 and that is the range once we deduct data from each other.

Therefore full range is start line 1 to end line 5

if this says 2 to 45 then full range is 45-2 = 43 and apply to each boxplot.

Step-by-step explanation:

6 0
2 years ago
In triangle $ABC$, let angle bisectors $BD$ and $CE$ intersect at $I$. The line through $I$ parallel to $BC$ intersects $AB$ and
Umnica [9.8K]

Answer:

41

Step-by-step explanation:

If you work through a series of obscure calculations involving area and the radius of the incircle, they boil down to a simple fact:

... For MN║BC, perimeter ΔAMN = perimeter ΔABC - BC = AB+AC

.. = 17+24 = 41

_____

Wow! Thank you for an interesting question with a not-so-obvious answer.

_____

<em>A little more detail</em>

The point I that you have defined is the incenter—the center of an inscribed circle in the triangle. Its radius is the distance from I to any side, such as BC, for example.

If we use "Δ" to represent the area of the triangle and "s" to represent the semi-perimeter, (AB+BC+AC)/2, then the incircle has radius Δ/s. The area Δ can be computed from Heron's formula by ...

... Δ = √(s(s-a)(s-b)(s-c)) . . . . where a, b, c are the side lengths

For this triangle, the area is Δ = √38480 ≈ 196.1632 units². That turns out to be irrelevant.

The altitude to BC will be 2Δ/(BC), so the altitude of ΔAMN = (2Δ/(BC) -Δ/s). Dividing this by the altitude to BC gives the ratio of the perimeter of ΔAMN to the perimeter of ΔABC, which is 2s.

Putting these ratios and perimeters together, we get ...

... perimeter ΔAMN = (2Δ/(BC) -Δ/s)/(2Δ/(BC)) × 2s

... = (2/(BC) -1/s) × BC × s = 2s -BC

... perimeter ΔAMN = AB +AC

8 0
3 years ago
Kathy is training for a 10 mile race. How many feet will Kathy run?
Julli [10]
5,280 feet per mile
10 miles
5,280x10= 52,800
5 0
3 years ago
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