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3 years ago

6

A construction crew is lengthening a road. Let L be the total length of the road (in miles). Let D be the number of days the cre

w has worked. Suppose that L=2D+200 gives L as a function of D. The crew can work for at most 90 days.

1 answer:

Inessa05 [86]3 years ago

7 0

L = 54 + 2d d represents the number of days

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On a treasure map, the last clue is that the treasure is buried is buried a certain number of feet away from a tree. the distanc

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The lcm of 20 and 8 is 40 feet.

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3 years ago

Write an equation in point-slope form of the line that passes through (-2,14) and is perpendicular to the line whose intercepts

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0.98

Step-by-step explanation:

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3 years ago

Using the graph below, what is the best estimate of month 7's average temperature if the equation for the line of best fit is y

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3 years ago

Read 2 more answers

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

Elena surveyed her classmates to see how many like to make jewelry. When she asked 20 of her classmates, 50 percent said they li

nordsb [41]

10 students of Elena's classmates like to make jewellery.

<u>Step-by-step explanation:</u>

Let total no.of classmates who liked the jewellery = x

Here, in the given problem, given she asked her classmates '20'. So, it clearly shows that total number of students/classmates = 20. Also, in that, 50% said like to make jewellery. Now find 'x' as below,

$x = 50 \% \text { of } 20$

$x = \frac{50}{100} \times 20$

$x = 10$

Hence, the total number of classmates who liked to make Jewellery = 10

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3 years ago