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Arturiano [62]
3 years ago
12

Which equation below has no solution?

Mathematics
1 answer:
soldier1979 [14.2K]3 years ago
8 0

Answer:

B

Step-by-step explanation:

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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
A sales associate is given a $500 hiring bonus with a new job. She earns an average commision of $250 per week for the next 12 w
allsm [11]
$3500 250 x 12 + 500
4 0
3 years ago
If you hike three miles every 5 minutes, how far are you going in 1 minute?
V125BC [204]

Answer:

0.6 miles per minute

Step-by-step explanation:

3/5 = 0.6/1 = 0.6

6 0
2 years ago
Read 2 more answers
Lim x-1 x2 - 1/ sin(x-2)
balu736 [363]

Answer:

           \lim_{x \to 1}\frac{x^2-1}{sin(x-2)}=0

Explanation:

Assuming the correct expression is to find the following limit:

         \lim_{x \to 1}\frac{x^2-1}{sin(x-2)}

Use the property the limit of the quotient is the quotient of the limits:

         \lim_{x \to 1}\frac{x^2-1}{sin(x-2)}=\frac{\lim_{x \to 1}x^2-1}{\lim_{x \to 1}sin(x-2)}

Evaluate the numerator:

          \frac{\lim_{x \to 1}x^2-1}{\lim_{x \to 1}sin(x-2)}=\frac{1^2-1}{\lim_{x \to1}sin(x-2)}=\frac{0}{\lim_{x \to 1}sin(x-2}

Evaluate the denominator:

  • Since         \lim_{x \to1}sin(x-2)\neq 0

                  \frac{0}{\lim_{x \to1}sin(x-2)}=0

4 0
2 years ago
The area of Ms. Anderson’s office floor measures 120
Maslowich

Answer:

72 square feet not covered

Step-by-step explanation:

100%-40% = 60%

60% of 120 = 0.6x120 = 72

8 0
2 years ago
Read 2 more answers
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