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lilavasa [31]
3 years ago
14

Kristin drew a triangle with tow congruent sides and one obtuse angle.whitch term accurately describes the triangle

Mathematics
2 answers:
kotykmax [81]3 years ago
6 0


Obtuse triangles are triangles which has an angle that is more then 90°.
Right triangles are triangles that has 90° angle
Acute triangles are triangles that has all angles below 90°
Artist 52 [7]3 years ago
5 0
Isosceles triangle where there are two equal sides
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I don't understand this.
sladkih [1.3K]
You mean the word or number
7 0
3 years ago
2<br> 6-2___ help please <br> 7<br> (30points)
serg [7]
6 - 2 2/7 = 
6 - 16/7 =
42/7 - 16/7 =
26/7 or 3 5/7
8 0
3 years ago
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Equipment was purchased for $85,000 on january 1, 2015. freight charges amounted to $3,500 and there was a cost of $10,000 for b
Elena-2011 [213]
85000+3500+10000=98500 is a historic price of equipment. Annual depreciation is (98500-15000)/5=83500/5=16700. At december 31 2016 will be gone 2 years-2015 and 2016. So the accumulated depreciation will be 2*16700=33400
3 0
3 years ago
I will give Brainiest if you are right
jok3333 [9.3K]

Answer:

B. 4x ≤ 320

Step-by-step explanation:

Not sure how to explain my thinking. However, 4x must be smaller than 320 because she cannot spend more than that.

6 0
3 years ago
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Can someone please explain this problem to me idk what I am doing wrong and can you draw the problem for me thanks
Bogdan [553]

Answer:

  you're not doing anything wrong

Step-by-step explanation:

In order for cos⁻¹ to be a function, its range must be restricted to [0, π]. The cosine value that is its argument is cos(-4π/3) = -1/2. You have properly identified cos⁻¹(-1/2) to be 2π/3.

__

Cos and cos⁻¹ are conceptually inverse functions. Hence, conceptually, cos⁻¹(cos(x)) = x, regardless of the value of x. The expected answer here may be -4π/3.

As we discussed above, that would be incorrect. Cos⁻¹ cannot produce output values in the range [-π, -2π] unless it is specifically defined to do so. That would be an unusual definition of cos⁻¹. Nothing in the problem statement suggests anything other than the usual definition of cos⁻¹ applies.

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This is a good one to discuss with your teacher.

4 0
3 years ago
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